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  • Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation Excercise 21.10 Question 37 subquestion (viii)

Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation Excercise 21.10 Question 37 subquestion (viii)

Answers (1)

Answer:  \begin{aligned} & y \sin x=2 x^{2}-\frac{\pi^{2}}{2}\\ & \end{aligned}

Give:  \frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x, y\left(\frac{\pi}{2}\right)=0\\

Hint: Using   \int \cot x d x \text { and } \int x d x\\

Explanation:  \frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x

 \frac{d y}{d x}+(\cot x) y=4 x \operatorname{cosec} x

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=4 x \operatorname{cosec} x \end{aligned}

The integrating factor If  of this differential equation is

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

=e^{\int \cot x d x} \\

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]

Hence, the solution is 

\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y(\sin x)=\int(4 x \operatorname{cosec} x) \sin x d x+C \\ &=y \sin x=4 \int x \frac{1}{\sin x} \sin x d x+C \quad\left[\operatorname{cosec} x=\frac{1}{\sin x}\right] \end{aligned}
\begin{aligned} &=y \sin x=4 \int x d x+C \\ &=y \sin x=4\left(\frac{x^{2}}{2}\right)+C \\ &=y \sin x=2 x^{2}+C \ldots(i) \end{aligned}

Now    \begin{aligned} &y\left(\frac{\pi}{2}\right)=0 \text { when } x=\frac{\pi}{2}, y=0 \\ & \end{aligned}

\quad=0 \sin x=2\left(\frac{\pi}{2}\right)^{2}+C \\

=0=2 \frac{\pi^{2}}{4}+C \\

=C=-\frac{\pi^{2}}{2}

Substituting in (i)

=y \sin x=2 x^{2}-\frac{\pi^{2}}{2} 

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