If the general solution of $\frac{d y}{d x}=\frac{y^2}{x y-y^2-x^2}$ is $ an ^{-1}\left(\frac{y}{x} ight)=f(y)+C$, then $f\left(e^3 ight)=$

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If the general solution of $\frac{d y}{d x}=\frac{y^2}{x y-y^2-x^2}$ is $	an ^{-1}\left(\frac{y}{x}ight)=f(y)+C$, then $f\left(e^3ight)=$

Aicharya · Accepted Answer

On putting $y=m x$ $ \begin{aligned} & \Rightarrow \quad m+x \cdot \frac{d m}{d x}=\frac{m^2}{m-m^2-1} \ & \Rightarrow \quad x \frac{d m}{d x}=\frac{m^3+m}{m-m^2-1} \ & 	herefore \quad \int \frac{m-m^2-1}{m^3+m} d m=\int \frac{d x}{x} \ & \Rightarrow \quad \int\left(\frac{1}{m^2+1}-\frac{1}{m}ight) d m=\ln |x|+C \ & \Rightarrow \quad 	an ^{-1}\left(\frac{y}{x}ight)-\ln \left|\frac{y}{x}ight|=\ln |x|+C \ & \Rightarrow \quad 	an ^{-}\left(\frac{y}{x}ight)=\ln |y|+C \ & 	herefore \quad f(y)=\ln |y| \ & 	herefore \quad f\left(e^3ight)=\ln \left|e^3ight|=3 \end{aligned} $

If the general solution of $\frac{d y}{d x}=\frac{y^2}{x y-y^2-x^2}$ is $\tan…

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