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

The solution for y^{''}-y^'-3y-4=0 is y=c_{1}e^{((1+sqrt(13))t)/2}+c_{2}e^{((1-sqrt(13))t)/2}-4/3

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

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

y = c_{1} e^{\frac{( 1 + 13 ) t}{2}} + c_{2} e^{\frac{( 1 - 13 ) t}{2}} - \frac{4}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( 1 + 13 ) t}{2}} + c_{2} e^{\frac{( 1 - 13 ) t}{2}} - \frac{4}{3}

y = c_{1} e^{\frac{( 1 + 13 ) t}{2}} + c_{2} e^{\frac{( 1 - 13 ) t}{2}} - \frac{4}{3}

x^{^{''}} + 0.2 x^{^{'}} - x = 0.3 cos (1 t)

y^{^{''}} - 8 y^{^{'}} + 16 y = x e^{4 x}

(d^{2} - 4 d + 3) y = 2 x e^{3 x} + 3 e^{x} cos (2 x)

x^{^{''}} + 6 x^{^{'}} + 23 x = 8 cos (4 t)

y^{^{''}} - 7 y^{^{'}} + 12 y = 4 e^{x}

What is the solution for y^{''}-y^'-3y-4=0 ?
The solution for y^{''}-y^'-3y-4=0 is y=c_{1}e^{((1+sqrt(13))t)/2}+c_{2}e^{((1-sqrt(13))t)/2}-4/3