2021-12-22

Find a solution of $x\frac{dy}{dx}={y}^{2}-y$ that passes through the indicated points.
a. $\left(0,0\right)$
b. $\left(\frac{1}{4},\frac{1}{4}\right)$
c. $\left(6,\frac{1}{8}\right)$

Melissa Moore

Expert

By separing variables in $x\frac{dy}{dx}={y}^{2}-y$; we have
$\frac{dy}{{y}^{2}-y}=\frac{dx}{x}$
$\frac{dy}{y\left(y-1\right)}=\frac{dx}{x}$
$\left[\frac{-1}{y}\right]+\frac{1}{y-1}\right]dy=\frac{dx}{x}$
On integration, we obtain
$-\mathrm{ln}\left(y\right)+\mathrm{ln}\left(y-1\right)=\mathrm{ln}\left(x\right)+\mathrm{ln}c$
$\mathrm{ln}\left(\frac{y-1}{y}\right)=\mathrm{ln}\left(xc\right)$
$\frac{y-1}{y}=xc$
$y-1=xyc$
$y-xyc$
$y\left(1-xc\right)=1$
$y=\frac{1}{1-xc}$
$y=\frac{1}{1-cx}$ is a solution and $y=0$ is another constant solution
a) At $\left(0,0\right)$
$y=\frac{1}{1-xc}$
$0=\frac{1}{1-0}$
$0=1$
This is not possible
Therefore in this case there is no solution.

karton

Expert

c) At $\left(6,\frac{1}{8}\right)$
Here, $x=6,y=\frac{1}{8}$
$\frac{1}{8}=\frac{1}{1-6c}$
$1-6c=8$
Substitute $c=\frac{-7}{6}$ in $y=\frac{1}{1-cx}$
Thus,
$y=\frac{1}{1+\frac{7}{6}x}$
$=\frac{6}{6+7x}$
The solution of the initial value problem is $y=\frac{6}{6+7x}$

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