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  • Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 19 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 19 textbook solution.

Answers (1)

Answer : y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C

Hint : To solve this we convert  \cot x \text { to } \frac{\cos x}{\sin x}  formula.

Give : \frac{d y}{d x}+y \tan x=x^{2} \cos ^{2} x

Solution : \frac{dy}{dx}+Py=Q

           \begin{aligned} &P=\tan x, Q=x^{2} \cos ^{2} x \\ \end{aligned}

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

         \begin{aligned} &=e^{\int \tan x d x} \\ \end{aligned}

         \begin{aligned} &=e^{\log \sec x} \\ \end{aligned}

          \begin{aligned} &=\sec x \end{aligned}

         \begin{aligned} &=y I f=\int Q I f d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(\sec x) d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(1 / \cos x) d x+C \\ &=y \sec x=\int x^{2} \cos x d x+C \end{aligned}

         =y \sec x=x^{2} \int \cos x d x-\int(2 x \cos x d x) d x+C

Using integration by parts

        \begin{aligned} &=y \sec x=x^{2} \sin x-2 \int x^{2} \sin x d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C \end{aligned}

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