What is the solution for y^{''}+y=1+2tsin(t) ?

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

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y^{''}+y=1+2tsin(t)

y^{^{''}} + y = 1 + 2 t sin (t)

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

Solution steps

y^{''} (t) + y = 1 + 2 t sin (t)

Solve linear ODE:

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

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

y^{^{''}} + 16 y = sec (4)

2 y^{^{''}} - 4 y^{^{'}} + 10 y = 2 x

0.25 y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

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

y^{^{''}} - 6 y^{^{'}} + 9 y = sin (2 x) + x

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