Verifying Solutions for Differential Equations

AP Course

AP Calculus AB/BC

Study the complete College Board sequence for AP Calculus AB and BC, from limits through infinite series.

Choose an official unit and topic to open its lecture, concept check, or focused practice.

AP Calculus AB/BC · Unit 7 · Topic 7.2

Verifying Solutions for Differential Equations

Differentiate a proposed function and substitute into the equation and initial condition.

1. Topic Focus

Model rates with differential equations, read slope fields, approximate solutions, solve separable equations, and interpret exponential or logistic models.

This topic: Differentiate a proposed function and substitute into the equation and initial condition.

2. Key Relationship

\(y=g(x)\Rightarrow y'=g'(x)\stackrel{?}{=}F(x,g(x))\)

Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.

3. Visual Connection

Model and verificationTranslate units and rate language into an equation, then substitute the proposed function and initial value.

4. Worked Example

Check y=Ce^(2x) by showing y′=2y.

Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.

5. Concept Development

A proposed function is a solution of a differential equation only when substituting the function and its required derivatives makes the equation true for every input in an appropriate interval. Verification is a differentiation-and-substitution argument; it is not the same as solving the differential equation.

\(y=f(x)\text{ solves }y'=F(x,y)\quad\Longleftrightarrow\quad f'(x)=F\bigl(x,f(x)\bigr)\)

An initial-value problem adds another requirement. A candidate must satisfy both the differential equation and every stated initial condition. Passing one test does not compensate for failing the other.

The verification protocol

State the candidate and interval. Note where the function and the differential equation are defined.
Differentiate independently. Compute every derivative appearing in the equation, using the candidate function rather than the differential equation.
Substitute completely. Replace \(y\), \(y'\), \(y''\), and other required expressions.
Compare the two sides. Simplify until they are visibly equal on the entire interval, or exhibit a mismatch.
Check initial conditions separately. Evaluate the candidate and any required derivatives at the specified input.
Write a precise conclusion. Distinguish “solves the differential equation” from “solves the initial-value problem.”

What counts as proof?

Evidence	What it establishes
Both sides simplify to the same expression	The candidate solves the equation on the stated interval.
The two sides differ at one allowed input	The candidate is not a solution on that interval.
The two sides agree at one or several sampled inputs	Only those samples agree; this does not prove an identity.
The differential equation holds but an initial value fails	The candidate solves the equation, but not the initial-value problem.
A parameter remains arbitrary after substitution	The formula may represent an infinite family of solutions.

Families and particular solutions

Differentiation removes additive constants in equations such as \(y'=2x\), so every function \(y=x^2+C\) is a solution. Other equations produce multiplicative families: every \(y=Ce^{kx}\) satisfies \(y'=ky\). An initial condition determines a particular member of the family when the problem has a unique solution.

Intervals and domain restrictions

A formula can satisfy a differential equation on one interval without defining a solution across a singularity. For example, \(y=1/(C-x)\) is undefined at \(x=C\). It solves \(y'=y^2\) on any interval lying entirely to one side of \(C\), but not on an interval containing \(C\). Check restrictions from denominators, logarithms, radicals, and the differential equation itself.

Explicit, implicit, and integral forms

A candidate need not arrive as \(y=f(x)\). For an implicit relation, differentiate the relation and determine where the resulting derivative is valid. For a function defined by an integral, the Fundamental Theorem of Calculus often supplies the derivative immediately. In every representation, the final task is the same: show that the required derivative relationship holds.

AP reasoning habit: Compute the left and right sides separately and conclude with a sentence naming both the interval and any initial condition. A numerical spot check is useful for finding an error, but an algebraic identity is needed to verify a solution.

6. Detailed Worked Example and Error Check

Example 1: Direct verification

Determine whether \(y=2e^{-3x}+x\) solves \(y'+3y=3x+1\).

\(\begin{aligned}y'&=-6e^{-3x}+1,\\y'+3y&=(-6e^{-3x}+1)+3(2e^{-3x}+x)=3x+1.\end{aligned}\)

The substituted left side equals the right side for every real \(x\), so the candidate is a solution on \(( -\infty,\infty )\).

Example 2: Equation passes, initial value fails

Test \(y=x^2+1\) for the initial-value problem \(y'=2x\), \(y(1)=3\).

\(y'=2x\quad\text{but}\quad y(1)=1^2+1=2\ne3.\)

The function solves the differential equation on all real numbers, but it does not solve the stated initial-value problem.

Example 3: Verifying an infinite family

For any real constant \(C\), let \(y=Ce^{2t}\). Then

\(y'=2Ce^{2t}=2y.\)

Thus every value of \(C\) gives a solution of \(y'=2y\). If \(y(0)=7\), then \(C=7\), selecting the particular solution \(y=7e^{2t}\).

Example 4: A nonlinear equation and its interval

Consider \(y=1/(2-x)\) and \(y'=y^2\).

\(y'=\frac{1}{(2-x)^2},\qquad y^2=\left(\frac{1}{2-x}\right)^2=\frac{1}{(2-x)^2}.\)

The identity holds wherever \(x\ne2\). The candidate is a solution on \(( -\infty,2 )\) or on \(( 2,\infty )\), but not on a single interval crossing \(x=2\). It also satisfies \(y(0)=1/2\) on the first interval.

