What is the solution for y^{''}-25y=e^{5t} ?

The solution for y^{''}-25y=e^{5t} is y=c_{1}e^{5t}+c_{2}e^{-5t}+(e^{5t}t)/(10)

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y^{''}-25y=e^{5t}

y^{^{''}} - 25 y = e^{5 t}

y = c_{1} e^{5 t} + c_{2} e^{- 5 t} + \frac{e ^{5 t} t}{10}

Solution steps

y^{''} (t) - 25 y = e^{5 t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 5 t} + \frac{e ^{5 t} t}{10}

y = c_{1} e^{5 t} + c_{2} e^{- 5 t} + \frac{e ^{5 t} t}{10}

y^{^{''}} + 12 y = 7 e^{2 x}

y^{^{''}} - 14 y^{^{'}} + 49 y = (x + 1) e^{x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 4 e^{5 x}

y^{^{''}} + 10 y^{^{'}} + 25 y = 50 sin (5 x)

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

What is the solution for y^{''}-25y=e^{5t} ?
The solution for y^{''}-25y=e^{5t} is y=c_{1}e^{5t}+c_{2}e^{-5t}+(e^{5t}t)/(10)