What is the solution for y^'-4y=2xy^2 ?

The solution for y^'-4y=2xy^2 is y=(8e^{4x})/(-4e^{4x)x+e^{4x}+c_{1}}

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y^'-4y=2xy^2

y^{^{'}} - 4 y = 2 x y^{2}

y = \frac{8 e ^{4 x}}{- 4 e ^{4 x} x + e ^{4 x} + c _{1}}

Solution steps

y^{'} (x) - 4 y = 2 x y^{2}

Solve Bernoulli ODE:

y = \frac{8 e ^{4 x}}{- 4 e ^{4 x} x + e ^{4 x} + c _{1}}

y = \frac{8 e ^{4 x}}{- 4 e ^{4 x} x + e ^{4 x} + c _{1}}

\frac{d}{d x} (\frac{( x + 3 )}{x - 3})

\frac{x}{x + 25}

y^{^{''}} + 2 y^{^{'}} + 5 y = x

\int \frac{5}{( x ^{2} - 1 ) ^{2}} d x

\frac{6}{x ^{4}}

What is the solution for y^'-4y=2xy^2 ?
The solution for y^'-4y=2xy^2 is y=(8e^{4x})/(-4e^{4x)x+e^{4x}+c_{1}}