Question

In Problems $35$ and $36$find values of m so that the function$y=m^x$ is a solution of the given differential equation.

$xy\text{'}\text{'}+2y\text{'}=0$

Short answer

The value of $m$ is $0$ or $-1$.

Step-by-step solution

Define a derivative of the function.

The derivatives of a function $y=f\left(x\right)$ of a quantity x is a measurement of the rate at which the function's value y changes when the variable x varies. The derivatives of

f with respect to x is what it's called.

Determine the derivative of the function.

Let the given function be$xy\text{'}\text{'}+2y\text{'}=0$.

Then, the first derivative of the function is$y\text{'}=m{x}^{m-1}$.

The second derivative of the function is$y\text{'}\text{'}=m(m-1)x^{m-2}$.

Determine the value.

Substitute$y$,$y\text{'}$ and$y"$into the differential equation.

$$\begin{gathered} x\left[m\right(m-1\left)x^{m-2}\right]+2\left[m{x}^{m-1}\right]=0 \\ m(m-1)x^{m-1}+2m{x}^{m-1}=0 \\ \left[m\right(m-1)+2m]x^{m-1}=0 \\ [m^2+m]x^{m-1}=0 \end{gathered}$$

This implies that$m^2+m=0$.

$$\begin{gathered} m^2+m=0 \\ m(m+1)=0 \\ m=0,-1 \end{gathered}$$

Thus, the value of $m$ is $0$ or $-1$.