Question

In Problems $31-34$find values of m so that the function$y=m{e}^{mx}$ is a solution of the given differential equation.

$5y\text{'}=2y$

Short answer

The value of $m$ is $\frac{2}{5}$.

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.

Let the given function be$5y\text{'}=2y$. Then, the first derivative of the function is$y\text{'}=m{e}^{mx}$.

Determine the value.

Substituteandinto the differential equation.

$$\begin{gathered} 5m{e}^{mx}=2e^{mx} \\ {e}^{mx}(5m-2)=0 \\ 5m-2=0 \\ m=\frac{2}{5} \end{gathered}$$

As$e^{mx}$will never equal to$0$, that implies that$5m-2=0$.

Thus, the value of $m$ is $\frac{2}{5}$.