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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 28 maths textbook solution.

Answers (1)

Answer : y e^{\sin x}=(\sin x-1) e^{\sin x}+C

Hint : To solve this equation we use \frac{d y}{d x}+P y=Qwhere P,Q are constants.

Give : \frac{d y}{d x}+y \cos x=\sin x \cos x

Solution :

\begin{aligned} &\frac{d y}{d x}+y \cos x=\sin x \cos x \\ &=\frac{d x}{d y}+P y=Q \\ &P=\cos x \operatorname{and} Q=\sin x \cos x \end{aligned}

I\; f of differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \cos x d x} \\ &=e^{\sin x} \end{aligned}

 

\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &y e^{\sin x}=\sin x \cos x e^{\sin x} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\sin x=t, \cos x d x=d t] \\ &=\int t e^{t} d t \\ &=t e^{t}-\int e^{t}+C \end{aligned}

\begin{aligned} &=t e^{t}-e^{t}+C \\ &=e^{t}(t-1)+C \\ &=y e^{\sin x}=(\sin x-1) e^{\sin x}+C \end{aligned}

 

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