What is the solution for 6y^{''}+40y^'+54y=0 ?

The solution for 6y^{''}+40y^'+54y=0 is y=c_{1}e^{((-10+sqrt(19))t)/3}+c_{2}e^{-((10+sqrt(19))t)/3}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6y^{''}+40y^'+54y=0

6 y^{^{''}} + 40 y^{^{'}} + 54 y = 0

y = c_{1} e^{\frac{( - 10 + 19 ) t}{3}} + c_{2} e^{- \frac{( 10 + 19 ) t}{3}}

Solution steps

6 y^{''} (t) + 40 y^{'} (t) + 54 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 10 + 19 ) t}{3}} + c_{2} e^{- \frac{( 10 + 19 ) t}{3}}

y = c_{1} e^{\frac{( - 10 + 19 ) t}{3}} + c_{2} e^{- \frac{( 10 + 19 ) t}{3}}

\frac{d}{d x} (- k x)

\int cos (x) - x sin (x) d x

e^{- x^{62}}

y^{^{'}} = x - 2 y

(1, 4), (- 5, 2)

What is the solution for 6y^{''}+40y^'+54y=0 ?
The solution for 6y^{''}+40y^'+54y=0 is y=c_{1}e^{((-10+sqrt(19))t)/3}+c_{2}e^{-((10+sqrt(19))t)/3}