Solving by seperation of variables.

[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.

 

[arildno]

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

 

[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.

 

[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.

<br /> \frac{{dy}}{{dx}} = \frac{{y^3 }}{{x^2 }}<br />

<br /> \int {\frac{{dy}}{{y^3 }}} = \int {\frac{{dx}}{{x^2 }}} <br />

<br /> \int {y^{ - 3} } dy = \int {x^{ - 2} } dx<br />

<br /> - \frac{1}{{2y^2 }} = - \frac{1}{x} + c<br />

 

[DrKareem]

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

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