DE-BOOK Verifying Solutions

Section 2.2 Verifying Solutions

In the previous section, we discussed what it means for a function to be a solution to a differential equation. Now, we turn our attention to how we can verify that a proposed function is indeed a solution.

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The process of verifying a solution is straightforward: you substitute the proposed function into the differential equation and check whether the equation holds true. This simple test helps ensure that the function satisfies the relationship defined by the differential equation.

Note 17. Tip: Before verifying, move all terms to one side.

y^{'} = 3 y and y^{'} - 3 y = 0 .

0 = 0 .

Let’s look at a few examples.

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Example 18.

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Verify the given solutions for each differential equation. A. Verify that

y = 2 x^{2}

is a solution to

x y^{'} - 2 x^{2} = y .

Solution 1. Solution

Let’s begin by moving all terms to the left-side of the equation.

x y^{'} - 2 x^{2} - y = 0

To reduce errors and break the problem into more manageable steps, we will compute and simplify the derivatives appearing in the equation.

y = 2 x^{2}, y^{'} = 4 x

Finally, we substitute in

y

and

y^{'}

and simplify.

\begin{aligned} x y^{'} - 2 x^{2} - y = & 0 \\ x (4 x) - 2 x^{2} - (2 x^{2}) = & 0 \\ 4 x^{2} - 2 x^{2} - 2 x^{2} = & 0 \\ 0 = & 0 \leftarrow true \end{aligned}

Therefore,

y = 2 x^{2}

is a solution to

x y^{'} - 2 x^{2} = y .

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B. Verify that

P = \sin t

is a solution to

2 P^{″} + P = \sin t .

Solution 2. Solution

As before, we compute

P^{″}

since it appears in the equation and move all terms to the left-side.

P = \sin t, P^{'} = \cos t, P^{″} = - \sin t

Plug

P

and

P^{″}

in and simplify.

\begin{aligned} 2 P^{″} + P - \sin t = & 0 \\ 2 (- \sin t) + \sin t - \sin t = & 0 \\ - 2 \sin t + \sin t - \sin t = & 0 \\ - 2 \sin t = & 0 \leftarrow false \end{aligned}

Therefore,

P = \sin t

is not a solution to

2 P^{″} + P = \sin t .

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Example 19.

y = 2 e^{- 3 t}

y = e^{4 t}

y^{″} - y^{'} - 12 y = 0 .

Solution 1.

y = 2 e^{- 3 t}

Find

y^{'}

and

y^{″}

since they appear in the equation.

\begin{aligned} y = & 2 e^{- 3 t} \\ y^{'} = & 2 e^{- 3 t} (- 3) = - 6 e^{- 3 t} \\ y^{″} = & - 6 e^{- 3 t} (- 3) = 18 e^{- 3 t} \end{aligned}

Plug

y, y^{'},

and

y^{″}

into the DE and simplify.

\begin{aligned} y^{″} - y^{'} - 12 y = & 0 \\ 18 e^{- 3 t} - (- 6 e^{- 3 t}) - 12 (2 e^{- 3 t}) = & 0 \\ 18 e^{- 3 t} + 6 e^{- 3 t} - 24 e^{- 3 t} = & 0 \\ 0 = & 0 \leftarrow true \end{aligned}

Therefore,

y = 2 e^{- 3 t}

is a solution to

y^{″} - y^{'} - 12 y = 0.

Solution 2.

y = e^{4 t}

Find

y^{'}

and

y^{″}

since they appear in the DE.

\begin{aligned} y = & e^{4 t} \\ y^{'} = & e^{4 t} (4) = 4 e^{4 t} \\ y^{″} = & 4 e^{4 t} (4) = 16 e^{4 t} \end{aligned}

Plug

y, y^{'},

and

y^{″}

into the DE and simplify.

\begin{aligned} y^{″} - y^{'} - 12 y = & 0 \\ 16 e^{4 t} - 4 e^{4 t} - 12 (e^{4 t}) = & 0 \\ 16 e^{4 t} - 4 e^{4 t} - 12 e^{4 t} = & 0 \\ 0 = & 0 \leftarrow true \end{aligned}

Therefore,

y = e^{4 t}

is a solution to

y^{″} - y^{'} - 12 y = 0.

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Example 20.

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Verify that

y = 3 \sin (x^{2})

is a solution to

y^{'} - x y^{″} = 12 x^{2} \sin (x^{2}) .

Solution. Solution

Find

y^{'}

and

y^{″}

since they appear in the equation and move all terms to the left-side.

\begin{aligned} y = & 3 \sin (x^{2}) \\ y^{'} = & 3 \cos (x^{2}) \cdot 2 x = 6 x \cos (x^{2}) \\ y^{″} = & 6 \cos (x^{2}) - 12 x \sin (x^{2}) = 6 \cos (x^{2}) - 12 x \sin (x^{2}) \end{aligned}

Plug in

y, y^{'},

and

y^{″}

and simplify.

\begin{aligned} y^{'} - x y^{″} - 12 x^{2} \sin (x^{2}) = & 0 \\ 6 x \cos (x^{2}) - x (6 \cos (x^{2}) - 12 x \sin (x^{2})) - 12 x^{2} \sin (x^{2}) = & 0 \\ 6 x \cos (x^{2}) - 6 x \cos (x^{2}) - 12 x^{2} \sin (x^{2}) - 12 x^{2} \sin (x^{2}) = & 0 \\ 0 = & 0 \end{aligned}

Therefore,

y = 3 \sin (x^{2})

is a solution to

y^{'} - 12 x^{2} \sin (x^{2}) = x y^{″} .

