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

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

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

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

y = e^{4 t} + 2 e^{- t}

Solution steps

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

Solve linear ODE:

y = e^{4 t} + 2 e^{- t}

y = e^{4 t} + 2 e^{- t}

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