feminizmiki

2021-12-30

Solve the following differential equations: $dy-{\left(y-1\right)}^{2}dx=0$

### Answer & Explanation

Steve Hirano

$dy-{\left(y-1\right)}^{2}\cdot dx=0$
$⇒dy={\left(y-1\right)}^{2}\cdot dx$
$⇒\frac{dy}{{\left(y-1\right)}^{2}}=dx$
$⇒\int {\left(y-1\right)}^{-2}\cdot dy=\int dx$
$⇒\frac{{\left(y-1\right)}^{-2+1}}{-2+1}=x+c$
$⇒-{\left(y-1\right)}^{-1}=x+c$
$⇒\frac{-1}{\left(y-1\right)}=x+c⇒y-1=\frac{-1}{x+c}$
Answer: $⇒y=\frac{-1}{x+c}+1$

Terry Ray

$dy-{\left(y-1\right)}^{2}dx=0$
$dy={\left(y-1\right)}^{2}dx$
$\frac{1}{{\left(y-1\right)}^{2}}dy=dx$
${\left(y-1\right)}^{-2}dy=dx$
$\int {\left(y-1\right)}^{-2}dy=\int dx$
$\frac{{\left(y-1\right)}^{-1}}{-1}=x+C$
$-\frac{1}{\left(y-1\right)}=x+C$
$\frac{1}{\left(y-1\right)}=-\left(x+C\right)$
$y-1=-\frac{1}{x+C}$
$y=1-\frac{1}{x+C}$
$y=\frac{x-1+C}{x+C}$

karton

$\begin{array}{}dy-\left(y-1{\right)}^{2}=dx\\ ⇒dy=\left(y-1{\right)}^{2}dx\\ ⇒\frac{dy}{\left(y-1{\right)}^{2}}=dx\\ \text{Integrating both sides}\\ -\frac{1}{y-1}=x+C\\ ⇒\frac{1}{y-1}=-x-C\\ ⇒y-1=-\frac{1}{x+C}\\ ⇒y=1-\frac{1}{x+C}\end{array}$

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