What is the solution for xy^2y^'=y^3-5x^3,y(1)=2 ?

The solution for xy^2y^'=y^3-5x^3,y(1)=2 is y=x\sqrt[3]{-15ln(x)+8}

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xy^2y^'=y^3-5x^3,y(1)=2

x y^{2} y^{^{'}} = y^{3} - 5 x^{3}, y (1) = 2

y = x 3 - 15 ln (x) + 8

Solution steps

x y^{2} y^{'} (x) = y^{3} - 5 x^{3}

Solve Bernoulli ODE:

y = x 3 - 15 ln (x) + 8

y = x 3 - 15 ln (x) + 8

t an g e n t f (x) = - x^{2} + 4

d er i v a t i v e sin (x) + \frac{4}{5} cot (x)

t an g e n t f (x) = 3 x^{2} - 4 x + 1, at x = 2

\int (2 u^{2} - 2 u^{- 1}) d u

3 y^{^{''}} + 2 y^{^{'}} + y = 0

What is the solution for xy^2y^'=y^3-5x^3,y(1)=2 ?
The solution for xy^2y^'=y^3-5x^3,y(1)=2 is y=x\sqrt[3]{-15ln(x)+8}