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

The solution for x^{''}+25x=t^2+t is x=c_{1}cos(5t)+c_{2}sin(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} cos (5 t) + c_{2} sin (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} cos (5 t) + c_{2} sin (5 t) + \frac{t ^{2}}{25} + \frac{t}{25} - \frac{2}{625}

x = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{t ^{2}}{25} + \frac{t}{25} - \frac{2}{625}

y^{^{''}} + 4 y = cos (x), y (0) = 6, y^{^{'}} (0) = 6

y^{^{''}} + 6 y^{^{'}} + 8 y = 3 t e^{- 4 t}

3 x^{^{''}} + 48 x = 120 e^{- 4 t} (cos (4 t))

D, D (t) = \frac{100 ln ( t + 2 )}{t - 2}

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 24 y = 4 e^{6 t}

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