What is the solution for x^2y^{''}+2xy^'-6y=0,y(1)=0 ?

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

x^{2} y^{^{''}} + 2 x y^{^{'}} - 6 y = 0, y (1) = 0

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (e^{5 x + 2 x^{2}})

\frac{d}{d x} (\frac{1}{2} arctan (\frac{x + 1}{2}))

\int sec (t) (sec (t) + tan (t)) d t

\int v^{- \frac{7}{8}} d v

y = 3 x - x^{2}, y = 0

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