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

The solution for y^{''}-2y^'+2y=3e^xsin(x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))-(3e^xxcos(x))/2

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y^{''}-2y^'+2y=3e^xsin(x)

y^{^{''}} - 2 y^{^{'}} + 2 y = 3 e^{x} sin (x)

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{3 e ^{x} x cos ( x )}{2}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 2 y = 3 e^{x} sin (x)

Solve linear ODE:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{3 e ^{x} x cos ( x )}{2}

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{3 e ^{x} x cos ( x )}{2}

What is the solution for y^{''}-2y^'+2y=3e^xsin(x) ?
The solution for y^{''}-2y^'+2y=3e^xsin(x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))-(3e^xxcos(x))/2