What is the solution for y^{''}-4y^'-5y=7e^2t ?

The solution for y^{''}-4y^'-5y=7e^2t is y=c_{1}e^{5t}+c_{2}e^{-t}-(7e^2t)/5+(28e^2)/(25)

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y^{''}-4y^'-5y=7e^2t

y^{^{''}} - 4 y^{^{'}} - 5 y = 7 e^{2} t

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{7 e ^{2} t}{5} + \frac{28 e ^{2}}{25}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 5 y = 7 e^{2} t

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{7 e ^{2} t}{5} + \frac{28 e ^{2}}{25}

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{7 e ^{2} t}{5} + \frac{28 e ^{2}}{25}

d, d (t) = (8 - 2.8 t)^{2}

y^{^{''}} + 6 \cdot y^{^{'}} + 25 y = e^{- 3 t} cos (4 t)

(D^{2} + 5 D + 6) y = e^{x} + e^{- x}

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

y^{^{''}} + 4 y = 2 csc^{2} (2 x)

What is the solution for y^{''}-4y^'-5y=7e^2t ?
The solution for y^{''}-4y^'-5y=7e^2t is y=c_{1}e^{5t}+c_{2}e^{-t}-(7e^2t)/5+(28e^2)/(25)