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

The solution for y^{''}-4y=tcos(2t) is y=c_{1}e^{2t}+c_{2}e^{-2t}-(tcos(2t))/8+1/16 sin(2t)

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y^{''}-4y=tcos(2t)

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

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{t cos ( 2 t )}{8} + \frac{1}{16} sin (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{t cos ( 2 t )}{8} + \frac{1}{16} sin (2 t)

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{t cos ( 2 t )}{8} + \frac{1}{16} sin (2 t)

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

\frac{d ^{2} y}{d x ^{2}} + 4 y = cos (\frac{9}{4} x)

\frac{d ^{2} y}{d x ^{2}} + a y^{^{'}} = b

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

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

What is the solution for y^{''}-4y=tcos(2t) ?
The solution for y^{''}-4y=tcos(2t) is y=c_{1}e^{2t}+c_{2}e^{-2t}-(tcos(2t))/8+1/16 sin(2t)