Example 5: Implicitly described solutions

The circle \(x^2+y^2=25\) implicitly describes upper and lower branches. Differentiate the relation:

\(2x+2y\frac{dy}{dx}=0\quad\Longrightarrow\quad\frac{dy}{dx}=-\frac{x}{y}.\)

Therefore each differentiable branch satisfies \(y'=-x/y\) wherever \(y\ne0\). The points \((\pm5,0)\) are excluded because the differential equation is undefined there and the circle has vertical tangents.

Example 6: A second-order equation

Verify \(y=3\cos x-2\sin x\) for \(y''+y=0\).

\(y'=-3\sin x-2\cos x,\qquad y''=-3\cos x+2\sin x=-y.\)

Hence \(y''+y=0\) for all real \(x\). A second-order equation requires computing through the second derivative; checking only \(y'\) is incomplete.

Example 7: A function defined by accumulation

Let

\(F(x)=4+\int_1^x(t^2+\cos t)\,dt.\)

By the Fundamental Theorem of Calculus, \(F'(x)=x^2+\cos x\). Also \(F(1)=4\) because the integral has equal limits. Therefore \(F\) solves the initial-value problem

\(y'=x^2+\cos x,\qquad y(1)=4.\)

Example 8: Why checking points is insufficient

For \(y=x^2\) and \(y'=x+y\), both sides agree at \(x=0\) and \(x=1\): \(2x=x+x^2\) at those two inputs. However,

\(2x-(x+x^2)=x-x^2\)

is not identically zero. At \(x=2\), the left side is \(4\) and the right side is \(6\), so \(y=x^2\) is not a solution on any interval containing \(2\).

Common verification errors

Using the differential equation itself to declare what \(y'\) is instead of differentiating the candidate.
Substituting for \(y'\) but forgetting to replace \(y\).
Checking only the initial condition.
Testing one convenient point and treating agreement there as proof.
Ignoring a singularity that splits the domain into separate solution intervals.
Concluding “not a solution” when only the initial condition failed.

7. AP Reasoning Routine

Translate the context into a rate equation, verify candidate solutions by substitution, carry constants through integration, and apply initial conditions last.

Identify the representation and requested quantity.
State the rule or theorem and verify its conditions.
Keep exact values until the final requested approximation.
Interpret sign, units, interval, and context.

AP Checkpoint

Verify or reject each claim. Show the derivative substitution, state a valid interval when relevant, and check all initial data.
(a) \(y=4e^{-3x}\) solves \(y'=-3y\).
(b) \(y=x^3\) solves \(y'=y/x\) on \(x>0\).
(c) \(y=x^2+2\) solves \(y'=2x\), \(y(1)=4\).
(d) Every \(y=Ce^{5t}\) solves \(y'=5y\). Which member satisfies \(y(0)=-2\)?
(e) \(y=\tan x\) solves \(y'=1+y^2\) on \((-\pi/2,\pi/2)\).
(f) \(y=1/(2-x)\) solves \(y'=y^2\), \(y(0)=1/2\). Give the largest solution interval containing \(0\).
(g) The relation \(x^2+y^2=9\) describes solutions of \(y'=-x/y\) wherever \(y\ne0\).
(h) \(y=\sin x\) solves \(y''+y=0\), \(y(0)=0\), \(y'(0)=1\).
(i) \(F(x)=2+\int_1^x(t^2+1)\,dt\) solves \(y'=x^2+1\), \(y(1)=2\).
(j) A student checks \(x=0\) and \(x=1\) and concludes that \(y=x^2\) solves \(y'=x+y\). Explain the flaw and give a counterexample input.

Check the solution

(a) \(y'=-12e^{-3x}\), while \(-3y=-12e^{-3x}\). Yes, on all real numbers.
(b) \(y'=3x^2\), but \(y/x=x^2\) for \(x>0\). These are not identical, so no.
(c) The differential equation holds because \(y'=2x\), but \(y(1)=3\ne4\). It is not a solution of the initial-value problem.
(d) \(y'=5Ce^{5t}=5y\), so every real \(C\) works. Since \(y(0)=C\), the initial condition gives \(C=-2\) and \(y=-2e^{5t}\).
(e) \(y'=\sec^2x=1+\tan^2x=1+y^2\). Yes, on the stated interval.
(f) \(y'=1/(2-x)^2=y^2\) and \(y(0)=1/2\). The largest interval containing \(0\) that avoids the singularity is \((-\infty,2)\).
(g) Implicit differentiation gives \(2x+2yy'=0\), hence \(y'=-x/y\) when \(y\ne0\). The claim is valid separately on differentiable arcs that avoid \((\pm3,0)\).
(h) \(y'=\cos x\) and \(y''=-\sin x=-y\), so the equation holds. Also \(y(0)=0\) and \(y'(0)=1\), so the complete initial-value problem is satisfied.
(i) The Fundamental Theorem gives \(F'(x)=x^2+1\), and \(F(1)=2+0=2\). Yes.
(j) Agreement at finitely many inputs does not prove equality for all inputs. At \(x=2\), \(y'=2x=4\), while \(x+y=2+4=6\), so the candidate is not a solution on an interval containing \(2\).