What is the solution for y^{''}-y=12t,y(0)=-12t ?

The solution for y^{''}-y=12t,y(0)=-12t is y=c_{1}e^t+c_{2}e^{-t}-12t

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y^{''}-y=12t,y(0)=-12t

y^{^{''}} - y = 12 t, y (0) = - 12 t

y = c_{1} e^{t} + c_{2} e^{- t} - 12 t

Solution steps

y^{''} (t) - y = 12 t

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} - 12 t

y = c_{1} e^{t} + c_{2} e^{- t} - 12 t

y^{^{''}} + 4 y^{^{'}} = e^{x} cos (2 x)

(D^{2} - 6 D + 13) y = 5^{x}

y^{^{''}} + 6 y^{^{'}} + 5 y = 10 cos (5 x)

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 4 y = sin (x)

What is the solution for y^{''}-y=12t,y(0)=-12t ?
The solution for y^{''}-y=12t,y(0)=-12t is y=c_{1}e^t+c_{2}e^{-t}-12t