What is the solution for y^{''}-8y^'+3y=0 ?

The solution for y^{''}-8y^'+3y=0 is y=c_{1}e^{(4+sqrt(13))t}+c_{2}e^{(4-sqrt(13))t}

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y^{''}-8y^'+3y=0

y^{^{''}} - 8 y^{^{'}} + 3 y = 0

y = c_{1} e^{(4 + 13) t} + c_{2} e^{(4 - 13) t}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = c_{1} e^{(4 + 13) t} + c_{2} e^{(4 - 13) t}

y = c_{1} e^{(4 + 13) t} + c_{2} e^{(4 - 13) t}

\frac{d ^{2} x}{d t ^{2}} + 6 \frac{d x}{d t} + 9 x = 0

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

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

y^{^{''}} - 7 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

4 x^{^{''}} + 2 x = 0

What is the solution for y^{''}-8y^'+3y=0 ?
The solution for y^{''}-8y^'+3y=0 is y=c_{1}e^{(4+sqrt(13))t}+c_{2}e^{(4-sqrt(13))t}