What is the solution for f^{''}(t)-f(t)=-8t,f(3)=-8 ?

The solution for f^{''}(t)-f(t)=-8t,f(3)=-8 is f(t)=e^{t-6}(-c_{2}-32e^3)+c_{2}e^{-t}+8t

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f^{''}(t)-f(t)=-8t,f(3)=-8

f^{^{''}} (t) - f (t) = - 8 t, f (3) = - 8

f (t) = e^{t - 6} (- c_{2} - 32 e^{3}) + c_{2} e^{- t} + 8 t

Solution steps

f^{''} (t) - f (t) = - 8 t

Solve linear ODE:

f (t) = e^{t - 6} (- c_{2} - 32 e^{3}) + c_{2} e^{- t} + 8 t

f (t) = e^{t - 6} (- c_{2} - 32 e^{3}) + c_{2} e^{- t} + 8 t

f (x) = \frac{2 x - 1}{4 x + 8}

\frac{1}{1 + x ^{3}}

\frac{\partial}{\partial y} (4 y - 2 x)

\int 4 - x^{2} x d x

\frac{\partial}{\partial x} (\frac{e ^{y}}{x + y ^{7}})

What is the solution for f^{''}(t)-f(t)=-8t,f(3)=-8 ?
The solution for f^{''}(t)-f(t)=-8t,f(3)=-8 is f(t)=e^{t-6}(-c_{2}-32e^3)+c_{2}e^{-t}+8t