What is the solution for x^2y^{''}=(y^')^2 ?

The solution for x^2y^{''}=(y^')^2 is y=-(-1+c_{1}x+ln(-1+c_{1}x))/(c_{1)^2}+c_{2}

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x^2y^{''}=(y^')^2

x^{2} y^{^{''}} = (y^{^{'}})^{2}

y = - \frac{- 1 + c _{1} x + ln ( - 1 + c _{1} x )}{c _{1}^{2}} + c_{2}

Solution steps

x^{2} y^{''} (x) = (y^{'} (x))^{2}

Solve using substitution:

y = - \frac{- 1 + c _{1} x + ln ( - 1 + c _{1} x )}{c _{1}^{2}} + c_{2}

y = - \frac{- 1 + c _{1} x + ln ( - 1 + c _{1} x )}{c _{1}^{2}} + c_{2}

\frac{\partial}{\partial x} (3 y^{3} - 4 y x^{2})

\int_{0}^{7} f (x) d x

θ \to 0 lim (\frac{sin ( 4 θ )}{θ})

f (x) = 8 x^{2} + 4 e^{- 0.5 x}

3 x

What is the solution for x^2y^{''}=(y^')^2 ?
The solution for x^2y^{''}=(y^')^2 is y=-(-1+c_{1}x+ln(-1+c_{1}x))/(c_{1)^2}+c_{2}