The general solution of the differential equa | vedclass

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(B) Given equation: $\sqrt{(1+x^{2})(1+y^{2})} + xy \frac{dy}{dx} = 0$$\Rightarrow \sqrt{1+x^{2}} \sqrt{1+y^{2}} = -xy \frac{dy}{dx}$$\Rightarrow \int

Vedclass · Accepted Answer

$\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C$

Vedclass · Answer

(B) Given equation: $\sqrt{(1+x^{2})(1+y^{2})} + xy \frac{dy}{dx} = 0$$\Rightarrow \sqrt{1+x^{2}} \sqrt{1+y^{2}} = -xy \frac{dy}{dx}$$\Rightarrow \int \frac{y}{\sqrt{1+y^{2}}} dy = -\int \frac{\sqrt{1+x^{2}}}{x} dx$Let $1+y^{2} = v^{2} \Rightarrow y dy = v dv$ and $1+x^{2} = u^{2} \Rightarrow x dx = u du \Rightarrow dx = \frac{u du}{x} = \frac{u du}{\sqrt{u^{2}-1}}$Substituting these into the integral:$\int \frac{v dv}{v} = -\int \frac{u}{\sqrt{u^{2}-1}} \cdot \frac{u du}{\sqrt{u^{2}-1}}$$\Rightarrow \int dv = -\int \frac{u^{2}}{u^{2}-1} du$$\Rightarrow v = -\int \left( 1 + \frac{1}{u^{2}-1} \right) du$$\Rightarrow v = -u - \frac{1}{2} \log_{e} \left| \frac{u-1}{u+1} \right| + C$$\Rightarrow v = -u + \frac{1}{2} \log_{e} \left| \frac{u+1}{u-1} \right| + C$Substituting back $u = \sqrt{1+x^{2}}$ and $v = \sqrt{1+y^{2}}$:$\sqrt{1+y^{2}} = -\sqrt{1+x^{2}} + \frac{1}{2} \log_{e} \left( \frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1} \right) + C$$\Rightarrow \sqrt{1+y^{2}} + \sqrt{1+x^{2}} = \frac{1}{2} \log_{e} \left( \frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1} \right) + C$

The general solution of the differential equation $\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0$ is (where $C$ is a constant of integration)

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