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Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (vi) textbook solution.

Answers (1)

Answer         :  y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+c

Hint               : You must know the rules of solving differential equation and integration

Given            :  \frac{d y}{d x}+2 y=\sin x

Solution         :   put in form \frac{d y}{d x}+P y=Q

                         \frac{d y}{d x}+2 y=\sin x                         ...(i)

Step : find P and Q

Compare (i) with \frac{d y}{d x}+P y=Q

P=2 \; \; \; \; \; \; \; \; \; \; \; Q=\sin x

Find integration factor I.F

\begin{aligned} &\text { I.F }=e^{\int P d x} \\ &\qquad \text { I.F }=e^{\int 2 d x}=e^{2 x} \\ &\text { So } I . F=e^{2 x} \end{aligned}

Step : 4

y \times I . F=\int Q \times I . F d x+C

Putting value

\begin{aligned} &y e^{2 x}=\int \sin x e^{2 x} d x+c\\ &\text { Let } I=\int \sin x e^{2 x} d x \end{aligned}

Solving I

\begin{aligned} &I=\int \sin x e^{2 x} d x \\ &=\sin x \int e^{2 x} d x-\int\left[\frac{d}{d x} \sin x \int e^{2 x} d x\right] d x \\ &=\sin x \frac{e^{2 x}}{2}-\int \cos x \frac{e^{2 x}}{2} d x \end{aligned}

\begin{aligned} &=\frac{1}{2} \sin x e^{2 x}-\frac{1}{2}\left[\cos x \int e^{2 x} d x-\int\left\{-\sin x \int \frac{e^{2 x}}{2} d x\right\} d x\right] \\ &=\frac{1}{2} \sin x e^{2 x}-\frac{1}{2}\left(\cos x \frac{e^{2 x}}{2}\right)-\frac{1}{4} \int \sin x e^{2 x} d x \\ &I=\frac{1}{2} \sin x e^{2 x}-\frac{1}{4} \cos x e^{2 x}-\frac{1}{4} I \end{aligned}

\begin{aligned} &I+\frac{1}{4} I=\frac{1}{4}\left(2 \sin x e^{2 x}-\cos x e^{2 x}\right) \\ &\frac{5 I}{4}=\frac{e^{2 x}}{4}(2 \sin x-\cos x) \\ &I=\frac{e^{2 x}}{5}(2 \sin x-\cos x) \end{aligned}

Substituting I in eq (ii)

y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+c

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