# Solving by seperation of variables.

1. Nov 20, 2004

### misogynisticfeminist

I've got a few 1st order ODEs which I have problems solving. I am new to the subject and self-taught so I may have a little difficulty absorbing. The question is.......

1. $$\frac {dy}{dx} = \frac {y^3}{x^2}$$

for 1. I put it in the form,

$$x^2 dy = y^3 dx$$

$$\frac {dy}{y^3} = \frac {dx}{x^2}$$

But I get stuck when i integrate both sides. I have not integrated dy/ y^n functions before.

2. Nov 20, 2004

### arildno

For ALL numbers except r=-1, we have:
$$\int{t}^{r}dt=\frac{1}{r+1}t^{r+1}+C$$

3. Nov 20, 2004

### misogynisticfeminist

hmmm, I'm quite familiar with the formula. But, usually when we have a

$$\frac {dy}{y}$$ function, we usually get $$ln y +c$$ for the integral. So, i was wondering if there is anything remotely close to ln which I can use. I can actually integrate $$x^3$$ but I don't know what to do with the dy on top.

thanks.

4. Nov 21, 2004

### Benny

When you have the dy on top you just need to rewrite the power of the y by putting a negative before the three. This is because dy/(x³) is just the same as (x^-3)dy.

$$\frac{{dy}}{{dx}} = \frac{{y^3 }}{{x^2 }}$$

$$\int {\frac{{dy}}{{y^3 }}} = \int {\frac{{dx}}{{x^2 }}}$$

$$\int {y^{ - 3} } dy = \int {x^{ - 2} } dx$$

$$- \frac{1}{{2y^2 }} = - \frac{1}{x} + c$$

Last edited: Nov 21, 2004
5. Nov 21, 2004

### DrKareem

That's is true. When you have r=-1, then the answer would be [tex]ln x [\tex]