Did you know? The same differential equation can have infinitely many solutions, but one correctly stated initial condition often identifies a unique curve. For example, dy/dx=2xy has the family y=Ce^(x²). The condition y(0)=3 gives y=3e^(x²), while the condition y(1)=3 gives y=3e^(x²−1). Both curves satisfy the differential equation, but only one passes through each specified point.
This JC2 H2 Mathematics Tutor Bukit Timah guide explains how to move from a general solution to a particular solution of a separable differential equation, including initial conditions away from x=0. The answer comes first, followed by three fully worked equations, a shifted equilibrium example and 36 original cases. Every worked result can be checked by differentiation and substitution of the stated point.
The 2027 SEAB H2 Mathematics syllabus 9758 includes general and particular solutions of differential equations of the form dy/dx=f(x)g(y), as well as modelling and interpreting differential equations. This page focuses on the role of the integration constant and the given point rather than repeating the separate lesson on equilibrium solutions lost through division. All contexts and numerical conditions are original teaching examples.
Quick answer: integrate, determine the constant, and check both conditions
If a separable differential equation gives a family y=F(x,C), a condition y(x₀)=y₀ requires F(x₀,C)=y₀. Solve that equation for the constant C, then substitute C back into F. The result must satisfy the original differential equation and return exactly y₀ at x=x₀. Do not stop at a family when the question requests a particular solution, and do not set x₀=0 unless it was actually supplied.
- Identify the original differential equation: determine which quantities are functions of x.
- Separate only where legal: dividing by y−L excludes y=L, which must be checked separately.
- Integrate carefully: ∫1/y dy=ln|y| on a branch where y≠0.
- Keep an arbitrary constant: C represents a family, not the particular solution.
- Use both parts of the initial condition: substitute x=x₀ and y=y₀.
- Check by differentiation: the resulting y(x) must satisfy the original right-hand side.
- Interpret the interval and sign: the differential equation and starting point control any growth or decline.
First complete example: a growth equation at x=0
Solve dy/dx=2xy with y(0)=3. For y≠0, separate dy/y=2x dx. Integrate to get ln|y|=x²+C. On a fixed sign branch, write y=Ae^(x²), where A is a signed constant. The initial condition becomes 3=Ae^0=A, so y(x)=3e^(x²).
Differentiate: y′=3e^(x²)×2x=2xy. This proves that the proposed curve satisfies the differential equation. At x=0, y=3, proving that it also satisfies the initial point. These two checks address different requirements, which is why both are useful.
Second example: the same equation at x=1
If the problem instead states y(1)=3, the general family remains y=Ae^(x²). At x=1, 3=Ae, so A=3e^(−1). The particular solution becomes y=3e^(x²−1). Substituting x=1 gives 3, whereas the previous curve 3e^(x²) would give 3e at that input and would miss the required point.
The difference arises from the initial x-coordinate. A common examination error is to copy the constant from a previous example with a zero-based condition. The safe method is to write A=y₀e^(−x₀²) for this particular equation and verify that the exponent difference x²−x₀² vanishes at the prescribed input.
Third example: a shifted equilibrium level
Solve dy/dx=2x(y−3) with y(1)=5. Separating the non-equilibrium branch gives dy/(y−3)=2x dx and ln|y−3|=x²+C. Therefore y=3+Ae^(x²). At x=1, 5=3+Ae, so A=2/e. The particular solution is y=3+2e^(x²−1). Its derivative is 4x e^(x²−1), matching 2x(y−3).
The shift of three is important. A student who writes y=2e^(x²−1) has found a solution of a different differential equation because the equilibrium level was lost. At the prescribed point x=1, the intended function must equal five, and the missing shift makes it equal two. Directly checking the original initial condition exposes that mistake.
A constant branch: do not divide away the answer
If the equation is dy/dx=2x(y−3) with initial condition y(1)=3, the constant function y≡3 satisfies the equation: its derivative is zero and the right side is 2x(3−3)=0. Dividing by y−3 during separation would exclude precisely this solution. It must be checked in the original equation and retained for that initial condition.
This point overlaps intentionally with the earlier narrow guide Don’t Lose Equilibrium Solutions, but serves a different purpose here. The central question in this article is selecting a particular solution from a stated initial value. The equilibrium case is one boundary of that selection problem, not the entire method.
The signed exponential constant
Integrating dy/(y−L)=kx dx gives ln|y−L|=kx²/2+C. Exponentiating produces |y−L|=e^C e^(kx²/2). On a connected solution branch, the sign of y−L is fixed, so it is convenient to write y−L=Ae^(kx²/2) with a real signed constant A. This accommodates particular solutions above or below L without pretending that the exponential factor itself can become negative.
