What is the solution for y^'+5x^4y=x^4,y(0)=y^0 ?

The solution for y^'+5x^4y=x^4,y(0)=y^0 is y=(e^{-x^5+c_{1}})/5+1/5

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y^'+5x^4y=x^4,y(0)=y^0

y^{^{'}} + 5 x^{4} y = x^{4}, y (0) = y^{0}

y = \frac{e ^{- x^{5} + c_{1}}}{5} + \frac{1}{5}

Solution steps

y^{'} (x) + 5 x^{4} y = x^{4}

Solve separable ODE:

y = \frac{e ^{- x^{5} + c_{1}}}{5} + \frac{1}{5}

y = \frac{e ^{- x^{5} + c_{1}}}{5} + \frac{1}{5}

d x = x^{5} y d y

y^{^{'}} = 1 + x + y^{2} + y^{2} x

\frac{d x}{d t} = 5000 + 0.04 x, x = 18000, t = 0

y + x \frac{d y}{d x} = 2 x^{2} y \frac{d y}{d x} + 2 x y^{2}

What is the solution for y^'+5x^4y=x^4,y(0)=y^0 ?
The solution for y^'+5x^4y=x^4,y(0)=y^0 is y=(e^{-x^5+c_{1}})/5+1/5