#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (i) textbook solution.

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

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

Hint: Use $\int e^{x}dx$

Explanation : $\frac{d y}{d x}+3 y=e^{m x}$

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

This is a first order linear differential equation of the form

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

Here $P=3$ and $Q=e^{mx}$

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 differenttial 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: $m+3=0 \; or \; m=-3$

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

\begin{aligned} &=e^{0}=1 \\ &\Rightarrow y e^{3 x}=\int d x+C \\ &\Rightarrow y e^{3 x}=x+C \end{aligned}

\begin{aligned} &\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} \end{aligned}

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

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 \\ &\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} \end{aligned}

\begin{aligned} &\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} \end{aligned}

Thus the solution of the given differential equation is

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