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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

3 y^{^{''}} - 15 y^{^{'}} + 4 y = 0

1 y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 12 y^{^{'}} + 27 y = 0, y (0) = 1, y^{^{'}} (0) = - 12

18 y^{^{''}} + 9 y^{^{'}} - 5 y = 0

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