What is the solution for y^{''}+4y^'+2y=0 ?

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

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

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

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 2 y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial u} (tan (\frac{u}{v}))

\int \frac{x}{( 3 x ^{2} - 4 ) ^{2}} d x

4 x^{2}, 16

f (x) = 7 x^{4} + 5

\frac{\partial}{\partial y} (4 - x^{2} - 9 y^{2})

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