Question

In Problems 21–24 verify that the indicated family of functions is a solution of the given differential equation. Assume an appropriate interval I of definition for each solution.

$\frac{d^{2}y}{d{x}^2}-4\frac{dy}{dx}+4y=0;y=c_1{e}^{2x}+c_{2}x{e}^{2x}$

Short answer

The indicated function is a solution of the given differential equation for all real values of $x$.

Step-by-step solution

Determine the derivatives of the function.

Let the given function be $y=c_1{e}^{2x}+c_{2}x{e}^{2x}$.

Then, the first derivative of the function is,

$\frac{dy}{dx}=2c_1{e}^{2x}+c_2{e}^{2x}+2c_{2}x{e}^{2x}$

The second derivative of the function is,

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=4c_1{e}^{2x}+2c_2{e}^{2x}+2c_2{e}^{2x}+4c_{2}x{e}^{2x} \\ \frac{d^{2}y}{d{x}^2}=4c_1{e}^{2x}+4c_2{e}^{2x}+4c_{2}x{e}^{2x} \end{gathered}$$

Determine the interval of the solution.

Substitute andinto the left-hand side of the differential equation.

$$\begin{gathered} \left(4c_1{e}^{2x}+4c_2{e}^{2x}+4c_{2}x{e}^{2x}\right)-4\left(2c_1{e}^{2x}+c_2{e}^{2x}+2c_{2}x{e}^{2x}\right)+4\left(c_1{e}^{2x}+c_{2}x{e}^{2x}\right)=0 \\ {e}^{2x}\left(4c_1+4c_2+4c_{2}x-8c_1-4c_2-8c_{2}x+4c_1+4c_{2}x\right)=0 \\ 0=0 \end{gathered}$$

That is same as the right-hand side of the differential equation for any real values of $x$. Thus, the indicated function is a solution of the given differential equation.