What is the solution for 4y^{''}-4y^'-5y=0 ?

The solution for 4y^{''}-4y^'-5y=0 is y=c_{1}e^{((1+sqrt(6))t)/2}+c_{2}e^{((1-sqrt(6))t)/2}

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4y^{''}-4y^'-5y=0

4 y^{^{''}} - 4 y^{^{'}} - 5 y = 0

y = c_{1} e^{\frac{( 1 + 6 ) t}{2}} + c_{2} e^{\frac{( 1 - 6 ) t}{2}}

Solution steps

4 y^{''} (t) - 4 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 1 + 6 ) t}{2}} + c_{2} e^{\frac{( 1 - 6 ) t}{2}}

y = c_{1} e^{\frac{( 1 + 6 ) t}{2}} + c_{2} e^{\frac{( 1 - 6 ) t}{2}}

(d^{3} - 3 d^{2} - 10 d) y = 0

0.2 y^{^{''}} + 14 y^{^{'}} + \frac{1}{0.004} y = 0

(D^{2} + 2 D + 5)^{3} y = 0

66 y^{^{'''}} + 17 y^{^{''}} + y^{^{'}} = 0

y^{^{'''}} - 15 y^{^{''}} + 27 y^{^{'}} + 243 y = 0

What is the solution for 4y^{''}-4y^'-5y=0 ?
The solution for 4y^{''}-4y^'-5y=0 is y=c_{1}e^{((1+sqrt(6))t)/2}+c_{2}e^{((1-sqrt(6))t)/2}