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

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

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

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

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

Solution steps

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

Solve using substitution:

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

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

\int \frac{9 + x}{x ^{2}} d x

f (x) = - 3 - 7 x^{2}, at x = 42

\frac{d}{d x} (e^{- 0.05 x})

f (x) = \frac{1}{x} + 4, at x = - 4

\frac{d y}{d x} + \frac{- x}{x + 1} = 0

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