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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + y^{^{''}} - 2 y = - e^{3 x} (9 + 67 x + 17 x^{2})

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = x + 6 e^{- x}

3 x^{^{''}} + 48 x = 5 cos (4 t)

8 y^{^{''}} + 6 y^{^{'}} - y = 11

y^{^{''}} - 6 y^{^{'}} = 37 sin (x)

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