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

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

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

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

y = - \frac{t cos ( t )}{2} + \frac{1}{2} sin (t)

Solution steps

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

Solve linear ODE:

y = - \frac{t cos ( t )}{2} + \frac{1}{2} sin (t)

y = - \frac{t cos ( t )}{2} + \frac{1}{2} sin (t)

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