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Calculus AB · Unit 7 · 7.2
AP Calculus · Differential Equations
Verifying Solutions for Differential Equations
Differentiate the candidate, simplify each side on its own, and see whether they agree everywhere.
key idea$\text{Differentiate the candidate, substitute both sides, and see whether they match}$
The procedure
Verifying is much easier than solving — the answer is handed to you and you only have to check it. Differentiate the proposed function, substitute it and its derivative into the equation, and simplify each side independently until they either agree or do not.
1
Differentiate the candidate
Take $\dfrac{dy}{dx}$ of the proposed solution. If it is given implicitly, differentiate implicitly and solve for $\dfrac{dy}{dx}$.
in a candidate $y$ out $y'$
2
Substitute into the LEFT side
Replace $\dfrac{dy}{dx}$ with what you just computed. Simplify as far as it goes, then stop.
out a simplified expression
3
Substitute into the RIGHT side, separately
Replace every $y$ with the candidate. Simplify this side on its own — do not move terms between the sides.
out a second expression
4
Compare
Identical for all $x$ in the domain? It is a solution. Equal at only a few points? It is not.
out a verdict
5
Check the initial condition, if there is one
A particular solution must satisfy the differential equation and pass through the given point. Both, not either.
out general or particular
what counts as verified
- A solution is a function, so the two sides must agree identically, not just at one convenient value of $x$.
- Verification never requires solving. Differentiation is the only operation involved.
- A general solution contains $C$ and satisfies the equation for every value of $C$.
- A particular solution satisfies the equation and the initial condition.
| Candidate | Its derivative | Equation | Verdict |
|---|
| $y=3e^{2x}$ | $y'=6e^{2x}$ | $y'=2y$ | solution — $2y=6e^{2x}$ ✓ |
| $y=e^{2x}+1$ | $y'=2e^{2x}$ | $y'=2y$ | no — $2y=2e^{2x}+2$ |
| $y=Ce^{2x}$ | $y'=2Ce^{2x}$ | $y'=2y$ | general solution, any $C$ |
| $y=0$ | $y'=0$ | $y'=2y$ | solution — the equilibrium, $C=0$ |
| $y=x^{2}$ | $y'=2x$ | $y'=2y$ | no — $2y=2x^{2}\neq2x$ |
Row 2 is the instructive failure: adding a constant to an exponential solution breaks it, even though adding a constant to an antiderivative is always fine. The two situations are not the same.
Worked examples
EX 1
Verify an explicit solution
substitute both sidesShow that $y=2e^{3x}$ satisfies $\dfrac{dy}{dx}=3y$.
1
Differentiate the candidate
$\dfrac{dy}{dx}=2\cdot3e^{3x}=6e^{3x}$.
2
Evaluate the left side
Left $=6e^{3x}$.
3
Evaluate the right side separately
Right $=3y=3\left(2e^{3x}\right)=6e^{3x}$.
4
Compare
$6e^{3x}=6e^{3x}$ for every $x$. ✓
5
State the conclusion
$y=2e^{3x}$ is a solution. Note that no integration occurred anywhere.
EX 2
Reject a near miss
agreement at a point is not enoughIs $y=x^{2}+1$ a solution of $\dfrac{dy}{dx}=2y$?
1
Differentiate
$\dfrac{dy}{dx}=2x$.
4
Compare
$2x=2x^{2}+2$ only where $x^{2}-x+1=0$, which has no real roots — so they never agree.
5
The general warning
Even if the two sides had matched at one value of $x$, that would not make it a solution. Identical means for all $x$.
EX 3
Verify a particular solution
two conditions to checkVerify that $y=4e^{-x}+3$ solves $\dfrac{dy}{dx}=3-y$ with $y(0)=7$.
1
Differentiate
$\dfrac{dy}{dx}=-4e^{-x}$.
2
Check the equation
Right $=3-y=3-\left(4e^{-x}+3\right)=-4e^{-x}$, which equals the left side. ✓
3
Check the initial condition
$y(0)=4e^{0}+3=4+3=7$. ✓
4
Conclude
Both requirements hold, so this is the particular solution through $(0,7)$.
5
Why both matter
$y=9e^{-x}+3$ also satisfies the equation but gives $y(0)=12$ — correct equation, wrong curve.
EX 4
An implicitly defined solution
differentiate implicitly firstVerify that $x^{2}+y^{2}=25$ is a solution of $\dfrac{dy}{dx}=-\dfrac{x}{y}$.
1
Differentiate implicitly
$2x+2y\dfrac{dy}{dx}=0$.
2
Solve for the derivative
$\dfrac{dy}{dx}=-\dfrac{2x}{2y}=-\dfrac{x}{y}$.
3
Compare with the equation
Exactly the given right-hand side. ✓
4
Note the domain restriction
The relation fails to define $y$ as a function at $y=0$, and the equation itself is undefined there. Verification holds wherever $y\neq0$.
5
The technique
For an implicit candidate there is nothing to substitute — implicit differentiation produces the derivative directly.
Where it goes wrong
Simplify each side independently
Moving terms across the equals sign assumes the very thing you are testing. Reduce the left, reduce the right, then compare.
Checking one value of $x$ proves nothing
Two different functions can cross. A solution must satisfy the equation at every point of the domain.
A particular solution has two hurdles
The equation and the initial condition. Passing only the first identifies a family member, not the member you were asked for.
$+C$ does not always survive
For $y'=2y$, the general solution is $Ce^{2x}$ — the constant multiplies. Adding a constant instead breaks the equation.