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  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 Subquestion (xiii) textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 Subquestion (xiii) textbook solution.

Answers (1)

Answer : y=x^{2}-\frac{\pi}{4} \operatorname{cosec} x+C

Give : \tan x \frac{d y}{d x}=2 x \tan x+x^{2}-y, \tan x \neq 0, y=0 \text { when } x=\frac{\pi}{2}

Hint : Using integration by parts and \int \cot x d x

Explanation : \tan x \frac{d y}{d x}=2 x \tan x+x^{2}-y

Divide by \tan\; x

               \begin{aligned} &=\frac{d y}{d x}=2 x+\frac{x^{2}}{\tan x}-\frac{y}{\tan x} \\ &=\frac{d y}{d x}=2 x+\cot x x^{2}-y \cot x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{1}{\tan x}=\cot x\right] \\ &=\frac{d y}{d x}+y \cot x=2 x+\cot x x^{2} \\ &=\frac{d y}{d x}+(\cot x) y=2 x+\cot x x^{2} \end{aligned}

This is a linear differential equation of the form

                  \begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=\cot x \text { and } Q=2 x+\cot x x^{2} \end{aligned}

The integrating factor If  of this differential equation is

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

Hence, the solution is

          \begin{aligned} &y I f=\int Q I f d x+C \\ &=y \sin x=\int\left(2 x+x^{2} \cot x\right) \sin x d x+C \\ &=y \sin x=\int 2 x \sin x+x^{2} \cot x \sin x d x+C \\ &=y \sin x=2 \int x \sin x d x+\int x^{2} \cot x \sin x d x+C \end{aligned}

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

Divide by \sin \; x

         \begin{aligned} &=y=x^{2}+\frac{c}{\sin x} \\ &=y=x^{2}+C \operatorname{cosec} x \ldots(i) \end{aligned}

Now y=0 \text { when } x=\frac{\pi}{2}

          \begin{aligned} &=0=\left(\frac{\pi}{2}\right)^{2}+C \operatorname{cosec} \frac{\pi}{2} \\ &=0=\frac{\pi^{2}}{4}+C(1) \quad\left[\operatorname{cosec} \frac{\pi}{2}=1\right] \\ &=C=-\frac{\pi^{2}}{4} \end{aligned}

Substituting in (i)

          =y=x^{2}-\frac{\pi^{2}}{4} \operatorname{cosec} x

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