How Do You Solve the Isobaric Equation (x+y)dx - (x-y)dy = 0?

[Mark Brewer]

Homework Statement

Find a one-parameter family of solutions to the isobaric equation

(x+y)dx - (x-y)dy = 0

The Attempt at a Solution

First I subtracted -(x-y) to the both sides
Step 1: (x+y)dx = (x-y dy

I then distributed dx into (x+y) and dy into (x-y) to get
Step 2: xdx +ydx = xdy -ydy

I then divided the y on the left side of the equal sign to both sides
Step 3: x/ydx + dx = x/ydy + dy

Step 4: using 1/u = x/y

I get
Step 5: 1/u dx + dx = 1/u dy - dy

I then factored out dx and dy to get
Step 6: (1/u +1)dx = (1/u - 1)dy

I then divided (1/u - 1) to both sides and dx to both sides to get
Step 7: (1/u +1) / (1/u - 1) = dy/dx

using the above 1/u = x/y and rearranging to the form y = xu and taking derivative of y in respect to x I get
Step 8: dy/dx = (du/dx)x + u

I then used dy/dx in step 8 to plug into step 7 dy/dx to get
Step 9: (1/u +1) / (1/u - 1) = (du/dx)x + u

I then subtracted the u on the right side of the equal sign to both sides to get
Step 10: (-u+u^3) / 1 - u = (du/dx)x

I then used separable equations to get
Step 11: (1/x)dx = (1-u) / (-u + u^3)du

This is where I stopped because I don't like (u cubed on the denominator).

Any help would be fantastic. I know my next steps will be integration but I feel there's another step that needs to be taken.

 

[Mark44]

Homework Statement

Find a one-parameter family of solutions to the isobaric equation

(x+y)dx - (x-y)dy = 0

The Attempt at a Solution

First I subtracted -(x-y) to the both sides
Step 1: (x+y)dx = (x-y dy

I then distributed dx into (x+y) and dy into (x-y) to get
Step 2: xdx +ydx = xdy -ydy

I then divided the y on the left side of the equal sign to both sides
Step 3: x/ydx + dx = x/ydy + dy

Step 4: using 1/u = x/y

Or u = y/x, which is a lot simpler. Use this substitution to get a separable differential equation.

I get
Step 5: 1/u dx + dx = 1/u dy - dy

I then factored out dx and dy to get
Step 6: (1/u +1)dx = (1/u - 1)dy

I then divided (1/u - 1) to both sides and dx to both sides to get
Step 7: (1/u +1) / (1/u - 1) = dy/dx

using the above 1/u = x/y and rearranging to the form y = xu and taking derivative of y in respect to x I get
Step 8: dy/dx = (du/dx)x + u

I then used dy/dx in step 8 to plug into step 7 dy/dx to get
Step 9: (1/u +1) / (1/u - 1) = (du/dx)x + u

I then subtracted the u on the right side of the equal sign to both sides to get
Step 10: (-u+u^3) / 1 - u = (du/dx)x

I then used separable equations to get
Step 11: (1/x)dx = (1-u) / (-u + u^3)du

This is where I stopped because I don't like (u cubed on the denominator).

Any help would be fantastic. I know my next steps will be integration but I feel there's another step that needs to be taken.

 

[Mark Brewer]

Or u = y/x, which is a lot simpler. Use this substitution to get a separable differential equation.

Thank you, Mark.

I'm going to work the problem from the beginning so I can use u = y/x.

Mark

 

[Mark Brewer]

Hi Mark,

Thank you again. I successfully finished the problem. My answer determined was arctan(1 + y²/x²) - 1/2ln|1 + y²/x²| = ln|x| + c

 

[Sammbuch]

should be
arctan(y/x) - 1/2 ln(1 + y^2/x^2) = ln(x) + c
http://www.wolframalpha.com/input/?i=solve+(x+y)dx-(x-y)dy=0The equation ##x+2=3## implies $$x=1.$$

 

[Sammbuch]

can someone explain to me how to post latex?
[tex]\int x^2 dx [/tex]

 

[Mark44]

should be
arctan(y/x) - 1/2 ln(1 + y^2/x^2) = ln(x) + c
http://www.wolframalpha.com/input/?i=solve+(x+y)dx-(x-y)dy=0

I didn't check the OP's answer. When you solve a DE, it's always a good idea to check that your solution satisifies the original diff. equation.

The equation ##x+2=3## implies $$x=1.$$

Does the above have anything to do with this problem?

can someone explain to me how to post latex?
[tex]\int x^2 dx [/tex]

See https://www.physicsforums.com/help/latexhelp/ under INFO --> Help/How-To, in the banner at the top of the page.[/QUOTE]