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

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

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y^{''}+y=2tcos(t)

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

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

Solution steps

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

Solve linear ODE:

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

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

y = (4 x^{2} + 5 x)^{2}

\int 2 t^{2} e^{- s t} d t

f (x) = 2 x^{3} - 4 x + 6, at x = 1

\int_{2}^{7} g (x) d x

\frac{d}{d x} (\frac{x ^{2} - 1}{3 x})

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