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  • Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 20 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 20 textbook solution.

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

              \begin{aligned} &=\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}} \\ \end{aligned}

            \begin{aligned} &I f=e^{\int P d x} \\ \end{aligned}

            \begin{aligned} &=e^{\int \frac{1}{1+x^{2}} d x} \\ \end{aligned}

            \begin{aligned} &=e^{\tan ^{-1} x} \end{aligned}                              .....(ii)

            \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}} \\ \end{aligned}

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

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

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

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