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  • Solve the differential equation (1+ y^{2}) tan-1x dx + 2y (1 + x^{2}) dy = 0

    Solve the differential equation (1+ y2) tan-1x dx + 2y (1 + x2) dy = 0

Answers (1)

Given:

(1+y^{2})\tan^{-1}x dx +2y(1+x^{2})dy=0

To find: Solution for differential equation that s given

Rewriting the given equation as.

(1+y^{2})\tan^{-1}x =-2y(1+x^{2})\frac{dy}{dx}\\ \frac{\tan^{-1}x}{(1+x^{2})}dx=-\frac{2y}{(1+y^{2})}dy

Integrate on the both sides

\int \frac{\tan^{-1}x}{(1+x^{2})}dx=-\int \frac{2y}{(1+y^{2})}dy

For LHS

Assume tan-1 =t

\frac{1}{1+x^{2}}dx=dt\\ Formula: \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^{2}}\\ \frac{d}{dx}(x^{n})=nx^{n-1}

For RHS

Assume 1+y2=z

2ydy=dz

Substituting and integrating on both the sides

\int t\; dt=-\int \frac{dz}{z}\\ \frac{t^{2}}{2}=-\ln z+c\\ Formula: \int \frac{dx}{x}=\ln x \\\int x^{n}dx=\frac{x^{n+1}}{n+1}

Substitute for t and z

Solution for the differential equation is

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

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infoexpert24

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