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  • Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 subquestion (i)

Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 subquestion (i)

Answers (1)

Answer:  y=\frac{e^{x}}{2} \\

Give:  \begin{aligned} & & y^{\prime}+y=e^{x} \end{aligned}

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 \text { QIfd } 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 

\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  \begin{aligned} y=& \frac{e^{x}}{2}+\frac{0}{e^{x}} \\ \end{aligned}

=y=\frac{e^{x}}{2}  

 

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