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  • Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise RE Question 66 Subquestion xii textbook solution

Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (xii) textbook solution.

Answers (1)

Answer :  y=\left(1+x^{2}\right)^{-1} \log |\sin x|+c\left(1+x^{2}\right)^{-1}

Hint :   integrate by applying integration of x^{n}

Given : \left(1+x^{2}\right) d y+2 x y d x=\cot x d x

Solution : given equation

\left(1+x^{2}\right) d y+2 x y d x=\cot x d x

Divide both sides by dx

\begin{gathered} \left(1+x^{2}\right) \frac{d y}{d x}+2 x y \frac{d x}{d x}=\cot x \frac{d x}{d x} \\ \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\cot x \end{gathered}

Divide by (1+x^{2})

\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}

Comparing above equation with \frac{d y}{d x}+P y=Q

    \begin{aligned} &\quad P=\frac{2 x}{1+x^{2}} \quad, Q=\frac{\cot x}{1+x^{2}} \\ &\text { I. } F=e^{\int P d x} \\ &\text { I. } F=e^{\int \frac{2 x}{1+x^{2}} d x} \\ &\text { Let } t=1+x^{2} \end{aligned}

\begin{gathered} d t=2 x d x \\ I . F=e^{\int \frac{d t}{t}}=e^{\log t}=t=\left(1+x^{2}\right) \end{gathered}

\begin{aligned} y \times I . F &=\int Q \times I . F d x+c \\ y\left(1+x^{2}\right) &=\int \frac{\cot x}{1+x^{2}} \times\left(1+x^{2}\right) d x+c \\ y\left(1+x^{2}\right) &=\int \cot x d x+c \end{aligned}

\begin{aligned} &y\left(1+x^{2}\right)=\log |\sin x|+c \\ &\text { Divide } 1+x^{2} \\ &y=\left(1+x^{2}\right)^{-1} \log |\sin x|+c\left(1+x^{2}\right)^{-1} \end{aligned}

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