7.2 - Verifying Solutions for Differential Equations

Particular vs. General Solutions

With differential equations, there are two types of solutions.

General solution: contains an arbitrary constant

C

from integration.

y = \int f(x)\,dx + C

Particular solution: a specific solution where

C

has been pinned down using an initial condition, which is a given point the function passes through.

Because

C

can be any number, a differential equation usually has infinitely many solutions, one for each value of

C

. Together they form a family of curves, and an initial condition picks out just one of them.

Verifying a Solution

To check whether a function is a solution, you don’t have to solve anything. Use derivatives:

Find the derivative of the proposed function (or the derivatives, if the equation has $\frac{d^2y}{dx^2}$ )

Substitute the function and its derivatives into both sides of the differential equation

Simplify. If the two sides match for every $x$ , it’s a solution

Example: Show that

y = 2x + 2 + Ce^x

is a solution to

\frac{dy}{dx} = y - 2x

for any constant

C

Left side:

\frac{dy}{dx} = 2 + Ce^x

Right side:

y - 2x = (2x + 2 + Ce^x) - 2x = 2 + Ce^x

Both sides equal

2 + Ce^x

✓. This works no matter what

C

is, so this one differential equation has infinitely many solutions.

A function that fails: try

y = 2x + e^x

. The left side is

\frac{dy}{dx} = 2 + e^x

, but the right side is

y - 2x = e^x

. They’re off by

2

, so it’s not a solution.

Example 1: Finding a Particular Solution with a First Derivative

Problem: Find the particular solution to

\frac{dy}{dx} = e^{3x} - 4x^2

that passes through the point

(0, 5)

Step 1: Integrate to find the general solution

We need to find

y

by taking the antiderivative:

y = \int \left(e^{3x} - 4x^2\right)\,dx

For the first term, use substitution with

u = 3x

, so

du = 3\,dx

, meaning

dx = \frac{du}{3}

y = \int e^u \cdot \frac{du}{3} - \int 4x^2\,dx

y = \frac{1}{3}e^{3x} - \frac{4x^3}{3} + C

Step 2: Use the initial condition to find $C$

Substitute the point

(0, 5)

, so

x = 0

and

y = 5

5 = \frac{1}{3}e^{0} - \frac{4(0)^3}{3} + C

5 = \frac{1}{3}(1) - 0 + C

5 = \frac{1}{3} + C

C = 5 - \frac{1}{3} = \frac{15}{3} - \frac{1}{3} = \frac{14}{3}

Step 3: Write the particular solution

y = \frac{1}{3}e^{3x} - \frac{4x^3}{3} + \frac{14}{3}

Verification: to check this, take the derivative:

\frac{dy}{dx} = \frac{1}{3} \cdot 3 \cdot e^{3x} - \frac{4 \cdot 3x^2}{3} + 0 = e^{3x} - 4x^2 \quad \checkmark

That matches the original differential equation, so the solution is correct.

Alternative Problem Wording

Differential equations get stated in a bunch of different ways. These are all equivalent:

Derivative notation: “Find $y$ if $\frac{dy}{dx} = e^{3x} - 4x^2$ and $y(0) = 5$ ”

Slope language: “A curve has slope $e^{3x} - 4x^2$ at each point. Find the equation of the curve if it passes through $(0, 5)$ ”

Rate of change language: “The rate of change of $y$ with respect to $x$ is $e^{3x} - 4x^2$ . What is $y$ if $y = 5$ when $x = 0$ ?”

All three describe the same problem.

Example 2: Second Derivative Problem

Problem: Find the particular solution to

\frac{d^2y}{dx^2} = x^{-2} + (1 - 3x)^2

, given that

\frac{dy}{dx} = 8

when

x = 1

, and

y = 0

when

x = 1

Step 1: Find the first derivative $\frac{dy}{dx}$

Integrate the second derivative:

\frac{dy}{dx} = \int x^{-2}\,dx + \int (1-3x)^2\,dx

For the second integral, use substitution with

u = 1 - 3x

, so

du = -3\,dx

, meaning

dx = -\frac{du}{3}

\frac{dy}{dx} = \frac{x^{-1}}{-1} + \int u^2 \cdot \left(-\frac{du}{3}\right)

