What is the solution for (1)y^{''}+(8)y^'-(7)y=0 ?

The solution for (1)y^{''}+(8)y^'-(7)y=0 is y=c_{1}e^{(-4+sqrt(23))t}+c_{2}e^{(-4-sqrt(23))t}

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(1)y^{''}+(8)y^'-(7)y=0

(1) y^{^{''}} + (8) y^{^{'}} - (7) y = 0

y = c_{1} e^{(- 4 + 23) t} + c_{2} e^{(- 4 - 23) t}

Solution steps

(1) y^{''} (t) + (8) y^{'} (t) - (7) y = 0

Solve linear ODE:

y = c_{1} e^{(- 4 + 23) t} + c_{2} e^{(- 4 - 23) t}

y = c_{1} e^{(- 4 + 23) t} + c_{2} e^{(- 4 - 23) t}

\frac{d ^{8} y}{d x ^{8}} + 8 \frac{d ^{4} y}{d x ^{4}} + 16 y = 0

4 y^{^{''}} + 4 y^{^{'}} + 17 y = 0, y (0) = - 1, y^{^{'}} (0) = - 2

x^{^{''}} + \frac{x}{a} = 0

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

15 y^{^{''}} + 26 y^{^{'}} + 7 y = 0, y (0) = 8, y^{^{'}} (0) = 1

What is the solution for (1)y^{''}+(8)y^'-(7)y=0 ?
The solution for (1)y^{''}+(8)y^'-(7)y=0 is y=c_{1}e^{(-4+sqrt(23))t}+c_{2}e^{(-4-sqrt(23))t}