What is the solution for y^{''}-2y^'-3y=0,y(0)=5,y^'(0)=-1 ?

The solution for y^{''}-2y^'-3y=0,y(0)=5,y^'(0)=-1 is y=e^{3t}+4e^{-t}

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y^{''}-2y^'-3y=0,y(0)=5,y^'(0)=-1

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 5, y^{^{'}} (0) = - 1

y = e^{3 t} + 4 e^{- t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = e^{3 t} + 4 e^{- t}

y = e^{3 t} + 4 e^{- t}

(d^{3} - 5 d^{2} + 8 d - 4) \cdot y = 0

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

y^{^{'''}} - y^{^{'}} = 0, y (0) = 4, y^{^{'}} (0) = 0, y^{^{''}} (0) = 9

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 8

4 y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} = 0

What is the solution for y^{''}-2y^'-3y=0,y(0)=5,y^'(0)=-1 ?
The solution for y^{''}-2y^'-3y=0,y(0)=5,y^'(0)=-1 is y=e^{3t}+4e^{-t}