Differential Equations: Solving Non-separable Equations

[rpgkevin]

Homework Statement

Solve The following Equations:
2(y+3)dx-xydy=0

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

xy³+e^x2dy=0

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

 

[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

 

[vela]

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?

 

[rpgkevin]

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?

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

 

[Mark44]

xy³+e^x2dy=0

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

 

[vela]

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

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.

 

[rpgkevin]

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

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

 

[rpgkevin]

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.

I tried what you said and plugged stuff back in and then I Tried separating things out and I can't seem to get it to separate out I am stuck at
X²(1-V-V²)dx=x²v²+(x³v)dv/dx)

 

[vela]

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

 

[HallsofIvy]

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.

 

[vela]

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

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

 

[Mark44]

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