What is the solution for y^{''}-4y^'+4y=e^{2x}sin(x) ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{- 3 t}

\frac{d ^{2} y}{d x ^{2}} - a^{2} y + b = 0

y^{^{''}} - y^{^{'}} = 3 cos (2 x)

y^{^{''}} - 25 y = cos (5 x)

y^{^{''}} - 9 y = (x^{2} - 5) sin (3 x)

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