What is the solution for y^{''}-y^'-6y=0,y(0)=3,y^'(0)=4 ?

The solution for y^{''}-y^'-6y=0,y(0)=3,y^'(0)=4 is y=2e^{3t}+e^{-2t}

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y^{''}-y^'-6y=0,y(0)=3,y^'(0)=4

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 3, y^{^{'}} (0) = 4

y = 2 e^{3 t} + e^{- 2 t}

Solution steps

y^{''} (t) - y^{'} (t) - 6 y = 0

Solve linear ODE:

y = 2 e^{3 t} + e^{- 2 t}

y = 2 e^{3 t} + e^{- 2 t}

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 9, y^{^{'}} (0) = 2

y^{^{''}} + 3 y^{^{'}} - 9 y = 0

100 y^{^{''}} + 90 y^{^{'}} - 162 y = 0

y^{^{''}} - 4 y^{^{'}} - 13 y = 0

\frac{3}{4} \frac{d ^{2} x}{d t ^{2}} + 72 x = 0, x (0) = - \frac{1}{2}

What is the solution for y^{''}-y^'-6y=0,y(0)=3,y^'(0)=4 ?
The solution for y^{''}-y^'-6y=0,y(0)=3,y^'(0)=4 is y=2e^{3t}+e^{-2t}