#### Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 Subquestion (i) maths textbook solution.

Answer : $y=\frac{e^{x}}{2}$

Give : $y'+y=e^{x}$

Hint : Using integration

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

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=1 \text { and } Q=e^{x} \end{aligned}

The integrating factor $If$  of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int 1 d x} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{x} d x=\log |x|+C\right] \\ &=e^{\int d x} \\ &=e^{x} \end{aligned}

Hence, the solution of different equation is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y e^{x}=\int e^{x} e^{x} d x+C \\ &=y e^{x}=\int e^{2 x} d x+C \\ &=y e^{x}=\frac{e^{2 x}}{2}+C \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int e^{2 x}=\frac{e^{2 x}}{2}+C\right] \end{aligned}

Divide by $e^{x}$

\begin{aligned} &=y=\frac{1}{e^{x}} \frac{e^{2 x}}{2}+\frac{1}{e^{x}} C \\ &=y=\frac{e^{x}}{2}+\frac{C}{e^{x}} \end{aligned}

Given when $y(0)=\frac{1}{2}, \text { when } x=0, y=\frac{1}{2}$

\begin{aligned} &=\frac{1}{2}=\frac{e^{0}}{2}+\frac{C}{e^{0}} \\ &=\frac{1}{2}=\frac{1}{2}+C \\ &=C=0 \end{aligned}

By $y=\frac{e^{x}}{2}+\frac{0}{e^{x}}$

$=y=\frac{e^{x}}{2}$

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