What is the solution for y^{''}+y=(t-2pi),y(0)=10,y^'(0)=0 ?

The solution for y^{''}+y=(t-2pi),y(0)=10,y^'(0)=0 is y=(10+2pi)cos(t)-sin(t)+t-2pi

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y^{''}+y=(t-2pi),y(0)=10,y^'(0)=0

y^{^{''}} + y = (t - 2 π), y (0) = 10, y^{^{'}} (0) = 0

y = (10 + 2 π) cos (t) - sin (t) + t - 2 π

Solution steps

y^{''} (t) + y = (t - 2 π)

Solve linear ODE:

y = (10 + 2 π) cos (t) - sin (t) + t - 2 π

y = (10 + 2 π) cos (t) - sin (t) + t - 2 π

y^{^{''}} + 16 y = 3.36

y^{^{''}} + 4 y^{^{'}} + 5 y = sin (x)

y^{^{''}} = 2 y + 20 tan^{3} (x), y (0) = 10 tan (x)

\frac{1}{6} y^{^{''}} + 6 y = 2 tan (6 t) - \frac{1}{4} e^{3 t}

y^{^{''}} + 4 y = 11 t^{3} sin (2 t)

What is the solution for y^{''}+y=(t-2pi),y(0)=10,y^'(0)=0 ?
The solution for y^{''}+y=(t-2pi),y(0)=10,y^'(0)=0 is y=(10+2pi)cos(t)-sin(t)+t-2pi