What is the solution for y^'+3y^{''}-y^'-3y=0 ?

The solution for y^'+3y^{''}-y^'-3y=0 is y=c_{1}e^t+c_{2}e^{-t}

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

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

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

Solution steps

y^{'} (t) + 3 y^{''} (t) - y^{'} (t) - 3 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 12 y^{^{'}} + 37 y = 4 y

y^{^{''}} - y^{^{'}} - 12 y = 0, y (1) = 0 y^{^{'}} (1)

y^{^{'''}} - 5 y^{^{''}} - 4 y^{^{'}} + 20 y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

\frac{d ^{2} y}{d x ^{2}} + 4 y = 0, y = - cos (2 x)

What is the solution for y^'+3y^{''}-y^'-3y=0 ?
The solution for y^'+3y^{''}-y^'-3y=0 is y=c_{1}e^t+c_{2}e^{-t}