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Name: Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21 point 10 question 36 subquestion (i)
Uploaded: Fri, 2021-08-27 22:21
Description: Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21 point 10 question 36 subquestion (i)

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

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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

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