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}

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{3 x}

d, d (p) = 480 - 3 p

y^{^{''}} + 4 y = 3 \cdot sin (2 x)

y^{^{''}} - y = 5 x + 2

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{2 x} - 2 x

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)