What is the solution for y^{''}-2y^'+2y=1,y^'0=0 ?

The solution for y^{''}-2y^'+2y=1,y^'0=0 is y=e^t(c_{1}cos(t)+c_{2}sin(t))+1/2

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y^{''}-2y^'+2y=1,y^'0=0

y^{^{''}} - 2 y^{^{'}} + 2 y = 1, y^{^{'}} 0 = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 2 y = 1

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{7 e ^{t}}{1 + t ^{2}}

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

y^{^{''}} + y^{^{'}} - 12 y = t, y (0) = 1, y^{^{'}} (0) = 1

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

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

What is the solution for y^{''}-2y^'+2y=1,y^'0=0 ?
The solution for y^{''}-2y^'+2y=1,y^'0=0 is y=e^t(c_{1}cos(t)+c_{2}sin(t))+1/2