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If the general solution of $\frac{d y}{d x}=\frac{y^2}{x y-y^2-x^2}$ is $\tan ^{-1}\left(\frac{y}{x}\right)=f(y)+C$, then $f\left(e^3\right)=$
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The correct answer is:
$3$
On putting $y=m x$
$$
\begin{aligned}
& \Rightarrow \quad m+x \cdot \frac{d m}{d x}=\frac{m^2}{m-m^2-1} \\
& \Rightarrow \quad x \frac{d m}{d x}=\frac{m^3+m}{m-m^2-1} \\
& \therefore \quad \int \frac{m-m^2-1}{m^3+m} d m=\int \frac{d x}{x} \\
& \Rightarrow \quad \int\left(\frac{1}{m^2+1}-\frac{1}{m}\right) d m=\ln |x|+C \\
& \Rightarrow \quad \tan ^{-1}\left(\frac{y}{x}\right)-\ln \left|\frac{y}{x}\right|=\ln |x|+C \\
& \Rightarrow \quad \tan ^{-}\left(\frac{y}{x}\right)=\ln |y|+C \\
& \therefore \quad f(y)=\ln |y| \\
& \therefore \quad f\left(e^3\right)=\ln \left|e^3\right|=3
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad m+x \cdot \frac{d m}{d x}=\frac{m^2}{m-m^2-1} \\
& \Rightarrow \quad x \frac{d m}{d x}=\frac{m^3+m}{m-m^2-1} \\
& \therefore \quad \int \frac{m-m^2-1}{m^3+m} d m=\int \frac{d x}{x} \\
& \Rightarrow \quad \int\left(\frac{1}{m^2+1}-\frac{1}{m}\right) d m=\ln |x|+C \\
& \Rightarrow \quad \tan ^{-1}\left(\frac{y}{x}\right)-\ln \left|\frac{y}{x}\right|=\ln |x|+C \\
& \Rightarrow \quad \tan ^{-}\left(\frac{y}{x}\right)=\ln |y|+C \\
& \therefore \quad f(y)=\ln |y| \\
& \therefore \quad f\left(e^3\right)=\ln \left|e^3\right|=3
\end{aligned}
$$
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