# Differential Equations, Subsitution Methods

1. Sep 16, 2009

### foobag

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

1. x(x+y)y' = y(x-y)

2.y' = (4x + y)^2

3. (x+y)y' = 1

2. Relevant equations

I have tried using the substitution method v =, the homogeneous method as well.

3. The attempt at a solution

For 1. i factored out the xy and got 1/xy y' = integral ( (1-y/x)/(1+(x/y)) but cant take it further to the final answer ln(xy) = (x/y) + C

For 2. i tried sub v = 4x+y, but didnt even come close to the answer

For 3. I tried homogenous equations making it (1+y/x)y' = 1/x, but i get stuck with 1/x^2 later on which i cant factor to make it a simple first order linear diff...

2. Sep 17, 2009

### Staff: Mentor

How did you factor out xy? I don't see that xy is a factor, so I'm puzzled about how you were able to factor it out.
Well, I tried the same substitution in #2 and was able to get a separable differential equation. Show me what you tried so I can see where you went wrong.
Why not try a substitution u = x + y? If you do it correctly, you should get a separable differential equation.

3. Sep 17, 2009

### foobag

ok so i redid problems 2 & 3 and they worked for me...

as for problem 1...

I can factor down to y' = (y/x) ((1-y/x)/(1+y/x))

but then i get the left side after using v = y/x to be (1/2)(1/v - lnv) = lnx + c but then i cant reduce it down to the answer in the back that is: ln(xy) = C + x/y....

I did: y' = y(x-y)/x(x+y)

then: y'/y = (x-y)/(x(x+y))

then: factor out x from top of right side... y'/y = x(1-y/x)/(x(x+y)), so x's cancel

i get: y'/y = (1-y/x)/(x+y) ok so then factor out x from bottom: y'/y = (1-y/x)/(x(1+y/x), then multiply y back....

so: y' = (y/x)((1-y/x)/(1+y/x) let me know if i made a mistake here

4. Sep 18, 2009

### Staff: Mentor

So far, so good.
Now multiply both sides by 2 to get
1/v - ln v = 2lnx + 2c = ln x2 + 2c
so 1/v = ln v + ln x2 + 2c = ln[v x2] + 2c
so x/y = ln(y/x * x2) + 2c = ln(xy) + C (where C = 2c)
That's essentially what your book shows for the answer. The answer also seems to assume that x > 0 and y > 0, otherwise you would have some absolute values after integrating and getting the ln terms.