What is the solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-15 ?

The solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-15 is y=e^{4t}(a+(-15-4a)/5)-(e^{-t}(-15-4a))/5

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y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-15

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = - 15

y = e^{4 t} (a + \frac{- 15 - 4 a}{5}) - \frac{e ^{- t} ( - 15 - 4 a )}{5}

Solution steps

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

Solve linear ODE:

y = e^{4 t} (a + \frac{- 15 - 4 a}{5}) - \frac{e ^{- t} ( - 15 - 4 a )}{5}

y = e^{4 t} (a + \frac{- 15 - 4 a}{5}) - \frac{e ^{- t} ( - 15 - 4 a )}{5}

2 y^{^{''}} + 128 y = 0

y^{^{''''}} - y^{^{'''}} - y^{^{''}} - y^{^{'}} - 2 y = 0

4 y^{^{''''}} + 8 y^{^{''}} + 4 y = 0

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 21 y = 0

(D^{2} + 2)^{3} (D^{2} - 3)^{2} y = 0

What is the solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-15 ?
The solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-15 is y=e^{4t}(a+(-15-4a)/5)-(e^{-t}(-15-4a))/5