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

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

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

y^{^{''}} + 4 y = t \cdot 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 \cdot 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}

2 + 5 arctan (\frac{x}{2})

\frac{1}{( x + 6 ) ^{2}}

x y^{^{'}} + y = yy

\frac{\partial}{\partial y} (x^{2} - y x)

\frac{2 x}{x + 1}

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