What is the solution for y^{''}-3y^'-4y=0,y(0)=1,y^'(0)=1 ?

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

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

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

y = (1 - \frac{3}{5}) e^{4 t} + \frac{3}{5} e^{- t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{3}{5}) e^{4 t} + \frac{3}{5} e^{- t}

y = (1 - \frac{3}{5}) e^{4 t} + \frac{3}{5} e^{- t}

y^{^{''''}} + 20 y^{^{''}} + 64 y = 0

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

y^{^{''}} - 11 y^{^{'}} + 18 y = 0

y^{^{'''}} + 9 y^{^{''}} + 27 y^{^{'}} + 27 y = 0

2 y^{^{''}} - 3 y^{^{'}} + 5 y = 0

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