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

The solution for y^{''}+4y^'-2y=0 is y=c_{1}e^{(-2+sqrt(6))t}+c_{2}e^{(-2-sqrt(6))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (\frac{x}{x ^{3} - y ^{6}})

\frac{\partial}{\partial x} (7 sec (x) - 4 x)

y^{^{'}} = \frac{4 x ^{2} + 3 y ^{2}}{2 x y}

x^{2} - 8, - x^{2} + 2 x + 4

\frac{d}{d x} (ln (\frac{2 - x}{1 - x}))

What is the solution for y^{''}+4y^'-2y=0 ?
The solution for y^{''}+4y^'-2y=0 is y=c_{1}e^{(-2+sqrt(6))t}+c_{2}e^{(-2-sqrt(6))t}