What is the solution for y^{''}+5y^'+3y=0 ?

The solution for y^{''}+5y^'+3y=0 is y=c_{1}e^{((-5+sqrt(13))t)/2}+c_{2}e^{((-5-sqrt(13))t)/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+5y^'+3y=0

y^{^{''}} + 5 y^{^{'}} + 3 y = 0

y = c_{1} e^{\frac{( - 5 + 13 ) t}{2}} + c_{2} e^{\frac{( - 5 - 13 ) t}{2}}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 5 + 13 ) t}{2}} + c_{2} e^{\frac{( - 5 - 13 ) t}{2}}

y = c_{1} e^{\frac{( - 5 + 13 ) t}{2}} + c_{2} e^{\frac{( - 5 - 13 ) t}{2}}

(6 x + 1) y^{2} \frac{d y}{d x} + 3 x^{2} + 2 y^{3} = 0

x \to - 1 + lim (3 x + 3 - 5)

y = cos (3 x^{2})

x \to 3 + lim (\frac{3 x + 4}{x - 3})

\frac{\partial}{\partial x} (\frac{( x + y )}{( 1 + x ^{2} )})

What is the solution for y^{''}+5y^'+3y=0 ?
The solution for y^{''}+5y^'+3y=0 is y=c_{1}e^{((-5+sqrt(13))t)/2}+c_{2}e^{((-5-sqrt(13))t)/2}