What is the solution for y^{''}+20y^'+y=12 ?

The solution for y^{''}+20y^'+y=12 is y=c_{1}e^{(-10+3sqrt(11))t}+c_{2}e^{(-10-3sqrt(11))t}+12

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y^{''}+20y^'+y=12

y^{^{''}} + 20 y^{^{'}} + y = 12

y = c_{1} e^{(- 10 + 311) t} + c_{2} e^{(- 10 - 311) t} + 12

Solution steps

y^{''} (t) + 20 y^{'} (t) + y = 12

Solve linear ODE:

y = c_{1} e^{(- 10 + 311) t} + c_{2} e^{(- 10 - 311) t} + 12

y = c_{1} e^{(- 10 + 311) t} + c_{2} e^{(- 10 - 311) t} + 12

x \to 0 lim (3 + x^{2} sin (\frac{1}{x}))

y = \frac{1}{x - 4}, x = 8

\frac{( x ^{2} + 2 x - 6 )}{( x + 1 ) ^{2}}

((\frac{2}{s}) - (\frac{1}{s ^{3}}))^{2}

f (x) = \frac{x ^{2}}{x + 8}

What is the solution for y^{''}+20y^'+y=12 ?
The solution for y^{''}+20y^'+y=12 is y=c_{1}e^{(-10+3sqrt(11))t}+c_{2}e^{(-10-3sqrt(11))t}+12