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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

Answers (1)

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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