\frac{dy}{dx} = -\frac{1}{x} - \frac{1}{3} \cdot \frac{u^3}{3} + C_1

\frac{dy}{dx} = -\frac{1}{x} - \frac{(1-3x)^3}{9} + C_1

Step 2: Use the first initial condition to find $C_1$

When

x = 1

\frac{dy}{dx} = 8

8 = -\frac{1}{1} - \frac{(1-3)^3}{9} + C_1

8 = -1 - \frac{(-2)^3}{9} + C_1

8 = -1 - \frac{-8}{9} + C_1

8 = -1 + \frac{8}{9} + C_1

C_1 = 8 + 1 - \frac{8}{9} = 9 - \frac{8}{9} = \frac{81 - 8}{9} = \frac{73}{9}

Step 3: Find $y$ by integrating again

y = \int \left(-\frac{1}{x} - \frac{(1-3x)^3}{9} + \frac{73}{9}\right)\,dx

y = -\ln|x| - \frac{1}{9} \cdot \frac{(1-3x)^4}{4} \cdot \frac{1}{-3} + \frac{73x}{9} + C_2

y = -\ln|x| + \frac{(1-3x)^4}{108} + \frac{73x}{9} + C_2

Step 4: Use the second initial condition to find $C_2$

When

x = 1

y = 0

0 = -\ln(1) + \frac{(1-3)^4}{108} + \frac{73}{9} + C_2

0 = 0 + \frac{16}{108} + \frac{73}{9} + C_2

0 = \frac{4}{27} + \frac{219}{27} + C_2

C_2 = -\frac{223}{27}

Step 5: Write the particular solution

y = -\ln|x| + \frac{(1-3x)^4}{108} + \frac{73x}{9} - \frac{223}{27}

Important: Trig Derivatives and Antiderivatives

When you’re working with differential equations that involve trig functions, keep these straight.

Derivatives

\frac{d}{dx}[\sin x] = \cos x

\frac{d}{dx}[\cos x] = -\sin x

Memory tip: if it starts with C (cosine), the derivative has a negative sign in front.

Antiderivatives

\int \sin x\,dx = -\cos x + C

\int \cos x\,dx = \sin x + C

Memory tip: if the answer starts with C (cosine), it has a negative sign in front.

Example 3: Finding an Unknown Constant

Problem: Given that

y = ke^{-2x} + 3\sin(4x)

is a solution to the differential equation

\frac{d^2y}{dx^2} + 16y = -68e^{-2x}

, find the value of

k

Step 1: Find the first derivative

\frac{dy}{dx} = k(-2)e^{-2x} + 3(4)\cos(4x)

\frac{dy}{dx} = -2ke^{-2x} + 12\cos(4x)

Step 2: Find the second derivative

\frac{d^2y}{dx^2} = -2k(-2)e^{-2x} + 12(-4)\sin(4x)

\frac{d^2y}{dx^2} = 4ke^{-2x} - 48\sin(4x)

Step 3: Substitute into the differential equation

\frac{d^2y}{dx^2} + 16y = -68e^{-2x}

\left[4ke^{-2x} - 48\sin(4x)\right] + 16\left[ke^{-2x} + 3\sin(4x)\right] = -68e^{-2x}

Step 4: Expand and simplify

4ke^{-2x} - 48\sin(4x) + 16ke^{-2x} + 48\sin(4x) = -68e^{-2x}

Notice the sine terms cancel:

20ke^{-2x} = -68e^{-2x}

Step 5: Solve for $k$

Divide both sides by

e^{-2x}

20k = -68

k = -\frac{68}{20} = -\frac{17}{5}

Key Takeaways

Particular solution: use initial conditions to solve for the constant $C$

Verify solutions by taking the derivatives and checking that you get back the original differential equation

Infinitely many solutions: a differential equation usually has a whole family of solutions, one for each value of $C$ , and an initial condition picks one

Second derivatives: integrate twice, using an initial condition after each integration

Multiple initial conditions: you need one condition for each integration you perform

Trig derivatives: $\frac{d}{dx}[\sin x] = \cos x$ and $\frac{d}{dx}[\cos x] = -\sin x$

Trig antiderivatives: $\int \sin x\,dx = -\cos x + C$ and $\int \cos x\,dx = \sin x + C$

Finding constants in solutions: substitute the given function and its derivatives into the differential equation, then solve algebraically

Back Next