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=(3-c_{2})e^{4t}+c_{2}e^{-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 = (3 - c_{2}) e^{4 t} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = (3 - c_{2}) e^{4 t} + c_{2} e^{- t}

y = (3 - c_{2}) e^{4 t} + c_{2} e^{- t}

y^{^{''}} - \frac{10}{3} y^{^{'}} + α y = 0

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

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

\frac{d ^{2} y}{d x ^{2}} + 0.25 \frac{d y}{d x} + 2 y = 0

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

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=(3-c_{2})e^{4t}+c_{2}e^{-t}