What is the solution for (0+1)y^{''}+(7+1)y^'-(6+1)y=0 ?

The solution for (0+1)y^{''}+(7+1)y^'-(6+1)y=0 is y=c_{1}e^{(-4+sqrt(23))t}+c_{2}e^{(-4-sqrt(23))t}

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(0+1)y^{''}+(7+1)y^'-(6+1)y=0

(0 + 1) y^{^{''}} + (7 + 1) y^{^{'}} - (6 + 1) y = 0

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

Solution steps

(0 + 1) y^{''} (t) + (7 + 1) y^{'} (t) - (6 + 1) 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}

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

y^{^{''}} - 2 y - 3 y^{^{'}} = 0

y^{^{''}} + 10 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

15 y^{^{''}} - 16 y^{^{'}} + y = 0, y (0) = 15, y^{^{'}} (0) = \frac{1}{15}

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

What is the solution for (0+1)y^{''}+(7+1)y^'-(6+1)y=0 ?
The solution for (0+1)y^{''}+(7+1)y^'-(6+1)y=0 is y=c_{1}e^{(-4+sqrt(23))t}+c_{2}e^{(-4-sqrt(23))t}