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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 20 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise  Revision Exercise Question 20 Maths Textbook Solution.

Answers (1)

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

Hint: Separate y & x and than integrate both sides

Given:\frac{dy}{dx}=y^{2}+2y+2

Solution:\frac{dy}{dx}=y^{2}+2y+2

\begin{aligned} &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}^{2}+2 \mathrm{y}+1+1 \\ &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{y}+1)^{2}+(1)^{2} \\ &\Rightarrow \frac{1}{(\mathrm{y}+1)^{2}+(1)^{2}} \mathrm{dy}=\mathrm{dx} \end{aligned}

Integrating both sides, we get

\begin{aligned} &\Rightarrow \int \frac{1}{(\mathrm{y}+1)^{2}+(1)^{2}} \mathrm{dy}=\int \mathrm{dx} \ldots \int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\ &\Rightarrow \tan ^{-1}\left(\frac{y+1}{1}\right)+\mathrm{c}=\mathrm{x} \\ &\Rightarrow \mathrm{x}=\tan ^{-1}(y+1)+\mathrm{c} \end{aligned}

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