What is the solution for y^{''}-4y^'+3y=2e^t ?

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

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y^{''}-4y^'+3y=2e^t

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{t}

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = 9 e^{- x}

y^{^{''}} + 4 y = - 20 sin (2 t)

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 e^{x}

x^{^{''}} + 2 x^{^{'}} + 5 x = 1, x^{^{'}} (0) = 0, x (0) = 0

y^{^{''}} + 4 y = 10 cos (2 x) + 15 sin (2 x)

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