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

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

Answers (1)

Answer:  \begin{aligned} & \cos x+y e^{2 x}=1 \\ & \end{aligned}

Give:  \frac{d y}{d x}+2 y=e^{-2 x} \sin x, y(0)=0

Hint: Using integrating factor

Explanation: \frac{d y}{d x}+2 y=e^{-2 x} \sin x

This is a linear differential equation of the form

 \begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=2 \text { and } Q=e^{-2 x} \sin x \end{aligned}

The integrating factor  If  of this differential equation is

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

 

Hence, the solution is

 \begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y\left(e^{2 x}\right)=\int e^{-2 x} \sin x e^{2 x} d x+C \\ &=y e^{2 x}=\int \sin x d x+C \quad\left[e^{-2 x} e^{2 x}=e^{-2 x+2 x}=e^{0}=1\right] \\ &=y e^{2 x}=-\cos x+C \ldots(i) \end{aligned}

Now  \begin{gathered} y(0)=0, \text { When } x=0, y=0 \\ \end{gathered}

 \quad=0 e^{2(0)}=-\cos 0+C

 \begin{aligned} &=0=-1+C \\ &=C=1 \\ & \end{aligned}          

By i

=y e^{2 x}=-\cos x+1 \\

=\cos x+y e^{2 x}=1 

Posted by

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