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)+3/2 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{3}{2} sin (t) - \frac{t cos ( t )}{2}

Solution steps

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

Solve linear ODE:

y = cos (t) + \frac{3}{2} sin (t) - \frac{t cos ( t )}{2}

y = cos (t) + \frac{3}{2} sin (t) - \frac{t cos ( t )}{2}

y^{^{''}} + 2 y^{^{'}} - 8 y = 5 e^{- 3 x} - e^{- x}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} = x^{2}

y^{^{''}} + 4 y^{^{'}} + 4 y = \frac{e ^{- 2 x}}{x ^{2} + 1}

y^{^{''}} - 3 y^{^{'}} + 2 y = 5 e^{3 x}

y^{^{''}} - 16 y = e^{4 x}

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)+3/2 sin(t)-(tcos(t))/2