For example, dy/dx=−2x(y−4) with y(0)=1 gives y−4=Ae^(−x²). At zero, 1−4=−3, so A=−3 and y=4−3e^(−x²). The function stays below the equilibrium level four and approaches it as |x| increases. The negative multiplier belongs to A; the exponential is always positive.
The family formula that makes the initial point visible
For dy/dx=kx(y−L), the non-equilibrium general family is y=L+Ae^(kx²/2). If y(x₀)=y₀, then A=(y₀−L)e^(−kx₀²/2). Substitution yields the compact particular solution y(x)=L+(y₀−L)e^[k(x²−x₀²)/2]. This is especially useful as an answer check because x=x₀ makes the exponent zero and recovers y₀ immediately.
Differentiating the compact expression gives y′=kx(y₀−L)e^[k(x²−x₀²)/2]=kx(y−L), so it meets the differential equation as well. If y₀=L, the formula reduces to y≡L, confirming the equilibrium case. Its role here is a transparent verification identity; the separation and integration steps remain important when a question requires mathematical working.
What if the differential equation uses time directly?
Consider a fictional quantity q(t) satisfying dq/dt=−0.5q and q(0)=80. Separating gives dq/q=−0.5dt and q=Ae^(−0.5t). The initial condition fixes A=80, so q(t)=80e^(−0.5t). The value at t=2 is 80/e≈29.43. This is a mathematical decay model under the stated rule, not an observed real-world population or financial forecast.
In an application, interpreting a negative derivative means the quantity is decreasing on the relevant time interval. The actual model and its parameters must be provided or justified by the question. The exponential formula alone cannot prove that every real system follows the same process.
A nonzero initial x value changes the constant exponentially
Let dy/dx=−2xy with y(2)=5. The general family is y=Ae^(−x²). Substituting x=2 gives 5=Ae^(−4), so A=5e⁴. Thus y=5e^(4−x²). A learner who writes 5e^(−x²) has used the y-value but not the initial x-value; at x=2 that incorrect expression is only 5e^(−4).
A quick mental audit is to substitute the prescribed input into the final exponent. The difference between the current x² and the reference x₀² must be zero. If it is not, the initial condition has probably been applied incorrectly. This audit works even when the resulting expression is too complicated for mental numerical evaluation.
Why the initial condition belongs after the integration
For dy/dx=2xy, if you substitute x=1 and y=3 into the equation immediately, you get y′(1)=6, a local derivative. That information alone does not produce the full function y(x). The differential equation describes rates at every point; you must solve that relationship before a single value can select one of its solutions. The condition is indispensable, but it does not replace solving the equation.
A question may ask only for the slope at an initial point, in which case direct substitution into dy/dx may suffice. But when it asks for the particular solution, a rate at one point is incomplete. Read the requested output carefully: a number, a derivative at a point, a general family and a particular function are different mathematical deliverables.
Two checks are stronger than one
A proposed y(x) can pass the given point but fail the differential equation, or satisfy the differential equation but pass through the wrong point. For instance y=3e^(x²) solves dy/dx=2xy but fails y(1)=3. The function y≡3 passes through (1,3) but fails dy/dx=2xy except at isolated inputs. Checking both conditions independently prevents either mistake from slipping through.
The best order is to differentiate the candidate and compare it with the right-hand side as a function, then substitute the initial point. If the formula contains a parameter, you can also verify how the integration constant was obtained. These audits are short and often reveal whether an error came from separation, integration or initial-value substitution.
A one-minute Bukit Timah diagnostic
Ask a learner for dy/dx=2xy and y(0)=3: the answer is 3e^(x²). Change only y(0)=3 to y(1)=3: the answer is 3e^(x²−1). Next change the differential equation to dy/dx=2x(y−3), y(1)=5: the answer is 3+2e^(x²−1). Finally change y(1)=5 to y(1)=3: the constant solution y=3 is valid. These changes isolate the starting x-coordinate, equilibrium shift and zero-displacement case.
Thirty-six original initial-condition clinics
Each clinic uses a fictional separable equation dy/dx=kx(y−L), with an exact initial condition y(x₀)=y₀. The constant k may be positive or negative, the initial x-value may be zero or nonzero, and the initial point may sit above, below or on L. Work through the general family, select the constant, then verify at the initial point and with the original derivative. The exercises are mathematically checkable and do not report real student outcomes.
