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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = sec^{2} (e^{x})

3 y^{^{''}} + y^{^{'}} - 2 y = x^{2} - 3 x + 7

y^{^{'''}} + y^{^{'}} - 10 y = x e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = e^{x} sin (x)

y^{^{''}} + 2 y^{^{'}} + y = 21 e^{- t}

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