What is the solution for y^'-2y=xy^3,y(0)=2sqrt(2) ?

The solution for y^'-2y=xy^3,y(0)=2sqrt(2) is y=sqrt(-8/(4x-1))

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y^'-2y=xy^3,y(0)=2sqrt(2)

y^{^{'}} - 2 y = x y^{3}, y (0) = 22

y = - \frac{8}{4 x - 1}

Solution steps

y^{'} (x) - 2 y = x y^{3}

Solve Bernoulli ODE:

y = - \frac{8}{4 x - 1}

y = - \frac{8}{4 x - 1}

(x \cdot y) d x + x d y = 0

x \to 0 lim (\frac{sin ( \frac{1}{x} )}{x})

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

y = x^{2} + 3, (1, 4)

\frac{d}{d x} (\frac{x ^{p (x)}}{x ^{m} - a ^{m}})

What is the solution for y^'-2y=xy^3,y(0)=2sqrt(2) ?
The solution for y^'-2y=xy^3,y(0)=2sqrt(2) is y=sqrt(-8/(4x-1))