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Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 8 textbook solution.

Answers (1)

Answer : y\left(x^{2}+1\right)^{2}=-x+C

Hint: To solve this equation we will use differentiate both terms.

Give: \frac{d y}{d x}+\frac{4 x}{x^{2}+1} y+\frac{1}{\left(x^{2}+1\right)^{2}}=0

Solution : \frac{d y}{d x}+\frac{4 x}{x^{2}+1} y=\frac{-1}{\left(x^{2}+1\right)^{2}}

                 \begin{aligned} &\frac{d y}{d x}+P y=Q \\ &I f=e^{\int P d x} \\ &=e^{\int \frac{4 x}{x^{2}+1} d x}\; \; \; \; \; \; \; \; \quad \quad\left[\text { Let } x^{2}+1=u, 2 x d x=d u\right. \end{aligned}

                 \begin{aligned} &=2 e^{\int \frac{d u}{u}} \\ &=e^{2} \ln \left|x^{2}+1\right| \\ &=\left(x^{2}+1\right)^{2} \end{aligned}

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

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