Can someone check my work here? Both ##f=f(x)## and ##y=y(x)##.(adsbygoogle = window.adsbygoogle || []).push({});

$$f'y'+\frac{fy''}{1+y'^2}=0\implies\\

\frac{y''}{y'(1+y'^2)}=-\frac{f'}{f}\\

\frac{y''}{y'(1+y'^2)}=-\ln(f)$$

Now let ##v=y'##, which implies

$$

\int\frac{1}{v(1+v^2)}\,dv=-\int\ln(f)\,dx\implies\\

\ln(v) - \frac{1}{2}\ln(1 + v^2)+C=-\int\ln(f)\,dx\implies\\

\frac{y'}{\sqrt{1 + y'^2}}=k\exp\left[-\int\ln(f)\,dx\right]$$

Am I correct to this point? Also, how would you proceed?

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# I 2nd order non-linear ODE

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