What is the solution for 4y^{''}+y=sin(t),y(0)=0,y^'(0)=1 ?

The solution for 4y^{''}+y=sin(t),y(0)=0,y^'(0)=1 is y= 8/3 sin(t/2)-1/3 sin(t)

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4y^{''}+y=sin(t),y(0)=0,y^'(0)=1

4 y^{^{''}} + y = sin (t), y (0) = 0, y^{^{'}} (0) = 1

y = \frac{8}{3} sin (\frac{t}{2}) - \frac{1}{3} sin (t)

Solution steps

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

Solve linear ODE:

y = \frac{8}{3} sin (\frac{t}{2}) - \frac{1}{3} sin (t)

y = \frac{8}{3} sin (\frac{t}{2}) - \frac{1}{3} sin (t)

y^{^{''}} + 16 y = cos (4 x) + sin (4 x), y^{0} = 3, y^{^{'}} 0 = 5

y^{^{'''}} + 289 y^{^{'}} = tan (17 x)

y^{^{''}} - 3 y^{^{'}} - 4 y = t^{2}

y^{^{''}} + y = 5 cos (x), y (0) = 0, y (0) = 1

y^{^{''}} - 2 y^{^{'}} + y = 2 e^{t}, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for 4y^{''}+y=sin(t),y(0)=0,y^'(0)=1 ?
The solution for 4y^{''}+y=sin(t),y(0)=0,y^'(0)=1 is y= 8/3 sin(t/2)-1/3 sin(t)