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  • Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 37 subquestion (v)

Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 37 subquestion (v)

Answers (1)

Answer:  x e^{\tan ^{-1} y}=\tan ^{-1} y

Give:  \begin{aligned} &\left(1+y^{2}\right) d x+\left(x-e^{\tan ^{-1} y}\right) d y=0 \\ & \end{aligned}

Hint: Using  \int \frac{1}{1+x^{2}} d x

Explanation:  \begin{aligned} &\left(1+y^{2}\right) d x+\left(x-e^{\tan ^{-1} y}\right) d y=0 \\ & \end{aligned}

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

 

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

=\frac{x-e^{\tan ^{-1} y}}{1+y^{2}}=-\frac{d x}{d y}

\begin{aligned} &=\frac{x}{1+y^{2}}-\frac{e^{\tan ^{-1} y}}{1+y^{2}}+\frac{d x}{d y}=0 \\ &\end{aligned}

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

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{e^{\tan ^{1} y}}{1+y^{2}} \end{aligned}

The integrating factor If  of this differential equation is 

\begin{aligned} &\text { If }=e^{\int \operatorname{Pdx}} \\ &=e^{\int \frac{1}{1+y^{2}} d x} \\ &=e^{\tan ^{-1} y} \quad\left[\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+C\right] \end{aligned}

Hence, the solution is

\begin{aligned} &x I f=\int Q I f d y+C \\ &=x\left(e^{\tan ^{-1} y}\right)=\int \frac{e^{-\tan ^{-1} y}}{1+y^{2}} e^{\tan ^{-1} y} d y+C \\ &=x e^{\tan ^{-1} y}=\int \frac{1}{1+y^{2}} d y+C \end{aligned}          \quad\left[e^{-\tan ^{-1} y+\tan ^{-1} y}=e^{0}=1\right]

\\ =x e^{\tan ^{-1} y}=\tan ^{-1} y+C \ldots \text { (i) }

\quad\left[\int \frac{1}{1+y^{2}} d y=\tan ^{-1} y\right]

Now  \begin{aligned} &y(0)=0 \\ & \end{aligned}

\quad=0 e^{\tan ^{-1}(0)}=\tan ^{-1}(0)+C \\

\quad=C=0

Substituting in (i)

 \begin{aligned} &=x e^{\tan ^{-1} y}=\tan ^{-1} y+0 \\ &=x e^{\tan ^{-1} y}=\tan ^{-1} y \end{aligned}

Posted by

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