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

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

Answers (1)

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

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 Q I f 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] \end{aligned}

        =y e^{2 x}=-\cos x+C \ldots(i)

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

        \begin{aligned} &=0 e^{2(0)}=-\cos 0+C \\ &=0=-1+C \\ &=C=1 \end{aligned}

By (i)

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

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