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

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

Answers (1)

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} 

 

So,

=-\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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