# Differential Equations Help

1. Aug 23, 2011

### rpgkevin

1. The problem statement, all variables and given/known data
Solve The following Equations:
2(y+3)dx-xydy=0

(x2-xy+y2)dx - xydy=0 use following assumption y=vx

xy3+ex2dy=0

3. The attempt at a solution

I am still a novice at diff eqs but here is what I got on the first one:
After seperating it I ended up with
(dx/x)=(ydy)/(2y+6) Then I get stuck with integrating the side with the Y

For the other two I believe they can not be separated and I am not sure what to do when this is the case

2. Aug 23, 2011

### rock.freak667

For the right side, you can rewrite it as

y/2(y+3) dy or ½(y+3-3)/(y+3), you can simply it even further i.e. polynomial division

3. Aug 23, 2011

### vela

Staff Emeritus
For the y-integral on the first one, you can do this:$$\frac{1}{2}\int \frac{y}{y+3}\,dy = \frac{1}{2}\int \frac{(y+3)-3}{y+3}\,dy = \frac{1}{2}\int \left[1-\frac{3}{y+3}\right]\,dy$$or you could use the substitution u=y+3.

On the second one, what did you get when you used the substitution y=vx?

4. Aug 23, 2011

### rpgkevin

I have never used the substitution method I have no clue how to use that I looked it up earlier because someone told me that but I was unable to use the examples to work that one out

5. Aug 23, 2011

### Staff: Mentor

I don't think you wrote this correctly - there seems to be a dx missing.

6. Aug 23, 2011

### vela

Staff Emeritus
If you differentiate y=vx with respect to x, you'll get
$$\frac{dy}{dx} = v + x\frac{dv}{dx}$$
Multiplying through by dx, you end up with
$$dy = v \,dx + x\, dv$$
Use this and the original substitution to eliminate y from the original equation. You should find it separates then, allowing you to solve for v, from which you can then find y.

7. Aug 23, 2011

### rpgkevin

ahh you are correct it is suppose to be a dx after the xy3

8. Aug 23, 2011

### rpgkevin

I tried what you said and plugged stuff back in and then I Tried separating things out and I cant seem to get it to separate out I am stuck at
X2(1-V-V2)dx=x2v2+(x3v)dv/dx)

9. Aug 24, 2011

### vela

Staff Emeritus
Please show your work. It's impossible to see what went wrong without seeing what you actually did.

10. Aug 24, 2011

### HallsofIvy

Staff Emeritus
For (2) you are told to let y= vx and from that dy= vdx+ xdv. Replace y and dy in the equation with those. It will reduce to a separable equation.

11. Aug 24, 2011

### vela

Staff Emeritus
In that case, it's pretty straightforward to see the equation separates. Why do you think it can't be separated?

12. Aug 24, 2011

### Staff: Mentor

Also, when the variables are separated, the integration is not very difficult.

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