What is the solution for y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=0 ?

The solution for y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=0 is y=(2-8/5)e^{4t}+8/5 e^{-t}

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y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y = (2 - \frac{8}{5}) e^{4 t} + \frac{8}{5} e^{- t}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 0

Solve linear ODE:

y = (2 - \frac{8}{5}) e^{4 t} + \frac{8}{5} e^{- t}

y = (2 - \frac{8}{5}) e^{4 t} + \frac{8}{5} e^{- t}

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (1) = 5

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

y^{^{''}} + 2 y^{^{'}} - 6 y = 0

\frac{d ^{4} y}{d x ^{4}} + 2 y^{^{'''}} - 2 y^{^{'}} - y = 0

y^{^{''}} - 6 y^{^{'}} + 13 y = 0

What is the solution for y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=0 ?
The solution for y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=0 is y=(2-8/5)e^{4t}+8/5 e^{-t}