Section 3.2 Verifying Solutions

Verifying that a given function is a solution to a differential equation amounts to showing that it satisfies the equation. This typically involves taking any necessary derivatives of the function, substituting them into the equation, and simplifying both sides to determine whether they match. Here are some examples to illustrate this process.

🌌 Example 24.

Verify that \(y = 2x^2\ \) is a solution to \(\ \ xy' - 2x^2 = y\text{.}\)

Solution.

First, compute \(y' = 4x\) and plug both \(y = 2x^2\) and \(y' = 4x\) into the equation:

\begin{align*} x\left( 4x \right) - 2x^2 \amp = \left( 2x^2 \right) \\ 4x^2 - 2x^2 \amp = 2x^2 \\ 2x^2 \amp = 2x^2 \quad ✅ \end{align*}

This shows that \(y = 2x^2\) satisfies the equation and proves it is a solution.

Checkpoint 25. 📖❓ What Verifying Means.

To verify that a given function is a solution of a differential equation, what do you do?

Substitute the function (and any required derivatives) into the equation, then check that both sides simplify to the same thing.
Correct — this is exactly what it means to satisfy the equation.
Solve the differential equation from scratch and check whether you get the same function back.
Verifying does not require solving; you only substitute and compare.
Plug in a single value of the independent variable and confirm the two sides agree there.
Agreement at one point is not enough; the two sides must match as functions.
Confirm the function passes through the origin.
Passing through the origin is unrelated to satisfying the equation.

🌌 Example 26.

Verify that \(P = \sin t\ \) is a solution to \(\ \ 2P'' + P = \sin t\text{.}\)

Solution.

Since \(P''\) appears, compute the derivatives first:

\begin{equation*} P = \sin t, \quad P' = \cos t, \quad P'' = -\sin t \end{equation*}

Now substitute in the dependent variable along with its derivatives:

\begin{align*} 2\left( -\sin t \right) + \left( \sin t \right) \amp = \sin t\\ -2\sin t + \sin t \amp = \sin t \\ -\sin t \amp \ne \sin t \quad ❌ \end{align*}

Since the function on the left is different from the function on the right, \(P = \sin t\) does not satisfy the equation and is not a solution.

Checkpoint 27. 📖❓ Does It Satisfy the Equation?

Is \(y = e^{-3x}\) a solution of \(y' + 3y = 0\text{?}\)

Yes — since \(y' = -3e^{-3x}\text{,}\) we get \(y' + 3y = -3e^{-3x} + 3e^{-3x} = 0\text{,}\) so both sides match.
The derivative and substitution both check out, giving \(0 = 0\text{.}\)
No — substituting gives \(6e^{-3x}\) on the left, not \(0\text{.}\)
That uses the wrong derivative \(y' = +3e^{-3x}\text{;}\) the correct derivative is \(-3e^{-3x}\text{.}\)
No — a solution must contain an arbitrary constant, and this one does not.
A specific function with no constant can still satisfy the equation, as the earlier examples show.
Yes, but only at \(x = 0\text{.}\)
The two sides must agree for all \(x\text{,}\) and here they do.

As you will soon learn, differential equations have many solutions and all of them share a common form, as the next example shows.

🌌 Example 28.

Show that the functions

\begin{equation*} y = e^{2x},\quad y = -5e^{2x},\quad y = \pi e^{2x},\quad y = 0 \end{equation*}

are all solutions to the differential equation

\begin{equation*} y' - 2y = 0\text{.} \end{equation*}

Solution.

Rather than verify each function separately, notice that they are all of the form \(y = c e^{2x}\text{,}\) where \(c\) is a constant. Let’s verify the general case.

Substitute \(y = c e^{2x}\) into the equation:

\begin{equation*} y' - 2y = (c e^{2x})' - 2(c e^{2x}) = 2c e^{2x} - 2c e^{2x} = 0 \quad ✅ \end{equation*}

So \(y = c e^{2x}\) is a solution for any constant \(c\text{.}\) In particular, this includes the functions

\begin{equation*} y = e^{2x},\quad y = -5e^{2x},\quad y = \pi e^{2x},\quad y = 0 \end{equation*}

for the constants \(c = 1, -5, \pi, 0\) respectively.

Checkpoint 29. 📖❓ Why the General Form Works.

In the example verifying that \(y = ce^{2x}\) solves \(y' - 2y = 0\text{,}\) why does checking the general form show that \(y = e^{2x}\text{,}\) \(y = -5e^{2x}\text{,}\) \(y = \pi e^{2x}\text{,}\) and \(y = 0\) are all solutions?

\(c\) is a placeholder for any constant, so verifying the general form verifies all of them at once.
One verification of the general form covers every choice of the constant.
Because every solution of any differential equation is an exponential function.
Not true in general; other examples in this section have polynomial and logarithmic solutions.
Because a differential equation can have at most four solutions.
Differential equations typically have infinitely many solutions.
Because it contains a constant \(c\text{.}\)
A constant alone is not enough, it must also match the functional part.

As you saw in the previous example, solutions can differ by a constant. Some solutions even involve multiple constants, as the next example shows.

🌌 Example 30.

Verify that \(\ y = c_1\ x^2 + c_2 - \ln x\ \) is a solution to

\begin{equation*} x^2y'' - xy' = 2\text{.} \end{equation*}

Solution.

Compute the needed derivatives:

\begin{equation*} y = c_1\ x^2 + c_2 - \ln x \ \ \Rightarrow \ \ y' = 2 c_1\ x - \frac{1}{x} \ \ \Rightarrow \ \ y'' = 2 c_1 + \frac{1}{x^2} \end{equation*}

and substitute into the equation:

\begin{align*} x^2\left( 2 c_1 + \frac{1}{x^2} \right) - x\left( 2 c_1\ x - \frac{1}{x} \right) \amp = 2\\ 2 c_1\ x^2 + 1 - 2 c_1\ x^2 + 1 \amp = 2\\ 2 \amp = 2 \quad ✅ \end{align*}

Thus, \(\ y = c_1\ x^2 + c_2 - \ln x\ \) is a solution to \(x^2y'' - xy' = 2\text{.}\)

You have attempted of activities on this page.

Prev Top Next