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

The solution for y^{''}-y=2t,y(0)=-2t is y=c_{1}e^t+c_{2}e^{-t}-2t

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

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

y = c_{1} e^{t} + c_{2} e^{- t} - 2 t

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} - 2 t

y = c_{1} e^{t} + c_{2} e^{- t} - 2 t

y^{^{''}} = - 3 y + cos (3 x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 sin (t)

y^{^{''}} - y^{^{'}} - 6 y = 62 cos (3 x)

\frac{d ^{2} x}{d t ^{2}} + k \frac{d x}{d t} + g = 0

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

What is the solution for y^{''}-y=2t,y(0)=-2t ?
The solution for y^{''}-y=2t,y(0)=-2t is y=c_{1}e^t+c_{2}e^{-t}-2t