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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