# Differential Equation

1. Sep 30, 2009

### KillerZ

1. The problem statement, all variables and given/known data

Find a solution of $$x\frac{dy}{dx} = y^{2} - y$$ that passes through the indicated points.

a) (0,1)
b) (0,0)
c) (1/2 , 1/2)
d) (2, 1/4)

2. Relevant equations

$$x\frac{dy}{dx} = y^{2} - y$$

3. The attempt at a solution

$$x\frac{dy}{dx} = y^{2} - y$$

$$\frac{dy}{y^{2} - y} = \frac{dx}{x}$$

$$\int\frac{dy}{y^{2} - y} = \int\frac{dx}{x}$$

$$\int\frac{dy}{y(y - 1)} = ln|x| + c$$

I used partial fractions to solve the left:

$$\int\left(\frac{1}{y - 1} - \frac{1}{y}\right)dy = ln|x| + c$$

$$\int\frac{1}{y - 1}dy - \int\frac{1}{y}dy = ln|x| + c$$

$$\int\frac{1}{y - 1}dy - ln|y| = ln|x| + c$$

u = y - 1
du = dy

$$\int\frac{1}{u}du - ln|y| = ln|x| + c$$

$$ln|u| - ln|y| = ln|x| + c$$

$$ln|y - 1| - ln|y| = ln|x| + c$$

$$ln|\frac{y - 1}{y}| = ln|x| + c$$

$$\frac{y - 1}{y} = e^{ln|x| + c} = (e^{ln|x|})(e^{c})$$

$$\frac{y - 1}{y} = |x|(e^{c})$$

$$\frac{y - 1}{y} = \pm(e^{c})(x)$$

$$\frac{y - 1}{y} = c(x)$$

Now this is where I am confused:

a) (0, 1)

$$\frac{1 - 1}{1} = c(0)$$

$$0 = 0$$

Does this mean c = 0 or does it mean that there is no solution through the point (0, 1) or that there is a solution through the point as left = right?

2. Sep 30, 2009

### tiny-tim

Hi KillerZ!
Nooo

ln|(y-1)/xy| = c,

so … ?

3. Sep 30, 2009

### KillerZ

I do not understand where the xy in ln|(y-1)/xy| came from.

4. Sep 30, 2009

### tiny-tim

lnA - lnB = ln(A/B)

5. Sep 30, 2009

### KillerZ

So.

$$ln|y - 1| - ln|y| = ln|x| + c$$

$$ln|y - 1| - ln|y| - ln|x| = c$$

becomes ?

$$ln|\frac{y - 1}{yx}| = c$$

6. Sep 30, 2009

### tiny-tim

:tongue2: go on …

7. Sep 30, 2009

### KillerZ

$$ln|\frac{y - 1}{yx}| = c$$

$$\frac{y - 1}{yx} = e^{c}$$

replace e^c with c as it is just a constant

$$\frac{y - 1}{yx} = c$$

8. Sep 30, 2009

### tiny-tim

ah, looking back, I see you got there anyway …

ok, you were puzzled about getting it to go through (0,1).

You have c|x| = |(y-1)/y|, so for any non-zero value of c, limy->1|x| = 0 …

they all go through (0,1) !

9. Sep 30, 2009

### KillerZ

ok thank you.