What is the solution for 4xy^'-(1+x)y=2xy^2 ?

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

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4xy^'-(1+x)y=2xy^2

4 x y^{^{'}} - (1 + x) y = 2 x y^{2}

y = \frac{e ^{\frac{l n ( x ) + x}{4}}}{\int - e ^{\frac{1}{4} (l n (x) + x)} d x}

Solution steps

4 x y^{'} (x) - (1 + x) y = 2 x y^{2}

Solve Bernoulli ODE:

y = \frac{e ^{\frac{l n ( x ) + x}{4}}}{\int - e ^{\frac{1}{4} (l n (x) + x)} d x}

y = \frac{e ^{\frac{l n ( x ) + x}{4}}}{\int - e ^{\frac{1}{4} (l n (x) + x)} d x}

\frac{d y}{d x} = \frac{( y ^{3} + x ^{3} )}{x y ^{2}}

f a c t or t^{3} - 6 t^{2} + 9 t

\int_{0}^{\infty} e^{- t} d t

d er i v a t i v e \frac{1}{2 y}

a re a y = \frac{40}{x}, y = - 9 x + 39

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