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  • tan-1x + tan-1 y = c is the general solution of the differential equation: dy/dx = 1+ y^2/1+x^2

\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\mathrm{c} is the general solution of the differential equation:
A. \frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}
B. \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1+\mathrm{x}^{2}}{1+\mathrm{y}^{2}}
C. \left(1+x^{2}\right) d y+\left(1+y^{2}\right) d x=0
D. \left(1+x^{2}\right) d x+\left(1+y^{2}\right) d y=0

Answers (1)

If \mathrm{y}=\mathrm{f}(\mathrm{x})$ is a solution of differential equation then differentiating it will give the same differential equation.
To find the differential equation differentiate with respect to x. 

$ $\tan ^{-1} x+\tan ^{-1} y=c$

$$ \\ \Rightarrow \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}\left(\frac{d y}{d x}\right)=0 \\\\ \left(1+y^{2}\right) d x+\left(1+x^{2}\right) d y=0 $$

Option C is correct..

Posted by

infoexpert22

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