1. Start at (-2,-7), multiplier -6, level -4
Question. Solve dy/dx=-6x(y−(-4)) subject to the initial condition y(-2)=-7. Decide whether the initial value is above, below or equal to the level L=-4. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-4+Ae^(-3x²). Substituting the entire initial point gives A=-3e^(12), hence the solution is y=-4+(-3)e^[-3(x²−4)]. The signed displacement at the start is y₀−L=-3; the exponential coefficient is not assumed positive. At x=-2, the exponent difference x²−4 equals zero, so the expression returns -7 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-6x(y−(-4)). At the initial point this is y′(-2)=-6×-2×-3=-36, matching the right side of the given equation. The integration constant changes when the initial x-coordinate changes, even if the differential equation stays the same. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
2. Start at (-1,-5), multiplier 2, level -3
Question. Solve dy/dx=2x(y−(-3)) subject to the initial condition y(-1)=-5. Decide whether the initial value is above, below or equal to the level L=-3. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-3+Ae^(1x²). Substituting the entire initial point gives A=-2e^(-1), hence the solution is y=-3+(-2)e^[1(x²−1)]. The signed displacement at the start is y₀−L=-2; the exponential coefficient is not assumed positive. At x=-1, the exponent difference x²−1 equals zero, so the expression returns -5 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=2x(y−(-3)). At the initial point this is y′(-1)=2×-1×-2=4, matching the right side of the given equation. A positive exponential factor multiplied by a negative coefficient can produce a solution below the equilibrium level. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
3. Start at (0,-3), multiplier -6, level -2
Question. Solve dy/dx=-6x(y−(-2)) subject to the initial condition y(0)=-3. Decide whether the initial value is above, below or equal to the level L=-2. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-2+Ae^(-3x²). Substituting the entire initial point gives A=-1e^(0), hence the solution is y=-2+(-1)e^[-3(x²−0)]. The signed displacement at the start is y₀−L=-1; the exponential coefficient is not assumed positive. At x=0, the exponent difference x²−0 equals zero, so the expression returns -3 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-6x(y−(-2)). At the initial point this is y′(0)=-6×0×-1=0, matching the right side of the given equation. The exponent x²−x₀² becomes zero at the initial point; use this as a fast independent check. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
4. Start at (1,-1), multiplier 2, level -1
Question. Solve dy/dx=2x(y−(-1)) subject to the initial condition y(1)=-1. Decide whether the initial value is above, below or equal to the level L=-1. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-1+Ae^(1x²). Substituting the entire initial point gives A=0e^(-1), hence the solution is y=-1. Because y₀=L, the solution is the constant equilibrium y≡-1, verified directly in the original equation. At x=1, the exponent difference x²−1 equals zero, so the expression returns -1 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=2x(y−(-1)). At the initial point this is y′(1)=2×1×0=0, matching the right side of the given equation. When y₀=L, check the equilibrium solution directly rather than dividing by y−L. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
5. Start at (2,1), multiplier -6, level 0
Question. Solve dy/dx=-6x(y−(0)) subject to the initial condition y(2)=1. Decide whether the initial value is above, below or equal to the level L=0. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=0+Ae^(-3x²). Substituting the entire initial point gives A=1e^(12), hence the solution is y=0+(1)e^[-3(x²−4)]. The signed displacement at the start is y₀−L=1; the exponential coefficient is not assumed positive. At x=2, the exponent difference x²−4 equals zero, so the expression returns 1 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-6x(y−(0)). At the initial point this is y′(2)=-6×2×1=-12, matching the right side of the given equation. The sign of the derivative at a point depends on k, x and y−L, not merely on whether the model contains an exponential. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
6. Start at (-2,3), multiplier 2, level 1
Question. Solve dy/dx=2x(y−(1)) subject to the initial condition y(-2)=3. Decide whether the initial value is above, below or equal to the level L=1. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=1+Ae^(1x²). Substituting the entire initial point gives A=2e^(-4), hence the solution is y=1+(2)e^[1(x²−4)]. The signed displacement at the start is y₀−L=2; the exponential coefficient is not assumed positive. At x=-2, the exponent difference x²−4 equals zero, so the expression returns 3 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=2x(y−(1)). At the initial point this is y′(-2)=2×-2×2=-8, matching the right side of the given equation. Differentiate the final expression as a second check of the original differential equation. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
7. Start at (-1,5), multiplier -4, level 2
Question. Solve dy/dx=-4x(y−(2)) subject to the initial condition y(-1)=5. Decide whether the initial value is above, below or equal to the level L=2. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=2+Ae^(-2x²). Substituting the entire initial point gives A=3e^(2), hence the solution is y=2+(3)e^[-2(x²−1)]. The signed displacement at the start is y₀−L=3; the exponential coefficient is not assumed positive. At x=-1, the exponent difference x²−1 equals zero, so the expression returns 5 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-4x(y−(2)). At the initial point this is y′(-1)=-4×-1×3=12, matching the right side of the given equation. The value of x₀ belongs in the exponent and must not be replaced automatically by zero. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
8. Start at (0,0), multiplier 4, level 3
