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  • Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 67 Subquestion (i) textbook solution.

Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 67 Subquestion (i) textbook solution.

Answers (1)

Answer :  y\left(1+x^{2}\right)=\tan ^{-1} x-\frac{\pi}{4}

Hint : Use the formula \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c

Given : \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}} ; y=0 \text { when } x=1

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

Divide by 1+x^{2}

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

This is a linear differential equation of the form

Here, P=\frac{2 x}{1+x^{2}} \quad ; Q=\frac{1}{\left(1+x^{2}\right)^{2}}

The integrating factor of this differential equation is

\text { I.F }=e^{\int P d x}=e^{\int \frac{2 x}{1+x^{2}} d x}

We have,      \begin{aligned} &\int \frac{2 x}{1+x^{2}}=\log \left|1+x^{2}\right|+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d x}{x}=\log |x|+c\right] \\ \end{aligned}

Therefore, \begin{aligned} &\text { I. } F=e^{\log \left|1+x^{2}\right|}=1+x^{2} \\ \end{aligned}

                  \begin{aligned} &\text { I. } F=1+x^{2} \end{aligned}

Hence the solution of the differential equation is

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

We know , \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c

Therefore, y\left(1+x^{2}\right)=\tan ^{-1} x+c                                      ....(ii)

Now y=0 when x=1

Therefore,  

\begin{aligned} 0\left(1+1^{2}\right) &=\tan ^{-1}(1)+c \\ \Rightarrow 0 &=\frac{\pi}{4}+c \Rightarrow c=-\frac{\pi}{4} \end{aligned}

By (ii)

y\left(1+x^{2}\right)=\tan ^{-1} x-\frac{\pi}{4}

 

               

Posted by

infoexpert23

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