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  • Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 44 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 44 Maths Textbook Solution.

Answers (1)

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

Hint: you must know the rules of solving differential equation and integrations.

Given:\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0

Solution:\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0

\begin{aligned} &x\left(1+x^{2}\right) d y=-\left[\left(1+y\left(1+x^{2}\right)\right] d x\right. \\ &\frac{\mathrm{dy}}{\mathrm{d} x}=\frac{-1-y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)} \\ &\frac{d y}{d x}=\frac{-1}{x} \cdot y-\frac{1}{x\left(1+x^{2}\right)} \\ &\frac{d y}{d x}+\frac{1}{x} \cdot y=-\frac{1}{x\left(1+x^{2}\right)} \end{aligned}

Comparing with  \frac{dy}{dx}+p\: y=q ,  we get,

\begin{aligned} &P=\frac{1}{x} \text { and } q=-\frac{1}{x\left(1+x^{2}\right)} \\ &\text { I.F }=e^{\int \frac{1}{x} d x} \\ &=e^{\log x}=x \quad\left[\because \mathrm{e}^{\log x}=x\right] \end{aligned}

The solution is,

\begin{aligned} &\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F}=\int-\frac{1}{\mathrm{x}\left(1+\mathrm{x}^{2}\right)} \cdot \mathrm{x} \mathrm{dx} \\ &\mathrm{y} \mathrm{x}=-\int \frac{1}{\left(1+\mathrm{x}^{2}\right)} \mathrm{dx} \\ &\mathrm{yx}=-\tan ^{-1} \mathrm{x}+\mathrm{C} \end{aligned}

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