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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 5 x \cdot sin (x)

3 x^{^{''}} + 48 x = 60 e^{- 4 t} cos (4 t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 2 e^{x}

- y^{^{''}} + y = e^{a \cdot x}, y (0) = 0, y (1) = 0

y^{^{''}} - 9 y = 9 e^{3} x - 6 x^{4}

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