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

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

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

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

\int 0.06 \cdot e^{\frac{t}{300}} d t

\frac{d}{d x} (ln (\frac{2 x + 1}{x - 1}))

n = 1 \sum \infty \frac{7}{( n + 1 ) !}

\frac{d}{d x} (x^{2} 6 x - 7)

ln (5 x^{2} + 7 y^{2})

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