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You can even verify that a function is a solution to a differential equation when the function contains constants as the following example shows.

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Example 21.

y = \frac{1}{2} x^{2} + c_{1} x + c_{2}

y^{″} = 1 .

Solution. Solution

Move terms to left:

y^{″} - 1 = 0

Find

y^{″}

since it appears in the DE.

\begin{aligned} y = & \frac{1}{2} x^{2} + c_{1} x + c_{2} \\ y^{'} = & x + c_{1} \\ y^{″} = & 1 \end{aligned}

Plug

y^{″}

into the DE and simplify.

\begin{aligned} y^{″} - 1 = & 0 \\ 1 - 1 = & 0 \\ 0 = & 0 \end{aligned}

Therefore,

y = \frac{1}{2} x^{2} + c_{1} x + c_{2}

is a solution to

y^{″} = 1 .

In fact, it is the general solution, but we don’t show that here.

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As you work through more complex differential equations, this verification process becomes a valuable tool. In the next section, we’ll discuss the different types of solutions you will encounter and how you can visualize them.

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Reading Questions Check-Point Questions

1. Moving all terms of a differential equation to one side of the equation is a required step for verifying the solution to a differential equation..

Moving all terms of a differential equation to one side of the equation is a required step for verifying the solution to a differential equation.

True
Incorrect.
Read the note provided in this section.
False
Correct! This is not a required step, but it can sometimes simplify the process.

2. $y = x^{2} + 3$ is a solution to $\frac{d y}{d x} - 3 = 2 x$ .

y = x^{2} + 3

\frac{d y}{d x} - 3 = 2 x .

True
Incorrect.

$y = x^{2} + 3$ is not a solution since

$\begin{aligned} \frac{d y}{d x} - 3 = & 2 x \\ \frac{d}{d x} [x^{2} + 3] - 3 = & 2 x \\ 2 x - 3 = & 2 x \leftarrow false \end{aligned}$
False
Correct!

$y = x^{2} + 3$ is not a solution since

$\begin{aligned} \frac{d y}{d x} - 3 = & 2 x \\ \frac{d}{d x} [x^{2} + 3] - 3 = & 2 x \\ 2 x - 3 = & 2 x \leftarrow false \end{aligned}$

3. List the steps needed to verify $s = \cos (3 x)$ is a solution to $s^{″} - 6 s^{'} + 9 s = 0$ .

Drag the step on the left to the appropriate step order on the right.

Review the examples above to see the steps in verifying a solution.

Compute $s^{'}$
Step 1
Compute $s^{″}$
Step 2
Plug in $s, s^{'}, s^{″}$
Step 3
Simplify both sides
Step 4
Check for $0 = 0$
Step 5

4. A differential equation has one solution.

True
Incorrect.
Example 19 shows a differential equation with two solutions.
False
Correct!

5. Consider the differential equation with missing right-hand side:
$y^{″} - \frac{4}{x} y^{'} = .$
Assuming $y = 2 x^{3}$ is a solution to this equation, which of the following is a possible right-hand side?

$2 x^{3} + 4 x^{2}$
Incorrect. Plug $y^{'}$ and $y^{″}$ into the left-side and simplify.
$0$
Incorrect. Plug $y^{'}$ and $y^{″}$ into the left-side and simplify.
$- 12 x$
Correct! Plugging $y^{'} = 6 x^{2}$ and $y^{″} = 12 x$ into the left-side gives

$\begin{aligned} (12 x) - \frac{4}{x} (6 x^{2}) = \\ 12 x - 24 x = \\ - 12 x = & \leftarrow true \end{aligned}$

and to get a true statement, the right-hand side must be $- 12 x .$

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Section 2.2 Verifying Solutions

Note 17. Tip: Before verifying, move all terms to one side.

Example 18.

Example 19.

Example 20.

Example 21.

Reading Questions Check-Point Questions

1. Moving all terms of a differential equation to one side of the equation is a required step for verifying the solution to a differential equation..

2. y=x2+3 is a solution to dydx−3=2x.

3. List the steps needed to verify s=cos⁡(3x) is a solution to s″−6s′+9s=0.

4. A differential equation has one solution.

5. Consider the differential equation with missing right-hand side: .y″−4xy′=XXXXX. Assuming y=2x3 is a solution to this equation, which of the following is a possible right-hand side?

2. $y = x^{2} + 3$ is a solution to $\frac{d y}{d x} - 3 = 2 x$ .

3. List the steps needed to verify $s = \cos (3 x)$ is a solution to $s^{″} - 6 s^{'} + 9 s = 0$ .

5. Consider the differential equation with missing right-hand side:
$y^{″} - \frac{4}{x} y^{'} = .$
Assuming $y = 2 x^{3}$ is a solution to this equation, which of the following is a possible right-hand side?