What is the solution for y^{''}-13y^'-40y=0 ?

The solution for y^{''}-13y^'-40y=0 is y=c_{1}e^{((13+sqrt(329))t)/2}+c_{2}e^{((13-sqrt(329))t)/2}

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y^{''}-13y^'-40y=0

y^{^{''}} - 13 y^{^{'}} - 40 y = 0

y = c_{1} e^{\frac{( 13 + 329 ) t}{2}} + c_{2} e^{\frac{( 13 - 329 ) t}{2}}

Solution steps

y^{''} (t) - 13 y^{'} (t) - 40 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 13 + 329 ) t}{2}} + c_{2} e^{\frac{( 13 - 329 ) t}{2}}

y = c_{1} e^{\frac{( 13 + 329 ) t}{2}} + c_{2} e^{\frac{( 13 - 329 ) t}{2}}

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y (0) = 3

x^{^{''}} = - 2 x^{^{'}}

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (1) x = e^{3 x}

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 5

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 4

What is the solution for y^{''}-13y^'-40y=0 ?
The solution for y^{''}-13y^'-40y=0 is y=c_{1}e^{((13+sqrt(329))t)/2}+c_{2}e^{((13-sqrt(329))t)/2}