What is the solution for y^{''}-4y^'+5y=xe^x ?

The solution for y^{''}-4y^'+5y=xe^x is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+(e^xx)/2+1/2 e^x

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y^{''}-4y^'+5y=xe^x

y^{^{''}} - 4 y^{^{'}} + 5 y = x e^{x}

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{e ^{x} x}{2} + \frac{1}{2} e^{x}

Solution steps

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

Solve linear ODE:

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{e ^{x} x}{2} + \frac{1}{2} e^{x}

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{e ^{x} x}{2} + \frac{1}{2} e^{x}

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 e^{- 2 t}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x}, y^{^{'}} (0) = - 1, y (0) = 1

y^{^{''}} + 4 y = 3 sin (t)

y^{^{''}} + 100 y^{^{'}} + 2500 y = 30

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

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