What is the solution for y^'+5y=xe^{-5x} ?

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

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y^'+5y=xe^{-5x}

y^{^{'}} + 5 y = x e^{- 5 x}

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

Solution steps

y^{'} (x) + 5 y = x e^{- 5 x}

Solve linear ODE:

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

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

\int sec^{7} (x) tan^{3} (x) d x

x \frac{d y}{d x} = \frac{1}{y ^{2}} + y

- 3 sin^{4} (x)

\frac{\partial}{\partial x} (5 y^{4} cos (4 x))

\int sin^{\frac{1}{2}} (x) cos (x) d x

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