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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-(x/(x-1))y^'+1/(x-1)y=0

y^{^{''}} - (\frac{x}{x - 1}) y^{^{'}} + \frac{1}{x - 1} y = 0

y = e^{x + c_{1} - 1} + c_{2} x

Solution steps

y^{''} (x) - (\frac{x}{x - 1}) y^{'} (x) + \frac{1}{x - 1} y = 0

Solve using substitution:

y = e^{x + c_{1} - 1} + c_{2} x

y = e^{x + c_{1} - 1} + c_{2} x

x \to 2 lim (3 x^{2} + 2 x - 16)

\frac{\partial}{\partial x} (x - 2 y^{3})

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

(1, 1), (6, - 2)

\int x d t

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