Question. Solve dy/dx=4x(y−(3)) subject to the initial condition y(0)=0. Decide whether the initial value is above, below or equal to the level L=3. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=3+Ae^(2x²). Substituting the entire initial point gives A=-3e^(0), hence the solution is y=3+(-3)e^[2(x²−0)]. The signed displacement at the start is y₀−L=-3; the exponential coefficient is not assumed positive. At x=0, the exponent difference x²−0 equals zero, so the expression returns 0 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=4x(y−(3)). At the initial point this is y′(0)=4×0×-3=0, matching the right side of the given equation. The exponent coefficient k/2 comes from integrating kx with respect to x. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
9. Start at (1,2), multiplier -4, level 4
Question. Solve dy/dx=-4x(y−(4)) subject to the initial condition y(1)=2. Decide whether the initial value is above, below or equal to the level L=4. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=4+Ae^(-2x²). Substituting the entire initial point gives A=-2e^(2), hence the solution is y=4+(-2)e^[-2(x²−1)]. The signed displacement at the start is y₀−L=-2; the exponential coefficient is not assumed positive. At x=1, the exponent difference x²−1 equals zero, so the expression returns 2 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-4x(y−(4)). At the initial point this is y′(1)=-4×1×-2=8, matching the right side of the given equation. Changing only y₀ changes the exponential multiplier, not the equilibrium level L. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
10. Start at (2,-5), multiplier 4, level -4
Question. Solve dy/dx=4x(y−(-4)) subject to the initial condition y(2)=-5. Decide whether the initial value is above, below or equal to the level L=-4. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-4+Ae^(2x²). Substituting the entire initial point gives A=-1e^(-8), hence the solution is y=-4+(-1)e^[2(x²−4)]. The signed displacement at the start is y₀−L=-1; the exponential coefficient is not assumed positive. At x=2, the exponent difference x²−4 equals zero, so the expression returns -5 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=4x(y−(-4)). At the initial point this is y′(2)=4×2×-1=-8, matching the right side of the given equation. Even if a calculator returns a plausible curve, the exact initial condition and derivative must both hold. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
11. Start at (-2,-3), multiplier -4, level -3
Question. Solve dy/dx=-4x(y−(-3)) subject to the initial condition y(-2)=-3. Decide whether the initial value is above, below or equal to the level L=-3. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-3+Ae^(-2x²). Substituting the entire initial point gives A=0e^(8), hence the solution is y=-3. Because y₀=L, the solution is the constant equilibrium y≡-3, verified directly in the original equation. At x=-2, the exponent difference x²−4 equals zero, so the expression returns -3 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-4x(y−(-3)). At the initial point this is y′(-2)=-4×-2×0=0, matching the right side of the given equation. The integration constant changes when the initial x-coordinate changes, even if the differential equation stays the same. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
12. Start at (-1,-1), multiplier 4, level -2
Question. Solve dy/dx=4x(y−(-2)) subject to the initial condition y(-1)=-1. Decide whether the initial value is above, below or equal to the level L=-2. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-2+Ae^(2x²). Substituting the entire initial point gives A=1e^(-2), hence the solution is y=-2+(1)e^[2(x²−1)]. The signed displacement at the start is y₀−L=1; the exponential coefficient is not assumed positive. At x=-1, the exponent difference x²−1 equals zero, so the expression returns -1 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=4x(y−(-2)). At the initial point this is y′(-1)=4×-1×1=-4, matching the right side of the given equation. A positive exponential factor multiplied by a negative coefficient can produce a solution below the equilibrium level. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
13. Start at (0,1), multiplier -2, level -1
Question. Solve dy/dx=-2x(y−(-1)) subject to the initial condition y(0)=1. Decide whether the initial value is above, below or equal to the level L=-1. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-1+Ae^(-1x²). Substituting the entire initial point gives A=2e^(0), hence the solution is y=-1+(2)e^[-1(x²−0)]. The signed displacement at the start is y₀−L=2; the exponential coefficient is not assumed positive. At x=0, the exponent difference x²−0 equals zero, so the expression returns 1 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-2x(y−(-1)). At the initial point this is y′(0)=-2×0×2=0, matching the right side of the given equation. The exponent x²−x₀² becomes zero at the initial point; use this as a fast independent check. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
14. Start at (1,3), multiplier 6, level 0
Question. Solve dy/dx=6x(y−(0)) subject to the initial condition y(1)=3. Decide whether the initial value is above, below or equal to the level L=0. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=0+Ae^(3x²). Substituting the entire initial point gives A=3e^(-3), hence the solution is y=0+(3)e^[3(x²−1)]. The signed displacement at the start is y₀−L=3; the exponential coefficient is not assumed positive. At x=1, the exponent difference x²−1 equals zero, so the expression returns 3 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=6x(y−(0)). At the initial point this is y′(1)=6×1×3=18, matching the right side of the given equation. When y₀=L, check the equilibrium solution directly rather than dividing by y−L. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
15. Start at (2,-2), multiplier -2, level 1
