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.

$x^{2}y\text{'}\text{'}-7xy\text{'}+15y=0$

Short answer

The value of $m$ is $3$ or $5$.

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$x^{2}y\text{'}\text{'}-7xy\text{'}+15y=0$.

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$y$,$y\text{'}$ androle="math" localid="1664213168578" $y"$into the differential equation.

$$\begin{gathered} x^2\left[m\right(m-1\left)x^{m-2}\right]-7x\left[m{x}^{m-1}\right]+15\left[x^m\right]=0 \\ m(m-1)x^m-7m{x}^m+15x^m=0 \\ \left[m\right(m-1)-7m+15]x^m=0 \\ [m^2-8m+15]x^m=0 \end{gathered}$$

This implies that$m^2-8m+15$.

$$\begin{gathered} m^2-8m+15=0 \\ (m-3)(m-5)=0 \\ m=3,5 \end{gathered}$$

Thus, the value of $m$ is $3$ or $5$.