What is the solution for y^{''}-4y^'+4y=3 ?

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

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y^{''}-4y^'+4y=3

y^{^{''}} - 4 y^{^{'}} + 4 y = 3

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{3}{4}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 3

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{3}{4}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{3}{4}

(D^{2} + 4) y = sinh (x) sin (2 x)

y^{^{''}} + 4 y^{^{'}} - 12 y = 3 x^{2} + x

2 y^{^{''}} + y^{^{'}} - 10 y = x e^{3 x} cos (π x)

y^{^{'''}} + 3 y^{^{''}} - 10 y^{^{'}} = x - 3

y^{^{'''}} - 24 y^{^{''}} + 2048 y = e^{16 x}

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