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

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

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y^{''}-2y^'+2y=2x

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} + y = 9 e^{2 t}

y^{^{''}} + 6 y = cos (7 t)

(d^{2} + 144) (d^{2} - 144) 2 (d + 12) 4 x = 144

\frac{d ^{2} x}{d t ^{2}} + 36 x = 5 sin (4 t), x (0) = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = x^{2}

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