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Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 42

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Answer:   \begin{aligned} & x+\cot ^{-1} y=1+C e^{\tan ^{-1} y}\\ & \end{aligned}

Give:  \begin{aligned} & \left(\cot ^{-1} y+x\right) d y=\left(1+y^{2}\right) d x\\ \end{aligned}

Hint: Using integration by parts and  \int \frac{1}{1+x^{2}} d x

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

 \begin{aligned} &=\frac{d x}{d y}=\frac{\cot ^{-1} y+x}{1+y^{2}} \\\\ &=\frac{d x}{d y}=\frac{\cot ^{-1} y}{1+y^{2}}+\frac{x}{1+y^{2}} \end{aligned}

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

This is a linear differential equation of the form

 \begin{aligned} &\frac{d x}{d y}+P x=Q \\ &P=\frac{-1}{1+y^{2}} \text { and } Q=\frac{\cot ^{-1} y}{1+y^{2}} \end{aligned}

The integrating factor  If  of this differential equation is

 \begin{aligned} &I f=e^{\int P d y} \\\\ &=e^{\int \frac{-1}{1+y^{2}} d y} \\\\ &=e^{-\int \frac{1}{1+y^{2}} d y} \end{aligned}

\begin{array}{ll} =e^{-\tan ^{-1} y} &\qquad\qquad {\left[\int \frac{1}{1+\mathrm{y}^{2}} d y=\tan ^{-1} y+C\right]} \\\\ =e^{\cot ^{-1} y} &\qquad {\left[\tan ^{-1} y=\cot ^{-1}\left(\frac{1}{y}\right)\right]} \end{array}

Hence, the solution is

x\: \text{If}=\int Q \: \text{If} \: d y+C

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

Put   \cot ^{-1} y=t \\

\begin{aligned} & &\qquad \begin{aligned} =&-\frac{d y}{1+y^{2}}=d t \\ =& \frac{d y}{1+y^{2}}=-d t \end{aligned} \end{aligned} 



=-\int t e^{t} d t  

Using integration by parts

\begin{aligned} &=-\left[t e^{t}-\int e^{t} d t\right] \\\\ &=-\left[t e^{t}-e^{t}-C\right] \\\\ &=-t e^{t}+e^{t}+C \\ \\&=x e^{\cot ^{-1} y}=-\cot ^{-1} y e^{\cot ^{-1} y}+e^{\cot ^{-1} y}+C \end{aligned}


Divide by  e^{cot^{-1}}y

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

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