What is the solution for y^{''}-3y^'+2y=12 ?

The solution for y^{''}-3y^'+2y=12 is y=c_{1}e^{2t}+c_{2}e^t+6

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

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

y = c_{1} e^{2 t} + c_{2} e^{t} + 6

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 12

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + 6

y = c_{1} e^{2 t} + c_{2} e^{t} + 6

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

(D^{2} - 4 D + 4) y = e^{x}

y^{^{''}} - 3 y^{^{'}} + 2 y = 12 e^{- 2 t}

3 y^{^{''}} + 12 y^{^{'}} + 39 y = 90 sin (3 t)

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

What is the solution for y^{''}-3y^'+2y=12 ?
The solution for y^{''}-3y^'+2y=12 is y=c_{1}e^{2t}+c_{2}e^t+6