Question. Solve dy/dx=-2x(y−(1)) subject to the initial condition y(2)=-2. Decide whether the initial value is above, below or equal to the level L=1. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=1+Ae^(-1x²). Substituting the entire initial point gives A=-3e^(4), hence the solution is y=1+(-3)e^[-1(x²−4)]. The signed displacement at the start is y₀−L=-3; the exponential coefficient is not assumed positive. At x=2, the exponent difference x²−4 equals zero, so the expression returns -2 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-2x(y−(1)). At the initial point this is y′(2)=-2×2×-3=12, matching the right side of the given equation. The sign of the derivative at a point depends on k, x and y−L, not merely on whether the model contains an exponential. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
16. Start at (-2,0), multiplier 6, level 2
Question. Solve dy/dx=6x(y−(2)) subject to the initial condition y(-2)=0. Decide whether the initial value is above, below or equal to the level L=2. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=2+Ae^(3x²). Substituting the entire initial point gives A=-2e^(-12), hence the solution is y=2+(-2)e^[3(x²−4)]. The signed displacement at the start is y₀−L=-2; the exponential coefficient is not assumed positive. At x=-2, the exponent difference x²−4 equals zero, so the expression returns 0 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=6x(y−(2)). At the initial point this is y′(-2)=6×-2×-2=24, matching the right side of the given equation. Differentiate the final expression as a second check of the original differential equation. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
17. Start at (-1,2), multiplier -2, level 3
Question. Solve dy/dx=-2x(y−(3)) subject to the initial condition y(-1)=2. Decide whether the initial value is above, below or equal to the level L=3. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=3+Ae^(-1x²). Substituting the entire initial point gives A=-1e^(1), hence the solution is y=3+(-1)e^[-1(x²−1)]. The signed displacement at the start is y₀−L=-1; the exponential coefficient is not assumed positive. At x=-1, the exponent difference x²−1 equals zero, so the expression returns 2 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-2x(y−(3)). At the initial point this is y′(-1)=-2×-1×-1=-2, matching the right side of the given equation. The value of x₀ belongs in the exponent and must not be replaced automatically by zero. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
18. Start at (0,4), multiplier 6, level 4
Question. Solve dy/dx=6x(y−(4)) subject to the initial condition y(0)=4. Decide whether the initial value is above, below or equal to the level L=4. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=4+Ae^(3x²). Substituting the entire initial point gives A=0e^(0), hence the solution is y=4. Because y₀=L, the solution is the constant equilibrium y≡4, verified directly in the original equation. At x=0, the exponent difference x²−0 equals zero, so the expression returns 4 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=6x(y−(4)). At the initial point this is y′(0)=6×0×0=0, matching the right side of the given equation. The exponent coefficient k/2 comes from integrating kx with respect to x. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
19. Start at (1,-3), multiplier 2, level -4
Question. Solve dy/dx=2x(y−(-4)) subject to the initial condition y(1)=-3. Decide whether the initial value is above, below or equal to the level L=-4. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-4+Ae^(1x²). Substituting the entire initial point gives A=1e^(-1), hence the solution is y=-4+(1)e^[1(x²−1)]. The signed displacement at the start is y₀−L=1; the exponential coefficient is not assumed positive. At x=1, the exponent difference x²−1 equals zero, so the expression returns -3 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=2x(y−(-4)). At the initial point this is y′(1)=2×1×1=2, matching the right side of the given equation. Changing only y₀ changes the exponential multiplier, not the equilibrium level L. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
20. Start at (2,-1), multiplier -6, level -3
Question. Solve dy/dx=-6x(y−(-3)) subject to the initial condition y(2)=-1. Decide whether the initial value is above, below or equal to the level L=-3. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-3+Ae^(-3x²). Substituting the entire initial point gives A=2e^(12), hence the solution is y=-3+(2)e^[-3(x²−4)]. The signed displacement at the start is y₀−L=2; the exponential coefficient is not assumed positive. At x=2, the exponent difference x²−4 equals zero, so the expression returns -1 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-6x(y−(-3)). At the initial point this is y′(2)=-6×2×2=-24, matching the right side of the given equation. Even if a calculator returns a plausible curve, the exact initial condition and derivative must both hold. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
21. Start at (-2,1), multiplier 2, level -2
Question. Solve dy/dx=2x(y−(-2)) subject to the initial condition y(-2)=1. Decide whether the initial value is above, below or equal to the level L=-2. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-2+Ae^(1x²). Substituting the entire initial point gives A=3e^(-4), hence the solution is y=-2+(3)e^[1(x²−4)]. The signed displacement at the start is y₀−L=3; the exponential coefficient is not assumed positive. At x=-2, the exponent difference x²−4 equals zero, so the expression returns 1 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=2x(y−(-2)). At the initial point this is y′(-2)=2×-2×3=-12, matching the right side of the given equation. The integration constant changes when the initial x-coordinate changes, even if the differential equation stays the same. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
22. Start at (-1,-4), multiplier -6, level -1
