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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 19 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 19 Maths Textbook Solution.

Answers (1)

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

Hint: Use the formula of \int \frac{1}{x^{2}+1}dx

Given:\frac{dy}{dx}=\frac{1}{x^{2}+4x+5}

Solution:\frac{dy}{dx}=\frac{1}{x^{2}+4x+5}

\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{1}{x^{2}+4 x+4+1} \\ &\Rightarrow \frac{d y}{d x}=\frac{1}{(x+2)^{2}+(1)^{2}} \\ &\Rightarrow d y=\frac{1}{(x+2)^{2}+(1)^{2}} d x \end{aligned}

Integrating both sides, we get

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

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