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^{-(3t)/4}

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

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

y = c_{1} e^{t} + c_{2} e^{- \frac{3 t}{4}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- \frac{3 t}{4}}

y = c_{1} e^{t} + c_{2} e^{- \frac{3 t}{4}}

(D^{2} - 3 D) y = 0

(D^{3} - D^{2} - D + 1) y = 0

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y^{0} = 0, y^{^{'}} 0 = 1

4 \frac{d ^{2} y}{d x ^{2}} + 32 \frac{d y}{d x} + 64 y = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 8, y^{^{'}} (0) = 9

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^{-(3t)/4}