Question. Solve dy/dx=-6x(y−(-1)) subject to the initial condition y(-1)=-4. Decide whether the initial value is above, below or equal to the level L=-1. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-1+Ae^(-3x²). Substituting the entire initial point gives A=-3e^(3), hence the solution is y=-1+(-3)e^[-3(x²−1)]. The signed displacement at the start is y₀−L=-3; the exponential coefficient is not assumed positive. At x=-1, the exponent difference x²−1 equals zero, so the expression returns -4 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-6x(y−(-1)). At the initial point this is y′(-1)=-6×-1×-3=-18, matching the right side of the given equation. A positive exponential factor multiplied by a negative coefficient can produce a solution below the equilibrium level. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
23. Start at (0,-2), multiplier 2, level 0
Question. Solve dy/dx=2x(y−(0)) subject to the initial condition y(0)=-2. Decide whether the initial value is above, below or equal to the level L=0. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=0+Ae^(1x²). Substituting the entire initial point gives A=-2e^(0), hence the solution is y=0+(-2)e^[1(x²−0)]. The signed displacement at the start is y₀−L=-2; the exponential coefficient is not assumed positive. At x=0, the exponent difference x²−0 equals zero, so the expression returns -2 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=2x(y−(0)). At the initial point this is y′(0)=2×0×-2=0, matching the right side of the given equation. The exponent x²−x₀² becomes zero at the initial point; use this as a fast independent check. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
24. Start at (1,0), multiplier -6, level 1
Question. Solve dy/dx=-6x(y−(1)) subject to the initial condition y(1)=0. Decide whether the initial value is above, below or equal to the level L=1. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=1+Ae^(-3x²). Substituting the entire initial point gives A=-1e^(3), hence the solution is y=1+(-1)e^[-3(x²−1)]. The signed displacement at the start is y₀−L=-1; the exponential coefficient is not assumed positive. At x=1, the exponent difference x²−1 equals zero, so the expression returns 0 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-6x(y−(1)). At the initial point this is y′(1)=-6×1×-1=6, matching the right side of the given equation. When y₀=L, check the equilibrium solution directly rather than dividing by y−L. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
25. Start at (2,2), multiplier 4, level 2
Question. Solve dy/dx=4x(y−(2)) subject to the initial condition y(2)=2. Decide whether the initial value is above, below or equal to the level L=2. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=2+Ae^(2x²). Substituting the entire initial point gives A=0e^(-8), hence the solution is y=2. Because y₀=L, the solution is the constant equilibrium y≡2, verified directly in the original equation. At x=2, the exponent difference x²−4 equals zero, so the expression returns 2 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=4x(y−(2)). At the initial point this is y′(2)=4×2×0=0, matching the right side of the given equation. The sign of the derivative at a point depends on k, x and y−L, not merely on whether the model contains an exponential. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
26. Start at (-2,4), multiplier -4, level 3
Question. Solve dy/dx=-4x(y−(3)) subject to the initial condition y(-2)=4. Decide whether the initial value is above, below or equal to the level L=3. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=3+Ae^(-2x²). Substituting the entire initial point gives A=1e^(8), hence the solution is y=3+(1)e^[-2(x²−4)]. The signed displacement at the start is y₀−L=1; the exponential coefficient is not assumed positive. At x=-2, the exponent difference x²−4 equals zero, so the expression returns 4 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-4x(y−(3)). At the initial point this is y′(-2)=-4×-2×1=8, matching the right side of the given equation. Differentiate the final expression as a second check of the original differential equation. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
27. Start at (-1,6), multiplier 4, level 4
Question. Solve dy/dx=4x(y−(4)) subject to the initial condition y(-1)=6. Decide whether the initial value is above, below or equal to the level L=4. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=4+Ae^(2x²). Substituting the entire initial point gives A=2e^(-2), hence the solution is y=4+(2)e^[2(x²−1)]. The signed displacement at the start is y₀−L=2; the exponential coefficient is not assumed positive. At x=-1, the exponent difference x²−1 equals zero, so the expression returns 6 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=4x(y−(4)). At the initial point this is y′(-1)=4×-1×2=-8, matching the right side of the given equation. The value of x₀ belongs in the exponent and must not be replaced automatically by zero. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
28. Start at (0,-1), multiplier -4, level -4
Question. Solve dy/dx=-4x(y−(-4)) subject to the initial condition y(0)=-1. Decide whether the initial value is above, below or equal to the level L=-4. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-4+Ae^(-2x²). Substituting the entire initial point gives A=3e^(0), hence the solution is y=-4+(3)e^[-2(x²−0)]. The signed displacement at the start is y₀−L=3; the exponential coefficient is not assumed positive. At x=0, the exponent difference x²−0 equals zero, so the expression returns -1 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-4x(y−(-4)). At the initial point this is y′(0)=-4×0×3=0, matching the right side of the given equation. The exponent coefficient k/2 comes from integrating kx with respect to x. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
29. Start at (1,-6), multiplier 4, level -3
