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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = 4 x^{2} - 2 x + 3

\frac{d}{d x} (lo g_{e} (tan (x)))

\frac{d}{d x} (\frac{x ^{2} - 8 x}{2 x - 3})

\int - \frac{1}{2 x} d x

\frac{6 x}{4 - x ^{2}}, at x = 0

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