# Linear non-exact differential equation made exact

1. Feb 9, 2014

### rsera

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

Find the general solution:
(y+1) dx + (4x - y) dy = 0

2. Relevant equations

dy/dx + P(x)y = Q(x) (standard form)
e^(∫ P(x) dx) (integrating factor)

3. The attempt at a solution

This exercise is in the chapter on linear equations, making non-exact equations exact.

So I know I need to put it into the form:
dy/dx + P(x)y = Q(x)

So that I can find the Integrating Factor:
e^( ∫ P(x) dx)

Then multiply through by the Integrating Factor and solve the resulting exact equation.

My problem with this one has been getting it into that standard form.

If I divide by dx and rearrange, I get:

(4x - y) dy/dx + (y+1) = 0

Then divide by (4x - y)

dy/dx + (y + 1) / (4x - y) = 0

However, this does not match the standard form, because it is not a function of x times y.

I know what to do once it's in standard form, I'm just having troubles getting to that point. Any clue as to what trick to use to get it to standard form would be appreciated. Thanks!

Last edited: Feb 9, 2014
2. Feb 9, 2014

### SammyS

Staff Emeritus
Hello rsera. Welcome to PF !

How about if you switch the roles of x & y? In other words, treat y as the independent variable and consider x to be a function of y .

3. Feb 9, 2014

### rsera

Thank you for the welcome and the reply!

You know, I had thought about doing that after I posted, but wasn't sure if that was the right way to approach it.

So, if I switch them, I get the standard form as:
dx/dy + [4/(y+1)] * x = y / (y+1)

So my integrating factor is:
e^∫P(y) dy ; P(y) = 4 / (y+1)
e^ ∫4/(y+1) dy = e^ 4ln|y+1| = (y+1)^4

I can multiply through by (y+1)^4 and get:
(y+1)^4 * dx/dy + (y+1)^4 [4/(y+1)] * x = (y+1)^4 * (y / (y+1))

Which simplifies to:
(y+1)^4 * dx/dy + (y+1)^4 [4/(y+1)] * x = y(y+1)^3

And then integrate both sides to get:
(y+1)^4 * x = ∫y(y+1)^3 dy

Am I on the right track here? I tried integrating y(y+1)^3 using integration by parts, but I didn't get an answer that looked anything like the one in the book. If I'm correct up to here, can you suggest a better method for integrating it?

4. Feb 9, 2014

### SammyS

Staff Emeritus
That looks good.

What did you get for the integral?

5. Feb 9, 2014

### rsera

The book has:
20x = 4y-1 + c(y+1)^-4

After working with it some more, I got:
20x = 5y-1 + c(y+1)^-4

For ∫y(y+1)^3 dy
I used u = y and dv = (y+1)^3; so, du = 1 dy and v = 1/4(y+1)^4
So I got:
∫u dv = y * 1/4 (y+1)^4 - 1/20 (y+1)^5 + c

Back to the DE, we have:
(y+1)^4 * x = (y)*(1/4)*(y+1)^4 - (1/20)*(y+1)^5 + c

I multiplied through by 20 to eliminate the fraction:
20*x*(y+1)^4 = 5*y*(y+1)^4 - (y+1)^5 + c

And factored out a (y+1)^4:
20*x*(y+1)^4 = (y+1)^4 (5y-1) +c

Finally, dividing by (y+1)^4 gives me:
20x = 5y-1 + c(y+1)^-4

So now my answer has a 5 where the answer in the book has a 4.

I really appreciate you looking over my work! It's nice to have someone to talk to about this. At my community college, the tutors don't go as high as DE, so getting help has been a real challenge.

6. Feb 9, 2014

### SammyS

Staff Emeritus
Check the algebra in that last step.
I get

20x(y+1)4 = (y+1)4 (5y - (y+1)) + c

7. Feb 9, 2014

### rsera

Yes! I fixed that and now my answer matches theirs! Thank you so much!