Question. Solve dy/dx=4x(y−(-3)) subject to the initial condition y(1)=-6. Decide whether the initial value is above, below or equal to the level L=-3. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-3+Ae^(2x²). Substituting the entire initial point gives A=-3e^(-2), hence the solution is y=-3+(-3)e^[2(x²−1)]. The signed displacement at the start is y₀−L=-3; the exponential coefficient is not assumed positive. At x=1, the exponent difference x²−1 equals zero, so the expression returns -6 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=4x(y−(-3)). At the initial point this is y′(1)=4×1×-3=-12, matching the right side of the given equation. Changing only y₀ changes the exponential multiplier, not the equilibrium level L. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
30. Start at (2,-4), multiplier -4, level -2
Question. Solve dy/dx=-4x(y−(-2)) subject to the initial condition y(2)=-4. Decide whether the initial value is above, below or equal to the level L=-2. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-2+Ae^(-2x²). Substituting the entire initial point gives A=-2e^(8), hence the solution is y=-2+(-2)e^[-2(x²−4)]. The signed displacement at the start is y₀−L=-2; the exponential coefficient is not assumed positive. At x=2, the exponent difference x²−4 equals zero, so the expression returns -4 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-4x(y−(-2)). At the initial point this is y′(2)=-4×2×-2=16, matching the right side of the given equation. Even if a calculator returns a plausible curve, the exact initial condition and derivative must both hold. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
31. Start at (-2,-2), multiplier 6, level -1
Question. Solve dy/dx=6x(y−(-1)) subject to the initial condition y(-2)=-2. Decide whether the initial value is above, below or equal to the level L=-1. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=-1+Ae^(3x²). Substituting the entire initial point gives A=-1e^(-12), hence the solution is y=-1+(-1)e^[3(x²−4)]. The signed displacement at the start is y₀−L=-1; the exponential coefficient is not assumed positive. At x=-2, the exponent difference x²−4 equals zero, so the expression returns -2 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=6x(y−(-1)). At the initial point this is y′(-2)=6×-2×-1=12, matching the right side of the given equation. The integration constant changes when the initial x-coordinate changes, even if the differential equation stays the same. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
32. Start at (-1,0), multiplier -2, level 0
Question. Solve dy/dx=-2x(y−(0)) subject to the initial condition y(-1)=0. Decide whether the initial value is above, below or equal to the level L=0. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=0+Ae^(-1x²). Substituting the entire initial point gives A=0e^(1), hence the solution is y=0. Because y₀=L, the solution is the constant equilibrium y≡0, verified directly in the original equation. At x=-1, the exponent difference x²−1 equals zero, so the expression returns 0 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-2x(y−(0)). At the initial point this is y′(-1)=-2×-1×0=0, matching the right side of the given equation. A positive exponential factor multiplied by a negative coefficient can produce a solution below the equilibrium level. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
33. Start at (0,2), multiplier 6, level 1
Question. Solve dy/dx=6x(y−(1)) subject to the initial condition y(0)=2. Decide whether the initial value is above, below or equal to the level L=1. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=1+Ae^(3x²). Substituting the entire initial point gives A=1e^(0), hence the solution is y=1+(1)e^[3(x²−0)]. The signed displacement at the start is y₀−L=1; the exponential coefficient is not assumed positive. At x=0, the exponent difference x²−0 equals zero, so the expression returns 2 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=6x(y−(1)). At the initial point this is y′(0)=6×0×1=0, matching the right side of the given equation. The exponent x²−x₀² becomes zero at the initial point; use this as a fast independent check. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
34. Start at (1,4), multiplier -2, level 2
Question. Solve dy/dx=-2x(y−(2)) subject to the initial condition y(1)=4. Decide whether the initial value is above, below or equal to the level L=2. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=2+Ae^(-1x²). Substituting the entire initial point gives A=2e^(1), hence the solution is y=2+(2)e^[-1(x²−1)]. The signed displacement at the start is y₀−L=2; the exponential coefficient is not assumed positive. At x=1, the exponent difference x²−1 equals zero, so the expression returns 4 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-2x(y−(2)). At the initial point this is y′(1)=-2×1×2=-4, matching the right side of the given equation. When y₀=L, check the equilibrium solution directly rather than dividing by y−L. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
35. Start at (2,6), multiplier 6, level 3
Question. Solve dy/dx=6x(y−(3)) subject to the initial condition y(2)=6. Decide whether the initial value is above, below or equal to the level L=3. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=3+Ae^(3x²). Substituting the entire initial point gives A=3e^(-12), hence the solution is y=3+(3)e^[3(x²−4)]. The signed displacement at the start is y₀−L=3; the exponential coefficient is not assumed positive. At x=2, the exponent difference x²−4 equals zero, so the expression returns 6 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=6x(y−(3)). At the initial point this is y′(2)=6×2×3=36, matching the right side of the given equation. The sign of the derivative at a point depends on k, x and y−L, not merely on whether the model contains an exponential. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
36. Start at (-2,1), multiplier -2, level 4
Question. Solve dy/dx=-2x(y−(4)) subject to the initial condition y(-2)=1. Decide whether the initial value is above, below or equal to the level L=4. Use separation to get a general family, but do not stop until the condition has selected the correct constant. Then calculate the instantaneous derivative at the supplied point.
