What is the solution for y^{''}+3y^'-18y=0,y(0)=-1,y^'(0)=24 ?

The solution for y^{''}+3y^'-18y=0,y(0)=-1,y^'(0)=24 is y=2e^{3t}-3e^{-6t}

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y^{''}+3y^'-18y=0,y(0)=-1,y^'(0)=24

y^{^{''}} + 3 y^{^{'}} - 18 y = 0, y (0) = - 1, y^{^{'}} (0) = 24

y = 2 e^{3 t} - 3 e^{- 6 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 18 y = 0

Solve linear ODE:

y = 2 e^{3 t} - 3 e^{- 6 t}

y = 2 e^{3 t} - 3 e^{- 6 t}

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 1

\frac{d ^{2} y}{d x ^{2}} (1) + 5 y = 0

\frac{d ^{2} x}{d t ^{2}} + 8 \frac{d x}{d t} + 16 x = 0

y^{^{'''}} + 2 y^{^{''}} - 10 y^{^{'}} + 4 y = 0

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} = 0

What is the solution for y^{''}+3y^'-18y=0,y(0)=-1,y^'(0)=24 ?
The solution for y^{''}+3y^'-18y=0,y(0)=-1,y^'(0)=24 is y=2e^{3t}-3e^{-6t}