What is the solution for y^{''}-4sqrt(3)y^'+12y=0 ?

The solution for y^{''}-4sqrt(3)y^'+12y=0 is y=c_{1}e^{2sqrt(3)t}+c_{2}te^{2sqrt(3)t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4sqrt(3)y^'+12y=0

y^{^{''}} - 43 y^{^{'}} + 12 y = 0

y = c_{1} e^{23 t} + c_{2} t e^{23 t}

Solution steps

y^{''} (t) - 43 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = c_{1} e^{23 t} + c_{2} t e^{23 t}

y = c_{1} e^{23 t} + c_{2} t e^{23 t}

\frac{d}{d x} (\frac{1}{3} ln (e^{x} + 1))

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

f (x) = - x^{2} sin (x) + 4 x cos (x) + 2 sin (x)

\frac{d}{d x} (\frac{3}{x ^{4}})

\frac{\partial}{\partial x} (7 - x^{5} - y^{3})

What is the solution for y^{''}-4sqrt(3)y^'+12y=0 ?
The solution for y^{''}-4sqrt(3)y^'+12y=0 is y=c_{1}e^{2sqrt(3)t}+c_{2}te^{2sqrt(3)t}