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  • Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 41 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 41 textbook solution.

Answers (1)

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

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

Hint : Using integration by parts and \int x^{n} d x

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

                    =\frac{d x}{d y}+ (\cot y)\; x=2 y+y^{2} \cot y

This is a linear differential equation of the form

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

The integrating factor If  of this differential equation is

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

Hence, the solution is

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

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

Using integration by parts

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

Substituting on (i)

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

Divide by \sin y

             \begin{aligned} &=x=y^{2}+\frac{c}{\sin y} \\ &=x=y^{2}+C \operatorname{cosec} y \ldots(i i) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{1}{\sin y}=\operatorname{cosec} y\right] \end{aligned}

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

              =0=\left(\frac{\pi}{2}\right)^{2}+C \operatorname{cosec} \frac{\pi}{2}

              \begin{aligned} &=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 (ii)

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

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