What is the solution for 3y^{''}+3y^'-y=0 ?

The solution for 3y^{''}+3y^'-y=0 is y=c_{1}e^{((-3+sqrt(21))t)/6}+c_{2}e^{((-3-sqrt(21))t)/6}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3y^{''}+3y^'-y=0

3 y^{^{''}} + 3 y^{^{'}} - y = 0

y = c_{1} e^{\frac{( - 3 + 21 ) t}{6}} + c_{2} e^{\frac{( - 3 - 21 ) t}{6}}

Solution steps

3 y^{''} (t) + 3 y^{'} (t) - y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 21 ) t}{6}} + c_{2} e^{\frac{( - 3 - 21 ) t}{6}}

y = c_{1} e^{\frac{( - 3 + 21 ) t}{6}} + c_{2} e^{\frac{( - 3 - 21 ) t}{6}}

x^{3} y^{2} = x y^{3} + 6, (2, 1)

y = 5 x + ln (3 x)

f^{^{'}} (x) = 4 x + 2

\int x^{2} x^{2} + 3 d x

\frac{\partial}{\partial x} (\frac{5 x}{( 3 + y )})

What is the solution for 3y^{''}+3y^'-y=0 ?
The solution for 3y^{''}+3y^'-y=0 is y=c_{1}e^{((-3+sqrt(21))t)/6}+c_{2}e^{((-3-sqrt(21))t)/6}