What is the solution for y^{''}-y-2=0,y(0)=3 ?

The solution for y^{''}-y-2=0,y(0)=3 is y=(-c_{2}+5)e^t+c_{2}e^{-t}-2

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y^{''}-y-2=0,y(0)=3

y^{^{''}} - y - 2 = 0, y (0) = 3

y = (- c_{2} + 5) e^{t} + c_{2} e^{- t} - 2

Solution steps

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

Solve linear ODE:

y = (- c_{2} + 5) e^{t} + c_{2} e^{- t} - 2

y = (- c_{2} + 5) e^{t} + c_{2} e^{- t} - 2

y^{^{''}} - 6 y^{^{'}} + 8 y = 2

y^{^{''}} + 3 y^{^{'}} - 4 = 0

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{12 e ^{x}}{x ^{3}}

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

What is the solution for y^{''}-y-2=0,y(0)=3 ?
The solution for y^{''}-y-2=0,y(0)=3 is y=(-c_{2}+5)e^t+c_{2}e^{-t}-2