Verifying Solutions to Differential Equations

A differential equation is an equation that links a function with its derivatives, like y′ = 2y. To verify a solution of a differential equation, find the derivatives the equation uses, substitute them and the function into both sides, and simplify. If the two sides are equal for every x, the function is a solution.

Before this lesson: Higher-Order Derivatives, Chain Rule and Derivatives of Exponential Functions

What a differential equation says

An ordinary equation like x² = 9 has numbers as solutions. A differential equation has functions as solutions. The equation y′ = 2y says: find a function whose slope is always twice its height.

Differential equations describe change: a population that grows in proportion to its size, a hot drink cooling toward room temperature, a spring bouncing. Solving one can be hard, but checking a proposed solution only takes derivatives and algebra.

Several solution curves of y′ + y = x, of the form y = Ce⁻ˣ + x − 1, for C = −2, 0, 2 and 4. As x grows, all of them approach the line y = x − 1. The curve with C = 4 passes through the point (0, 3) and is drawn thicker. — Each value of C gives a different solution; the initial condition y(0) = 3 picks out C = 4.

How to verify a solution

Find each derivative that appears in the equation (y′, y″, …).
Substitute y and its derivatives into the left side and the right side.
Simplify both sides. If they match for every x in the domain, y is a solution. If not, it isn’t.
For an initial value problem, also check the given value, such as y(0) = 3.

Term	Meaning
Order	The highest derivative in the equation (y′ is first order, y″ is second order)
General solution	A family of solutions with constants, like y = Ce²ˣ
Particular solution	One member of the family, with the constants chosen
Initial condition	A given value, like y(0) = 3, that picks the particular solution

Verifying solutions examples

Example 1: A growth equation

Show that $y = 3e^{2x}$ is a solution of y′ = 2y.

y' = 3e^{2x}\cdot 2 = 6e^{2x}

Differentiate with the chain rule.

\text{Right side: } 2y = 2\cdot 3e^{2x} = 6e^{2x}

Substitute y into the right side.

6e^{2x} = 6e^{2x}

Compare the two sides.

Answer: the sides match for every x, so $y = 3e^{2x}$ is a solution.

Example 2: A family of solutions and an initial condition

Show that $y = Ce^{-x} + x - 1$ solves y′ + y = x for any constant C. Then find the solution with y(0) = 3.

y' = -Ce^{-x} + 1

Differentiate; the derivative of x − 1 is 1.

y' + y = (-Ce^{-x} + 1) + (Ce^{-x} + x - 1)

Substitute into the left side.

= x

The Ce⁻ˣ terms cancel, and so do 1 and −1.

y(0) = Ce^{0} + 0 - 1 = C - 1

Use the initial condition; e⁰ = 1.

C - 1 = 3 \Rightarrow C = 4

Set it equal to 3 and add 1.

Answer: every function in the family is a solution, and the particular solution is $y = 4e^{-x} + x - 1$ .

Example 3: A second-order equation

Show that $y = 2\sin 3x - \cos 3x$ is a solution of y″ + 9y = 0.

y' = 6\cos 3x + 3\sin 3x

Chain rule on each term (see derivatives of trig functions).

y'' = -18\sin 3x + 9\cos 3x

Differentiate again.

9y = 18\sin 3x - 9\cos 3x

Multiply y by 9.

y'' + 9y = (-18 + 18)\sin 3x + (9 - 9)\cos 3x = 0

Add and combine like terms.

Answer: y″ + 9y = 0 for every x, so the function is a solution.

Example 4: A function that fails

Which of y = x² and y = x³ is a solution of xy′ = 2y?

y = x^2: \; xy' = x(2x) = 2x^2, \; 2y = 2x^2

Test the first function.

2x^2 = 2x^2

The sides match.

y = x^3: \; xy' = x(3x^2) = 3x^3, \; 2y = 2x^3

Test the second function.

3x^3 \ne 2x^3

These differ whenever x ≠ 0.

Answer: y = x² is a solution, but y = x³ is not. One point where the sides agree (x = 0) is not enough.

Example 5: A cooling cup of coffee

A cup of coffee cools in a 20°C room. Its temperature after t minutes is modeled by $T(t) = 20 + 70e^{-0.1t}$ . Show that T solves dT/dt = −0.1(T − 20) with T(0) = 90.

\dfrac{dT}{dt} = 70e^{-0.1t}\cdot(-0.1) = -7e^{-0.1t}

Differentiate; the derivative of 20 is 0.

-0.1(T - 20) = -0.1(70e^{-0.1t}) = -7e^{-0.1t}

Substitute T into the right side.

-7e^{-0.1t} = -7e^{-0.1t}

The sides match.

T(0) = 20 + 70e^{0} = 90

Check the initial condition.

Answer: T(t) solves the equation and starts at 90°C. The equation says the coffee cools faster when it is much hotter than the room.

