What is the solution for y^'-3y=xy^{-4} ?

The solution for y^'-3y=xy^{-4} is y=\sqrt[5]{(-15x-1+45c_{1}e^{15x})/(45)}

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y^'-3y=xy^{-4}

y^{^{'}} - 3 y = x y^{- 4}

y = 5 \frac{- 15 x - 1 + 45 c _{1} e ^{15 x}}{45}

Solution steps

y^{'} (x) - 3 y = x y^{- 4}

Solve Bernoulli ODE:

y = 5 \frac{- 15 x - 1 + 45 c _{1} e ^{15 x}}{45}

y = 5 \frac{- 15 x - 1 + 45 c _{1} e ^{15 x}}{45}

d er i v a t i v e T 4 \frac{7}{9}

s l o p e in t erce pt (3.1) (4.2)

\frac{d}{d x} (\frac{e ^{x}}{6 x ^{2} + 5})

\frac{\partial}{\partial x} (x^{2} + y^{2} + x^{2} y + 9)

d er i v a t i v e f (x) = (x e^{2 x - 4})

What is the solution for y^'-3y=xy^{-4} ?
The solution for y^'-3y=xy^{-4} is y=\sqrt[5]{(-15x-1+45c_{1}e^{15x})/(45)}