What is the solution for xy^'+5y=(2e^{-2x})/(x^3) ?

The solution for xy^'+5y=(2e^{-2x})/(x^3) is y=-(e^{-2x})/(x^4)-(e^{-2x})/(2x^5)+(c_{1})/(x^5)

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xy^'+5y=(2e^{-2x})/(x^3)

x y^{^{'}} + 5 y = \frac{2 e ^{- 2 x}}{x ^{3}}

y = - \frac{e ^{- 2 x}}{x ^{4}} - \frac{e ^{- 2 x}}{2 x ^{5}} + \frac{c _{1}}{x ^{5}}

Solution steps

x y^{'} (x) + 5 y = \frac{2 e ^{- 2 x}}{x ^{3}}

Solve linear ODE:

y = - \frac{e ^{- 2 x}}{x ^{4}} - \frac{e ^{- 2 x}}{2 x ^{5}} + \frac{c _{1}}{x ^{5}}

y = - \frac{e ^{- 2 x}}{x ^{4}} - \frac{e ^{- 2 x}}{2 x ^{5}} + \frac{c _{1}}{x ^{5}}

4^{x + 3} \times 8^{2 x - 1} = 2

5^{2 x + 1} + 3 \cdot 5^{6 x - 3} = 500

3 e^{3 x + 4} + 8 = e

y, 5^{x + 3} = 2 5^{2 y - 1}

1 = 2^{\frac{1}{x}}

What is the solution for xy^'+5y=(2e^{-2x})/(x^3) ?
The solution for xy^'+5y=(2e^{-2x})/(x^3) is y=-(e^{-2x})/(x^4)-(e^{-2x})/(2x^5)+(c_{1})/(x^5)