What is the solution for y^{''}+6y^'-4y=0 ?

The solution for y^{''}+6y^'-4y=0 is y=c_{1}e^{(-3+sqrt(13))t}+c_{2}e^{(-3-sqrt(13))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+6y^'-4y=0

y^{^{''}} + 6 y^{^{'}} - 4 y = 0

y = c_{1} e^{(- 3 + 13) t} + c_{2} e^{(- 3 - 13) t}

Solution steps

y^{''} (t) + 6 y^{'} (t) - 4 y = 0

Solve linear ODE:

y = c_{1} e^{(- 3 + 13) t} + c_{2} e^{(- 3 - 13) t}

y = c_{1} e^{(- 3 + 13) t} + c_{2} e^{(- 3 - 13) t}

ln (1 0^{x})

f (x) = e^{x^{2} + 2}

\frac{d}{d y} (x^{2} y^{4})

n = 0 \sum \infty (\frac{1}{1.01})^{n}

\frac{d y}{d t} t + 5 y = 8 t

What is the solution for y^{''}+6y^'-4y=0 ?
The solution for y^{''}+6y^'-4y=0 is y=c_{1}e^{(-3+sqrt(13))t}+c_{2}e^{(-3-sqrt(13))t}