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

The solution for y^{''}+4y=tsin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-(t^2cos(2t))/8+(tsin(2t))/(16)

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{t ^{2} cos ( 2 t )}{8} + \frac{t sin ( 2 t )}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{t ^{2} cos ( 2 t )}{8} + \frac{t sin ( 2 t )}{16}

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{t ^{2} cos ( 2 t )}{8} + \frac{t sin ( 2 t )}{16}

\frac{d y}{d x} + 12 y = 7

\frac{\partial}{\partial x} (\frac{x}{y} + \frac{y}{z})

\int 3 x 3 - 3 x d x

\frac{d y}{d x} = \frac{2 x}{3 y ^{2}}

\frac{\partial}{\partial x} (e^{4 x y})

What is the solution for y^{''}+4y=tsin(2t) ?
The solution for y^{''}+4y=tsin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-(t^2cos(2t))/8+(tsin(2t))/(16)