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  • Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 20

Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 20

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Answer:  2 y e^{\tan ^{-1} x}=e^{2 \tan ^{-1} y}+C

Hint: To solve this equation we use  I\int f\left ( x \right )dx  formula.

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

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

\begin{aligned} & \\ &\quad=\frac{d y}{d x}+\frac{y}{\left(1+x^{2}\right)}=\frac{e^{\tan ^{-1} x}}{1+x^{2}} \ldots(i) \\\\ &\quad=\frac{d y}{d x}+P(x) y=Q x \end{aligned}

\begin{aligned} &P=\frac{1}{1+x^{2}}, Q=\frac{e^{\tan ^{-1} x}}{1+x^{2}} \\\\ &I f=e^{\int P d x} \\\\ &=e^{\int \frac{1}{1+x^{2}} d x} \\\\ &=e^{\tan ^{-1} x} \ldots(i i) \end{aligned}

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

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

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

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