What is the solution for y^{''}-4y^'+4y=4t^2 ?

The solution for y^{''}-4y^'+4y=4t^2 is y=c_{1}e^{2t}+c_{2}te^{2t}+t^2+2t+3/2

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y^{''}-4y^'+4y=4t^2

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 t^{2}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + t^{2} + 2 t + \frac{3}{2}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 4 t^{2}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + t^{2} + 2 t + \frac{3}{2}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + t^{2} + 2 t + \frac{3}{2}

y^{^{''}} + y^{^{'}} - 2 y = 4 x^{2}

y^{^{''}} - 5 y^{^{'}} + 6 y = 4 x e^{2 x}

y^{^{''}} - y^{^{'}} - 6 y = - e^{x} + 12 x

y^{^{''}} - 9 y = sinh (3 x)

y^{^{''''}} - 2 y^{^{'''}} + y^{^{''}} = e^{x} + 1

What is the solution for y^{''}-4y^'+4y=4t^2 ?
The solution for y^{''}-4y^'+4y=4t^2 is y=c_{1}e^{2t}+c_{2}te^{2t}+t^2+2t+3/2