What is the solution for x^{''}-25x=t^2+t ?

The solution for x^{''}-25x=t^2+t is x=c_{1}e^{5t}+c_{2}e^{-5t}-(t^2)/(25)-t/(25)-2/625

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x^{''}-25x=t^2+t

x^{^{''}} - 25 x = t^{2} + t

x = c_{1} e^{5 t} + c_{2} e^{- 5 t} - \frac{t ^{2}}{25} - \frac{t}{25} - \frac{2}{625}

Solution steps

x^{''} (t) - 25 x = t^{2} + t

Solve linear ODE:

x = c_{1} e^{5 t} + c_{2} e^{- 5 t} - \frac{t ^{2}}{25} - \frac{t}{25} - \frac{2}{625}

x = c_{1} e^{5 t} + c_{2} e^{- 5 t} - \frac{t ^{2}}{25} - \frac{t}{25} - \frac{2}{625}

y^{^{''}} + 2 y^{^{'}} - 3 y = 2 x sin (x) sin (2 x)

4 y^{^{''}} - 4 y^{^{'}} + y = 48 e^{\frac{t}{2}}

y^{^{''}} + 3 y^{^{'}} + 5 y = sin (t), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - y = 5 sin (2 t)

y^{^{''}} + y^{^{'}} - 6 y = 3 e^{2 t}

What is the solution for x^{''}-25x=t^2+t ?
The solution for x^{''}-25x=t^2+t is x=c_{1}e^{5t}+c_{2}e^{-5t}-(t^2)/(25)-t/(25)-2/625