What is the solution for x^2y^{''}+5xy^'+3y=0 ?

The solution for x^2y^{''}+5xy^'+3y=0 is y=(c_{1})/x+(c_{2})/(x^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^2y^{''}+5xy^'+3y=0

x^{2} y^{^{''}} + 5 x y^{^{'}} + 3 y = 0

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

Solution steps

x^{2} y^{''} (x) + 5 x y^{'} (x) + 3 y = 0

Solve linear ODE:

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

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

d er i v a t i v e \frac{7 + 2 x}{9 - 8 x}

\int 15 e^{- 5 x} d x

d u = \frac{1}{x} d x

\int (12 e^{x} - 5 x^{- 2}) d x

\frac{d}{d x} ((x^{2} - \frac{1}{x ^{2}})^{6})

What is the solution for x^2y^{''}+5xy^'+3y=0 ?
The solution for x^2y^{''}+5xy^'+3y=0 is y=(c_{1})/x+(c_{2})/(x^3)