What is the solution for y^{''}+12y^'+37y=4y ?

The solution for y^{''}+12y^'+37y=4y is y=c_{1}e^{(-6+sqrt(3))t}+c_{2}e^{(-6-sqrt(3))t}

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y^{''}+12y^'+37y=4y

y^{^{''}} + 12 y^{^{'}} + 37 y = 4 y

y = c_{1} e^{(- 6 + 3) t} + c_{2} e^{(- 6 - 3) t}

Solution steps

y^{''} (t) + 12 y^{'} (t) + 37 y = 4 y

Solve linear ODE:

y = c_{1} e^{(- 6 + 3) t} + c_{2} e^{(- 6 - 3) t}

y = c_{1} e^{(- 6 + 3) t} + c_{2} e^{(- 6 - 3) t}

y^{^{''}} - y^{^{'}} - 12 y = 0, y (1) = 0 y^{^{'}} (1)

y^{^{'''}} - 5 y^{^{''}} - 4 y^{^{'}} + 20 y = 0

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

\frac{d ^{2} y}{d x ^{2}} + 4 y = 0, y = - cos (2 x)

y^{^{'''}} + 2 y^{^{''}} - 11 y^{^{'}} - 12 y = 0

What is the solution for y^{''}+12y^'+37y=4y ?
The solution for y^{''}+12y^'+37y=4y is y=c_{1}e^{(-6+sqrt(3))t}+c_{2}e^{(-6-sqrt(3))t}