What is the solution for 4y^{''}+y=t-cos(t/2) ?

The solution for 4y^{''}+y=t-cos(t/2) is y=c_{1}cos(t/2)+c_{2}sin(t/2)+t-(tsin(t/2))/4

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4y^{''}+y=t-cos(t/2)

4 y^{^{''}} + y = t - cos (\frac{t}{2})

y = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t - \frac{t sin ( \frac{t}{2} )}{4}

Solution steps

4 y^{''} (t) + y = t - cos (\frac{t}{2})

Solve linear ODE:

y = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t - \frac{t sin ( \frac{t}{2} )}{4}

y = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t - \frac{t sin ( \frac{t}{2} )}{4}

What is the solution for 4y^{''}+y=t-cos(t/2) ?
The solution for 4y^{''}+y=t-cos(t/2) is y=c_{1}cos(t/2)+c_{2}sin(t/2)+t-(tsin(t/2))/4