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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = \frac{e ^{- 2 x + 2 + c_{1}}}{x ^{3} - 3 x ^{2} + 3 x - 1}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{- 2 x + 2 + c_{1}}}{x ^{3} - 3 x ^{2} + 3 x - 1}

y = \frac{e ^{- 2 x + 2 + c_{1}}}{x ^{3} - 3 x ^{2} + 3 x - 1}

\frac{d}{d s} (ln (\frac{s - 1}{s ^{2} + 2 s + 5}))

f (x) = x^{2} - x + 2, at x = - 1

9 y^{^{''}} + 18 y^{^{'}} + 17 y = 0

n = 1 \sum \infty \frac{2 n}{5 n + 1}

\frac{\partial}{\partial y} (\frac{12 x}{x + y})

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