What is the solution for y^{''}+y=cos(t),y(0)=1,y^'(0)=-3 ?

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

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y^{''}+y=cos(t),y(0)=1,y^'(0)=-3

y^{^{''}} + y = cos (t), y (0) = 1, y^{^{'}} (0) = - 3

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} + 27 y = e^{t} - e^{- t}, y^{t} (0) = 2, y (0) = 0

y^{^{''}} - \frac{2}{5} y = - x^{2}

3 y^{^{'}} + 2 y^{^{''}} = - cos (3.14 \cdot \frac{t}{0.2})

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

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

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