#### Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21 point 10 question 36 subquestion (i)

Answer:  $y=\left\{\begin{array}{l} (x+C) c^{-3 x}, \quad m=-3 \\ \frac{e^{m x}}{m+3}+C e^{-3 x}, \text { otherwise } \end{array}\right.$ ,otherwise

Give:  $\frac{d y}{d x}+3 y=e^{m x}, m$  is a given real number.

Hint: Use  $\int e^{x} d x$

Explanation:  \begin{aligned} & \frac{d y}{d x}+3 y=e^{m x} \\ & \end{aligned}

$=\frac{d y}{d x}+(3) y=e^{m x}$

This is a first order linear differential equation of the form

$=\frac{d y}{d x}+P y=Q$

Here  $P=3 \text { and } Q=e^{m x}$

The integrating factor  $If$  of the differential equation is

\begin{aligned} &I f=e^{\int P d x}\\ &=e^{\int 3 d x}\\ &=e^{3 \int d x}\\ &=e^{3 x} \quad\left[\int d c=x+C\right] \end{aligned}

Hence, the solution of differential equation is

\begin{aligned} &y(I f)=\int Q I f d x+C \\ &=y\left(e^{3 x}\right)=\int e^{m x} e^{3 x} d x+C \\ &=y\left(e^{3 x}\right)=\int e^{m x+3 x} d x+C \\ &=y\left(e^{3 x}\right)=\int e^{x(m+3)} d x+C \end{aligned}

Case 1: $m+3=0 \text { or } m=-3$

When $m+3=0$ , we have  $e^{x\left ( m+3 \right )}$

\begin{aligned} &=e^{0}=1 \\ & \end{aligned}

$\Rightarrow y e^{3 x}=\int d x+C \\$

$\Rightarrow y e^{3 x}=x+C \\$

$\Rightarrow y e^{3 x} e^{-3 x}=(x+C) e^{-3 x} \\$

$\Rightarrow y e^{3 x-3 x}=(x+C) e^{-3 x}$

$\Rightarrow y=(x+C) e^{-3 x} \quad\left[e^{3 x-3 x}=e^{0}=1\right]$

Case 2:  $m+3 \neq 0 \text { or } m \neq-3$

When  $m+3 \neq 0$  we have

\begin{aligned} &y e^{3 x}=\int e^{x(m+3)} d x+C \\ & \end{aligned}

$\Rightarrow y e^{3 x}=\frac{e^{(m+3) x}}{m+3}+C \\$

$\Rightarrow y e^{3 x} e^{-3 x}=\left(\frac{e^{(m+3) x}}{m+3}+C\right) e^{-3 x} \\$

$\Rightarrow y e^{3 x} e^{-3 x}=\frac{\left(e^{m x} e^{3 x}\right) e^{-3 x}}{m+3}+C e^{-3 x} \\$

$\Rightarrow y=\frac{e^{m x}}{m+3}+C e^{-3 x}$

Thus the solution of the given differential equation is

$y=\left\{\begin{array}{l} (x+C) c^{-3 x}, \quad m=-3 \\ \frac{e^{m x}}{m+3}+C e^{-3 x}, \text { otherwise } \end{array}\right.$