What is the solution for x^2y^{''}+6xy^'+6y=0 ?

The solution for x^2y^{''}+6xy^'+6y=0 is y=(c_{1})/(x^2)+(c_{2})/(x^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^{''}+6xy^'+6y=0

x^{2} y^{^{''}} + 6 x y^{^{'}} + 6 y = 0

y = \frac{c _{1}}{x ^{2}} + \frac{c _{2}}{x ^{3}}

Solution steps

x^{2} y^{''} (x) + 6 x y^{'} (x) + 6 y = 0

Solve linear ODE:

y = \frac{c _{1}}{x ^{2}} + \frac{c _{2}}{x ^{3}}

y = \frac{c _{1}}{x ^{2}} + \frac{c _{2}}{x ^{3}}

\int_{1}^{\infty} e^{- x} d x

\frac{\partial}{\partial y} (y^{2} - cos (x) y)

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

\frac{d}{d x} (∣ x ∣ + ∣ x - 1 ∣)

\frac{d}{d x} (x tan (2 x))

What is the solution for x^2y^{''}+6xy^'+6y=0 ?
The solution for x^2y^{''}+6xy^'+6y=0 is y=(c_{1})/(x^2)+(c_{2})/(x^3)