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]