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

The solution for y^{''}-6y^'+7y=0 is y=c_{1}e^{(3+sqrt(2))t}+c_{2}e^{(3-sqrt(2))t}

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y^{''}-6y^'+7y=0

y^{^{''}} - 6 y^{^{'}} + 7 y = 0

y = c_{1} e^{(3 + 2) t} + c_{2} e^{(3 - 2) t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(3 + 2) t} + c_{2} e^{(3 - 2) t}

y = c_{1} e^{(3 + 2) t} + c_{2} e^{(3 - 2) t}

\frac{\partial}{\partial x} (arctan (x) y)

(x - 1)^{3}, x - 1

\frac{d}{d x} (x^{2} - 2)

\int - x - 1 d x

\frac{d}{d t} (9 sin (t))

What is the solution for y^{''}-6y^'+7y=0 ?
The solution for y^{''}-6y^'+7y=0 is y=c_{1}e^{(3+sqrt(2))t}+c_{2}e^{(3-sqrt(2))t}