Question

Let $y=(A+Bx)e^{3x}$ is a solution of the differential equation $\frac{d^{2}y}{d{x}^2}+m\frac{dy}{dx}+ny=0;m,n\in \text{I}$, then find $m$ and $n$.

Answer 1

Step-$1:$ Obtain a differential equation for the given function:

A function $y=(A+Bx)e^{3x}$ is given.

Differentiate both sides with respect to $x$.

$$\begin{gathered} \frac{dy}{dx}=3(A+Bx)e^{3x}+B{e}^{3x} \\ \Rightarrow \frac{dy}{dx}=3y+B{e}^{3x}...\left(i\right)\left[\because y=(A+Bx)e^{3x}\right] \end{gathered}$$

Again differentiate both sides with respect to $x$.

$\frac{d^{2}y}{d{x}^2}=3\frac{dy}{dx}+3B{e}^{3x}...\left(ii\right)$

From the equation $\left(i\right)$ we get,

$B{e}^{3x}=\frac{dy}{dx}-3y$

Substitute $B{e}^{3x}=\frac{dy}{dx}-3y$ in equation $\left(ii\right)$.

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=3\frac{dy}{dx}+3\left(\frac{dy}{dx}-3y\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}-6\frac{dy}{dx}+9y=0...\left(iii\right) \end{gathered}$$

Step- $2:$ Find the value of $m$ and $n$:

Since $y=(A+Bx)e^{3x}$ is the solution of the differential equation $\frac{d^{2}y}{d{x}^2}+m\frac{dy}{dx}+ny=0;m,n\in \text{I}$.

$\frac{d^{2}y}{d{x}^2}+m\frac{dy}{dx}+ny=0...\left(iv\right)$

On comparing the equation $\left(iii\right)$ and equation $\left(iv\right)$. we get,

$m=-6,n=9$

Therefore, the value of $m$ and $n$ are $-6$ and $9$ respectively.