What is the solution for y^{''}-(6/x)y^'-(8/(x^2))y=0 ?

The solution for y^{''}-(6/x)y^'-(8/(x^2))y=0 is y=c_{1}x^8+(c_{2})/x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-(6/x)y^'-(8/(x^2))y=0

y^{^{''}} - (\frac{6}{x}) y^{^{'}} - (\frac{8}{x ^{2}}) y = 0

y = c_{1} x^{8} + \frac{c _{2}}{x}

Solution steps

y^{''} (x) - (\frac{6}{x}) y^{'} (x) - (\frac{8}{x ^{2}}) y = 0

Solve linear ODE:

y = c_{1} x^{8} + \frac{c _{2}}{x}

y = c_{1} x^{8} + \frac{c _{2}}{x}

What is the solution for y^{''}-(6/x)y^'-(8/(x^2))y=0 ?
The solution for y^{''}-(6/x)y^'-(8/(x^2))y=0 is y=c_{1}x^8+(c_{2})/x