What is the solution for y^{''}-2y^'+2y=2x-2,y(0)=0 ?

The solution for y^{''}-2y^'+2y=2x-2,y(0)=0 is y=c_{2}e^xsin(x)+x

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y^{''}-2y^'+2y=2x-2,y(0)=0

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x - 2, y (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 25 y = x^{2} e^{x} - x e^{5 x}

y^{^{''}} + 2 y^{^{'}} - 3 y = 8 e^{x}

y^{^{''}} + 16 y = 4 cos (4 x) + 6 sin (4 x)

y^{^{''}} + 3 y^{^{'}} + 2 y = e - t, y (4) = 1, y^{^{'}} (4) = 3

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 e^{2 x}

What is the solution for y^{''}-2y^'+2y=2x-2,y(0)=0 ?
The solution for y^{''}-2y^'+2y=2x-2,y(0)=0 is y=c_{2}e^xsin(x)+x