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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 4 x

2 y^{^{''}} + 6 y^{^{'}} + 1 = 95

x^{^{''}} + 2 x^{^{'}} + 41 x = 11 cos (6 t)

y^{^{''}} + 3 y = 9

y^{^{''}} + y^{^{'}} = sin (x) + cos (x)

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