What is the solution for y^{''}-3y^'+2y=10 ?

The solution for y^{''}-3y^'+2y=10 is y=c_{1}e^{2t}+c_{2}e^t+5

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y^{''}-3y^'+2y=10

y^{^{''}} - 3 y^{^{'}} + 2 y = 10

y = c_{1} e^{2 t} + c_{2} e^{t} + 5

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 10

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + 5

y = c_{1} e^{2 t} + c_{2} e^{t} + 5

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 (x + 1)

y^{^{''}} - 6 y + 9 y = e^{- 2 x} sin (3 x)

y^{^{''}} - y^{^{'}} + 2 y = 3 sin (3 x)

y^{^{''''}} + 2 y^{^{''}} + y = 9 t + 6

y^{^{''''}} + 2 y^{^{''}} + y = 9 t + 2

What is the solution for y^{''}-3y^'+2y=10 ?
The solution for y^{''}-3y^'+2y=10 is y=c_{1}e^{2t}+c_{2}e^t+5