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Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (vii) maths textbook solution.

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Answer : y=\frac{1}{2} e^{\sin x}+\frac{C}{e^{\sin x}}

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

Hint : Using \int \cos\; x\; dx

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

This is a first order linear differential equation of the form

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

The integrating factor If  of this differential equation is

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

Hence, the solution of different equation is

     \begin{aligned} &y I f=\int \text { QIfd } d x+C \\ &=y e^{\sin x}=\int e^{\sin x} \cos x e^{\sin x} d x+C \\ &=y e^{\sin x}=\int e^{2 \sin x} \cos x d x+C \end{aligned}

      \begin{aligned} &=y e^{\sin x}=\int e^{2 t} d t+C \quad[\sin x=t, \cos x d x=d t] \\ &=y e^{\sin x}=\left[\frac{e^{2 t}}{2}\right]+C \\ &=y e^{\sin x}=\frac{e^{2 \sin x}}{2}+C \end{aligned}

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

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