What is the solution for y^{''}-8y^'-33y=0,y(0)=0,y^'(0)=-7 ?

The solution for y^{''}-8y^'-33y=0,y(0)=0,y^'(0)=-7 is y=-1/2 e^{11t}+1/2 e^{-3t}

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y^{''}-8y^'-33y=0,y(0)=0,y^'(0)=-7

y^{^{''}} - 8 y^{^{'}} - 33 y = 0, y (0) = 0, y^{^{'}} (0) = - 7

y = - \frac{1}{2} e^{11 t} + \frac{1}{2} e^{- 3 t}

Solution steps

y^{''} (t) - 8 y^{'} (t) - 33 y = 0

Solve linear ODE:

y = - \frac{1}{2} e^{11 t} + \frac{1}{2} e^{- 3 t}

y = - \frac{1}{2} e^{11 t} + \frac{1}{2} e^{- 3 t}

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

3 y^{^{'''}} + 10 y^{^{''}} + 15 y^{^{'}} + 4 y = 0

y^{^{''}} + 0.25 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''''}} = 16 y

y^{^{'''}} + y^{^{''}} - 32 y^{^{'}} - 60 y = 0

What is the solution for y^{''}-8y^'-33y=0,y(0)=0,y^'(0)=-7 ?
The solution for y^{''}-8y^'-33y=0,y(0)=0,y^'(0)=-7 is y=-1/2 e^{11t}+1/2 e^{-3t}