What is the solution for y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=1 ?

The solution for y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=1 is y=(2-7/5)e^{4t}+7/5 e^{-t}

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y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=1

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y = (2 - \frac{7}{5}) e^{4 t} + \frac{7}{5} e^{- t}

Solution steps

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

Solve linear ODE:

y = (2 - \frac{7}{5}) e^{4 t} + \frac{7}{5} e^{- t}

y = (2 - \frac{7}{5}) e^{4 t} + \frac{7}{5} e^{- t}

y^{^{''}} - 25 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} + 25 y = 0, y (\frac{π}{10}) = - 3, y (π) = - 8

\frac{d ^{3} y}{d x ^{3}} - 5 y = 0

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

15 y^{^{''''}} + 68 y^{^{''}} + 64 y = 0

What is the solution for y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=1 ?
The solution for y^{''}-3y^'-4y=0,y(0)=2,y^'(0)=1 is y=(2-7/5)e^{4t}+7/5 e^{-t}