What is the solution for y^{''}-2y^'-7y=0 ?

The solution for y^{''}-2y^'-7y=0 is y=c_{1}e^{(1+2^{1+1/2})t}+c_{2}e^{(1-2^{1+1/2})t}

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y^{''}-2y^'-7y=0

y^{^{''}} - 2 y^{^{'}} - 7 y = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 7 y = 0

Solve linear ODE:

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

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

x^{^{''}} + \frac{15}{\frac{1}{32}} x^{^{'}} + \frac{13}{\frac{1}{32}} x = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 4, y^{^{'}} (- 1) = 2

(D^{3} - 5 D^{2} + 8 D - 4) y = 0

y^{^{''}} + b \cdot y = 0, y (0) - 9 y^{^{'}} (0) = 0, y^{^{'}} (9) = 0

y^{^{'''}} + 6 y^{^{''}} + 9 y^{^{'}} + 4 y = 0

What is the solution for y^{''}-2y^'-7y=0 ?
The solution for y^{''}-2y^'-7y=0 is y=c_{1}e^{(1+2^{1+1/2})t}+c_{2}e^{(1-2^{1+1/2})t}