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

The solution for y^{''}+4y=cos(2t)+sin(t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(tsin(2t))/4+1/3 sin(t)

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t sin ( 2 t )}{4} + \frac{1}{3} sin (t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t sin ( 2 t )}{4} + \frac{1}{3} sin (t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t sin ( 2 t )}{4} + \frac{1}{3} sin (t)

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

y^{^{''}} - 6 y + 9 = e^{3 x}

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = 4 cos (x)

y^{^{''}} + 81 y = 2 sin (9 t)

y^{^{''}} - 4 y = 12 x, y (0) = 4, y^{^{'}} (0) = 1

What is the solution for y^{''}+4y=cos(2t)+sin(t) ?
The solution for y^{''}+4y=cos(2t)+sin(t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(tsin(2t))/4+1/3 sin(t)