What is the solution for y^{''}+4y=t^2 ?

The solution for y^{''}+4y=t^2 is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8

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y^{''}+4y=t^2

y^{^{''}} + 4 y = t^{2}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8}

Solution steps

y^{''} (t) + 4 y = t^{2}

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8}

f (x) = e^{\infty}

(5 x^{2} - 2 x + 4) (4 x^{2} + 3 x - 6)

\int - x^{2} + 6 d x

x \to 5 lim (\frac{1}{( x - 5 )})

\frac{\partial}{\partial x} (x^{2} \frac{1}{y + 1})

What is the solution for y^{''}+4y=t^2 ?
The solution for y^{''}+4y=t^2 is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8