Particular solution. The separated non-equilibrium family is y=4+Ae^(-1x²). Substituting the entire initial point gives A=-3e^(4), hence the solution is y=4+(-3)e^[-1(x²−4)]. The signed displacement at the start is y₀−L=-3; the exponential coefficient is not assumed positive. At x=-2, the exponent difference x²−4 equals zero, so the expression returns 1 as required.
Derivative and interpretation check. Differentiating the particular solution gives dy/dx=-2x(y−(4)). At the initial point this is y′(-2)=-2×-2×-3=-12, matching the right side of the given equation. Differentiate the final expression as a second check of the original differential equation. As a changed-case test, move only the starting x-coordinate while keeping y₀ fixed. The constant A changes, even though the differential equation and equilibrium level stay the same.
Ten quick exact checks
- dy/dx=2xy, y(0)=3 ⇒ y=3e^(x²).
- dy/dx=2xy, y(1)=3 ⇒ y=3e^(x²−1).
- dy/dx=2x(y−3), y(1)=5 ⇒ y=3+2e^(x²−1).
- dy/dx=2x(y−3), y(1)=3 ⇒ y≡3.
- dy/dx=−2xy, y(2)=5 ⇒ y=5e^(4−x²).
- dy/dx=−0.5y, y(0)=80 ⇒ y=80e^(−0.5x).
- The general family for dy/dx=kx(y−L) is y=L+Ae^(kx²/2).
- The initial condition y(x₀)=y₀ gives y=L+(y₀−L)e^[k(x²−x₀²)/2].
- At x=x₀, the exponent difference is zero, so the initial value is recovered.
- Differentiating the particular solution checks the differential equation independently of the point condition.
A five-day study and teaching plan
Day one: separate and integrate dy/dx=kxy, keeping an arbitrary constant before using a condition. Day two: practise initial points where x₀ is one or two rather than zero. Day three: introduce shifted equations dy/dx=kx(y−L) and retain the level L during separation. Day four: handle negative exponential coefficients and the constant solution selected when y₀=L. Day five: mix equations and starting points without a title revealing the intended trick, then verify every answer by differentiation and by substitution into the condition.
In a small-group lesson of three learners, one can separate variables, one select the constant using the initial point, and the third verify the derivative and condition independently. Rotate those roles. A learner has mastered the method when they can explain why two different initial points on the same differential equation give two different particular curves without changing the general solution family.
Frequently asked questions
What is a particular solution?
It is one solution of a differential equation chosen from the general family by a stated condition, such as y(x₀)=y₀.
Why keep C during integration?
The constant represents the family of functions before a particular initial value is given.
Must the initial point be at zero?
No. The point can be at any allowed x₀, and the full ordered pair must be used to determine the constant.
Why does ln|y−L| appear?
Separating dy/(y−L) produces a logarithm on a branch where y−L≠0. The absolute value covers positive or negative displacement.
What if the initial value equals L?
The constant equilibrium solution y≡L should be checked directly, especially if a separation step divided by y−L.
Can a curve satisfy the initial point but fail the equation?
Yes. Matching one point alone does not guarantee its derivative satisfies the differential equation for all x.
How do I check an exponential solution?
Differentiate it using the chain rule and compare with the stated right-hand side, then evaluate at the initial point.
Is this included in 9758 for 2027?
Yes. Section 5.5 includes general and particular solutions of separable differential equations.
Further study and official source
- Discrete Probability Tables: Missing p and Expected Value
- Differential Equations: Initial Conditions and Particular Solutions: this guide
- Vectors: Line of Intersection Between Two Planes
- Committee Selection: At Least One From Each Group
Related articles: Don’t Lose Equilibrium Solutions, the Bukit Timah JC2 tuition hub, and the immutable Clementi three-student Mathematics reference for the small-group teaching format at a different level.
Official reference: SEAB H2 Mathematics 9758, 2027 examination syllabus, §5.5 Differential equations. Every example is original mathematical teaching content, not an official examination item or a real student record.