# Is this the correct answer for a partial differentiation question?

1. Jul 9, 2008

### laura_a

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

Find the general solution of

y' + (2/x)y = 3/(x^2)

3. The attempt at a solution

xy' + 2y = 3/x
d/dx (x * 2y) = 3/x
integrating both sides (using product rule for LHS) I end up with

y= (3lnx + C)/2x

Then I am supposed to find the solution for which y(2) = -1 so I subbed that into the above and got c = -4/3 - ln2

So then I can say that y = 3lnx/(2x) - 4/3 -ln2

Hope thats right!
Thanks guys!

2. Jul 9, 2008

### vipulsilwal

u shud have repeated that
(x^2)y' + 2xy = 3
d/dx ((x ^ 2)*y) = 3
integrating both sides

y(x^2)+c=3x
this is the general solution now u can find c by given values of x and y and not supposing.

3. Jul 9, 2008

### HallsofIvy

Staff Emeritus
In what sense is this a partial differential equation?

These are NOT the same. differentiating (x*2y) gives 2xy'+ 2y. there is no "2" multipying xy' in the equation.

There is no "product rule" for integration. You are simply using the "fundamental theorem of calculus, that the integral of f ' is f.

If y= (3lnx+ C)/2x, then y'= [2x(3/x)- 2(3lnx+C)]/4x2
Then xy'+ 2y= [6- 6lnx+2C]/4x+ [6lnx+ 2C/4x= 3/2x, not 3/x.

As I pointed out d/dx[2xy] is NOT equal to xy'+ 2y. There is NO function of x and y whose derivative with respect to x is xy'+ 2y, it is not an exact derivative. However, multiplying the entire equation by the "integrating factor", x, wil make it exact.

Multiplying the equation xy'+ 2y= 3/x by x gives x2y'+ 2xy= 3. The left side is equal to d[x2y]/dx so the equation is the same as d(x2y)/dx= 3 and integrating both sides of that x2y= 3x+ C.

Last edited: Jul 9, 2008
4. Jul 9, 2008

### laura_a

THanks guys, I know I typed partial but it is not what I meant... I've read about 16 text books today I think and I've made myself more confused.... I understand the part that I missed now and I didn't even think to differentiate and check if my answer was right, I was more surprised that I got an answer... I'm studying very hard mathematics in order to teach high school maths and I'm struggling... thanks heaps for all your info