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Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 61 textbook solution.

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Answer : y \sec x=\frac{x^{n+1}}{n+1}+c

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

 Given             :   \frac{d y}{d x}+y \tan x=x^{n} \cos x

Solution          : \frac{d y}{d x}+y \tan x=x^{n} \cos x

Comparing with, \frac{d y}{d x}+Py=Q

P=\tan x \quad, Q=x^{n} \cos x

Now,

       \begin{aligned} \text { I.F } &=e^{\int \tan x d x} \\ &=e^{\log (\sec x)} \\ &=\sec x \end{aligned}

So the solution is ,

       \begin{aligned} &y \times \text { I.F }=\int(Q \times I . F) d x+c \\ &y \sec x=\int x^{n} \cos x \sec x d x+c \\ &y \sec x=\int x^{n} d x+c \\ &y \sec x=\frac{x^{n+1}}{n+1}+c \end{aligned}

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