1. Mar 2, 2015

### Math10

1. The problem statement, all variables and given/known data
Find a second solution y2 for x^2*y"+xy'-y=0; y1=x that isn't a constant multiple of the solution y1.

2. Relevant equations
None.

3. The attempt at a solution
Here's my work:
I divided by x^2,
y"+(1/x)y'-(1/x^2)y=0
P(x)=1/x and Q(x)=-1/x^2
Let y(x)=v(x)*x
y'(x)=v'(x)*x+v(x)
(1/x)y'(x)=v'(x)+v(x)/x
y"(x)=v'(x)+v"(x)*x+v'(x)=xv"(x)+2v'(x)
xv"(x)+2v'(x)+v'(x)+v(x)/x-v(x)*x/(x^2)=0
xv"(x)+3v'(x)=0
Let w=v'
w'=v"
xw'+3w=0
w'=-3w/x
dw/dx=-3w/x
dw/w=-3/x dx
integrate
ln abs(w)=-3ln abs(x)+C
Don't count C, the constant.
w=1/x^3
w=v'=1/x^3
dv/dx=1/x^2
don't count c, the constant.
v=-1/2x^2
y=v*x
y=-1/2x^2*x=-1/2x
But the answer in the book is y2=1/x. I got y=-1/2x, which answer is right?

2. Mar 2, 2015

### HallsofIvy

Staff Emeritus
Both are correct. If "y" is a solution to a "linear homogenous differential equation", which this is, then "Ay" is also a solution for any constant, A.

Last edited: Mar 3, 2015
3. Mar 2, 2015

### SammyS

Staff Emeritus
If you include a much needed set of parentheses in your solution, y = -1/(2x) , then both are correct.

Instead of throwing out C, use C = ln(2) .

4. Mar 2, 2015

### LCKurtz

What stops you from plugging the answers back in the DE and checking for yourself?

5. Mar 3, 2015

### Staff: Mentor

I agree. @Math10, at this point in your studies, any problem that you can check, you should check. You have already done the hard work of find the second solution. It's a simple matter to verify that what you have found actually is a solution.

6. Mar 3, 2015

### Math10

But when do I use C=ln(2)? After integrating dw/w=-3/x dx? But how did you get C=ln(2)?

7. Mar 3, 2015

### SammyS

Staff Emeritus
I assume that you are referring to my response somewhere above.

Notice that most of the previous respondents used the 'Reply' feature which includes a quote from the post they reference.

C is an arbitrary constant, Right?

Suppose you keep it, rather than discarding it.

You then have $\displaystyle e^{\,\ln|w|}=e^{\,\ln(|x|^{-3})+C}\$.

So that $\displaystyle w=\pm e^Cx^{-3}\$ .

Carry that on through.

Then, $\displaystyle v=\pm\frac{1}{2}e^Cx^{-3}+c_2$ .

In your solution, you let c2 = 0. That's fine.

You can choose C = ln(2), so that $e^C$ cancels the $\displaystyle \frac{1}{2}\$.

This together with choosing the - sign from the ± gives the book solution.