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

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

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

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

y = - ln (- x^{2} + c_{1}) + c_{2}

Solution steps

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

Solve using substitution:

y = - ln (- x^{2} + c_{1}) + c_{2}

y = - ln (- x^{2} + c_{1}) + c_{2}

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

x \to + 0 + lim (x^{x^{8}})

5 x y^{^{'}} - 2 y = - \frac{x ^{2}}{y ^{4}}

\int \frac{7}{4 - x ^{2}} d x

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