## Homework Statement

Find a second solution y2 for x^2*y"+xy'-y=0; y1=x that isn't a constant multiple of the solution y1.

None.

## 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?

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HallsofIvy
Homework Helper
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.

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SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Find a second solution y2 for x^2*y"+xy'-y=0; y1=x that isn't a constant multiple of the solution y1.

None.

## 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?
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) .

LCKurtz
Homework Helper
Gold Member
But the answer in the book is y2=1/x. I got y=-1/2x, which answer is right?
What stops you from plugging the answers back in the DE and checking for yourself?

Mark44
Mentor
What stops you from plugging the answers back in the DE and checking for yourself?
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.

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
But when do I use C=ln(2)? After integrating dw/w=-3/x dx? But how did you get C=ln(2)?
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.

...
integrate
ln abs(w)=-3ln abs(x)+C
Don't count C, the constant.
w=1/x^3
...
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.