What is the solution for y^{''}+4y^'-5y=3e^t ?

The solution for y^{''}+4y^'-5y=3e^t is y=c_{1}e^t+c_{2}e^{-5t}+(e^tt)/2

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y^{''}+4y^'-5y=3e^t

y^{^{''}} + 4 y^{^{'}} - 5 y = 3 e^{t}

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

Solution steps

y^{''} (t) + 4 y^{'} (t) - 5 y = 3 e^{t}

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{- 4} \cdot e^{2 x}

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + 3 + 2 cos (2 x)

2 y^{^{''}} - 3 y^{^{'}} + y = 64

y^{^{''}} - 3 y + 2 = x^{2} + 2 x + 1

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

What is the solution for y^{''}+4y^'-5y=3e^t ?
The solution for y^{''}+4y^'-5y=3e^t is y=c_{1}e^t+c_{2}e^{-5t}+(e^tt)/2