What is the solution for y^{''}-4y^'+5y=xe^{2x} ?

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

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y^{''}-4y^'+5y=xe^{2x}

y^{^{''}} - 4 y^{^{'}} + 5 y = x e^{2 x}

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 5 y = x e^{2 x}

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} + 80 y = - 2 sin (3 t)

0.25 y^{^{''}} + 0.25 y^{^{'}} + 16 y = 3 cos (2 t)

y^{^{''}} - 4 y^{^{'}} + 40 y = t e^{2 t} sin (6 t)

y^{^{''}} + 3 y^{^{'}} - 7 y = t^{4} e^{t}

y^{^{''}} + 25 y = 5 (csc (5 t))^{2}

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