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)

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)