# Solving Differential Equations

1. Oct 20, 2010

### mrmonkah

1. The problem statement, all variables and given/known data
Solve the following: xdx = y$$^{2}$$dy

2. Relevant equations
Fundamental theorem of calculus - thanks gabbagabbahey

3. The attempt at a solution
$$\frac{dy}{dx}$$=$$\frac{x}{y^{2}}$$

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

=$$\frac{x^{2}}{2y^{2}}$$

So with the question, ive integrated both sides to find y^2 on the bottom, so should i rearrange to find y on the LHS and the x on the RHS?

Last edited: Oct 20, 2010
2. Oct 20, 2010

### gabbagabbahey

Unless NA is your abbreviation for the fundamental theorem of calculus, you are mistaken

What happens if you integrate both sides of the equation $x=y^2\frac{dy}{dx}$ with respect to $x$?

3. Oct 20, 2010

### mrmonkah

Hi gabbagabbahey,

Sorry, i am just getting to grips with the funky math features on the site, so my translation from paper to web isn't vry good. First of all i put in a monster mistake on the web, working on correcting this now.

4. Oct 20, 2010

### mrmonkah

Okay, more or less fixed now..... thanks again Gabb

5. Oct 20, 2010

### mrmonkah

Right, so carrying on from my first post,

if i rearrange for y, i get:

y = $$\sqrt{\frac{x^{2}}{2}}$$

6. Oct 20, 2010

### Char. Limit

$$x dx = y^2 dy$$

Right?

7. Oct 20, 2010

### mrmonkah

Ok, so if i integrate both sides i get:

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

and y = $$\sqrt[3]{\frac{3x^{2}}{2}}$$

Surely this isn't right is it? I don't recall coming across cubic roots in 'this particular' module. (I am simply looking for familiarity with the rest of the course)

8. Oct 20, 2010

### Char. Limit

You forgot the plus C.

9. Oct 20, 2010

### mrmonkah

Ahh okay fair enough Char.Limit, i am confused as to why re-arranging the initial equation (as i did earlier) yields such a different result?

10. Oct 20, 2010

### Char. Limit

It's mainly because you can't integrate x/y^2 dx, because y also depends on x.

So you can see why you couldn't integrate...

$$\int \frac{x}{y^2(x)} dx$$

11. Oct 20, 2010

### mrmonkah

Oh i see now. So with questions like these, i should generally keep all the y's on one side and the x's on the other? Ill attempt another question and post it to see if i have my head in the right place. Thank you Char.Limit, and as ever, you make a good point.

12. Oct 20, 2010

### Char. Limit

No problem. Glad I could help.

13. Oct 20, 2010

### mrmonkah

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

Solve the following: $$\frac{dy}{dx}$$ = $$\frac{1 + y}{1 + x}$$

2. Relevant equations
Fundamental theorem of calculus - thanks gabbagabbahey

3. The attempt at a solution

Re-arranging to get y terms and x terms on opposite sides:

$$\int$$$$\frac{dy}{1 + y}$$ = $$\int$$$$\frac{dx}{1 + x}$$

Which gives:

ln(x + 1) = ln(y + 1)

and so:

y = x

Does this work out?

14. Oct 20, 2010

### Char. Limit

Close, but you still need to remember the +C.

15. Oct 20, 2010

### mrmonkah

Damn, every time. A habit i need to get used to.

16. Oct 20, 2010

### Char. Limit

No real problem. You just have to make sure you do it.