What is the solution for y^{''}-4y^'+4y=t ?

The solution for y^{''}-4y^'+4y=t is y=c_{1}e^{2t}+c_{2}te^{2t}+t/4+1/4

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y^{''}-4y^'+4y=t

y^{^{''}} - 4 y^{^{'}} + 4 y = t

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t}{4} + \frac{1}{4}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t}{4} + \frac{1}{4}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t}{4} + \frac{1}{4}

y^{^{''''}} + 4 y^{^{''}} = sin (2 t) + t e^{3 t} + 4

y^{^{''}} - y^{^{'}} - 12 = 0

y^{^{'''}} + y = sin (x)

y^{^{''}} + 4 y = g (t), y (0) = 4, y^{^{'}} (0) = 0

y^{^{''}} - 3 + y^{^{'}} + 2 y = 0

What is the solution for y^{''}-4y^'+4y=t ?
The solution for y^{''}-4y^'+4y=t is y=c_{1}e^{2t}+c_{2}te^{2t}+t/4+1/4