What is the solution for (x-1)y^'-2+3x+y=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-1)y^'-2+3x+y=0

(x - 1) y^{^{'}} - 2 + 3 x + y = 0

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

Solution steps

(x - 1) y^{'} (x) - 2 + 3 x + y = 0

Solve linear ODE:

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

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

\int_{0}^{1} \frac{1}{e ^{x} + e ^{- x}} d x

\frac{d}{d x} (- 3 ln (3 + 4 x^{2}) + 5 x^{3} - 9)

y = 6

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 10

\frac{d}{d x} (6 x^{6})

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