Common mistakes

Mistake	Example of the error	Fix
Checking only one point	y = x³ works in xy′ = 2y at x = 0, so it’s a solution	The sides must match for every x, not just one.
Missing the chain rule	(3e²ˣ)′ = 3e²ˣ	(3e²ˣ)′ = 6e²ˣ.
Substituting into one side only	Finding y′ and stopping	Substitute into both sides and compare.
Forgetting the initial condition	Showing y = Ce⁻ˣ + x − 1 works, then stopping	For an initial value problem, also solve for C.
Confusing order and degree	Calling y″ + (y′)³ = 0 third order	Order is the highest derivative: this is second order.

Related methods

Picturing solutions

Even without a formula, a differential equation tells you the slope at every point. Drawing those slopes gives a slope field, the next lesson.

Finding solutions

Later in this unit you’ll solve equations like dy/dx = ky directly by separating the variables. Verifying is the way to check those answers.

Guess-and-check with a parameter

To find which exponentials solve an equation like y″ + y′ − 6y = 0, try y = e^(kx) and solve for k. Practice problem 9 does this.

Practice problems

Decide whether each function is a solution of the differential equation. Show your work.

$y = 5e^{-3x}$ ; y′ = −3y
y = x³ + C; y′ = 3x²
$y = e^{x} + e^{-x}$ ; y″ = y
y = cos 2x; y″ + 4y = 0
y = sin x; y″ + 4y = 0
y = Cx²; xy′ = 2y. Then find the solution with y(1) = 5.
$y = \dfrac{1}{x + C}$ ; y′ = −y². Then find the solution with y(0) = 1/2.
Which of e²ˣ, e³ˣ and eˣ solve y″ − 5y′ + 6y = 0?
Find every k so that $y = e^{kx}$ solves y″ + y′ − 6y = 0.
A town’s population is $P(t) = 500e^{0.04t}$ . Show it solves dP/dt = 0.04P with P(0) = 500, and find P(10).

Show answers

y' = 5e^{-3x}\cdot(-3) = -15e^{-3x}

Chain rule.

-3y = -15e^{-3x}

Right side.

Answer: yes, the sides match

y' = 3x^2

The constant’s derivative is 0.

Answer: yes, for every C

y' = e^{x} - e^{-x}, \; y'' = e^{x} + e^{-x}

Differentiate twice.

y'' = e^{x} + e^{-x} = y

Compare.

Answer: yes

y' = -2\sin 2x, \; y'' = -4\cos 2x

Differentiate twice.

y'' + 4y = -4\cos 2x + 4\cos 2x = 0

Substitute and add.

Answer: yes

y'' = -\sin x

Differentiate twice.

y'' + 4y = -\sin x + 4\sin x = 3\sin x

Substitute and add.

3\sin x \ne 0

It is not 0 for most x.

Answer: no

xy' = x(2Cx) = 2Cx^2 = 2y

Substitute y′ = 2Cx.

C(1)^2 = 5 \Rightarrow C = 5

Use y(1) = 5.

Answer: yes; the particular solution is y = 5x²

y' = -\dfrac{1}{(x + C)^2} = -y^2

Write y = (x + C)⁻¹ and use the power and chain rules.

\dfrac{1}{0 + C} = \dfrac{1}{2} \Rightarrow C = 2

Use y(0) = 1/2.

Answer: yes; the particular solution is y = 1/(x + 2)

y = e^{rx}: \; y'' - 5y' + 6y = (r^2 - 5r + 6)e^{rx}

Each derivative brings down a factor of r.

r = 2: \; 4 - 10 + 6 = 0; \; r = 3: \; 9 - 15 + 6 = 0

Test 2 and 3.

r = 1: \; 1 - 5 + 6 = 2 \ne 0

Test 1.

Answer: e²ˣ and e³ˣ are solutions; eˣ is not

(k^2 + k - 6)e^{kx} = 0

Substitute y′ = ke^(kx) and y″ = k²e^(kx).

k^2 + k - 6 = 0

e^(kx) is never 0, so divide it out.

(k + 3)(k - 2) = 0

Factor the trinomial (see factoring trinomials).

k = -3 \text{ or } k = 2

Set each factor equal to 0.

Answer: k = 2 or k = −3

10.

\dfrac{dP}{dt} = 500e^{0.04t}(0.04) = 0.04P

Chain rule.

P(0) = 500e^{0} = 500

Check the initial value.

P(10) = 500e^{0.4} \approx 745.9

Substitute t = 10.

Answer: yes; about 746 people after 10 years

Verifying solutions FAQ

What is a solution to a differential equation?

It is a function that makes the equation true for every x in its domain when you substitute it and its derivatives.

How do you check if a function is a solution to a differential equation?

Find the derivatives the equation needs, substitute everything into both sides, and simplify. Equal sides for every x mean it is a solution.

What is the difference between a general and a particular solution?

A general solution has arbitrary constants and describes a whole family of curves. A particular solution has the constants fixed, usually by an initial condition.

What is the order of a differential equation?

The order is the highest derivative that appears. For example, y″ + 9y = 0 is second order.

Summary

A differential equation links a function with its derivatives; its solutions are functions.
To verify: differentiate, substitute into both sides and simplify.
The sides must match for every x, not just one point.
A general solution has constants; an initial condition picks a particular solution.
The order is the highest derivative in the equation.

Previous lesson: Center of Mass

Next lesson: Slope Fields

Back to the Calculus 2 Index