# Homework Help: Second order diff eq with two variables

1. Mar 8, 2007

### glid02

Here's the problem:
x^2y''-3xy'-12y=0

with initial conditions y(1)=0 and y'(1)=7

I'm supposed to solve for y in the form y=c1y1+c2y2

y1 = x^6 by inspection

Now to solve for y2

y2=y1v

v can be solved for by the equation
y1v''+(2y1'+py1)v'=0
where p is the function in front of y' when there is no function in front of y'':
x^2y''-3xy'-12y=0 divide by x^2
y''-3/xy'-12/x^2y=0
so p=-3/x

x^6v''+(2*6x^5+(-3/x)x^6)v'=0
x^6v''+(12x^5-3x^5)v'=0
x^6v''=-9x^5v'
so v'=e^(-x^2/18)
and v= -x/9(e^(-x^2/18)+c

so y2=x^6(-x/9(e^(-x^2/18)))
y2=-x^(7)/9*e^(-x^2/18)

Now y=c1y1+c2y2 and my initial conditions are y(1)=0 and y'(1)=7

c1*x^6+c2*-x^(7)/9*e^((-x^2)/18)
y'=c1y1'+c2y2'
= c1*6*x^5+
c2*((-7/9*x^6*e^((-x^2)/18)+((-x^7)/9*(-x/9)*e^((-x^2)/18)

Subbing in 1 for x for both equations I get
c1+c2*-1/9*e^(-1/18)
and
6c1+c2*(-7/9*e^(-1/18)+(-1/9*-1/9*e^(-1/18))
=
6c1+c2*(-63/81*e^(-1/18)-1/81*e^(-1/18)
=
6c1-64/81*e^(-1/18)c2

I solved for c1 in the first equation:
c1=c2*1/9*e^(-1/18)
and plugged that into the second equation
6*c2*1/9*e^(-1/18)-64/81*e^(-1/18)*c2=7
c2*e^(-1/18)*(54/81-64/81)=7
c2*e^(-1/18)*-10/81=7
c2=-81/10*7*e^(1/18)

so c1=(-81/10*7*e^(1/18))*1/9*e^(-1/18)
c1=-567/10*e^(1/18)*1/9*e^(-1/18)
=-567/90

Now I plug those back in the equation c1y1+c2y2=y and that's not the right answer. There's a good chance I messed up in there somehow, although I've double checked everything. Does anyone see what I did wrong, or better yet, an easier way to solve this problem. This way is just ridiculous.

Thanks a lot.

2. Mar 8, 2007

### Dick

If the first solution is a power law (x^6) it's a good guess that there may be another power law solution. Put y=x^n and solve for n.

3. Mar 8, 2007

### glid02

I think the two different solutions have to be independent of each other - or not be multiples of each other. I'll give it a shot and see what happens though. Thanks.

4. Mar 8, 2007

### glid02

wow, yeah, that's all I had to do. Thanks a lot.

5. Mar 8, 2007

### Dick

x^n and x^m where n and m are different are independent.

6. Mar 9, 2007

### HallsofIvy

In other words, with u= v', u'= (-9/x)u
du/u= -9dx/x so ln(u)= -9 ln x+ C, u= v'= Cx-9
v= Cx-8 and so vx6= x-2
NOT what you give:

7. Mar 9, 2007

### D H

Staff Emeritus
You were too quick on the draw with "y1 = x^6 by inspection".
An easy way to solve this is to assume each $y_i$ is of the form $y_i=x^n$. Then

$$x^2y_i^{\prime\prime}-3xy_i^{\prime}-12y_i = n(n-1)x^2x^{n-2} -3nxx^{n-1}-12x^n = (n(n-1)-3n-12)x^n$$

The characteristic polynomial

$$n(n-1)-3n-12 = n^2-4n-12$$

has zeros $n=6$ and $n=-2$. Thus

$$y=c_1x^6 + c_2x^{-2}$$

Applying the given conditions $y(1)=0$, $y'(1)=7$ to the above leads to a unique solution for $c_1$ and $c_2$.

8. Mar 9, 2007

### HallsofIvy

Yes, that's the simplest way to do it, but I suspect he was told to find a second solution by reducing the order- and that "x6 is a solution by inspection" was in the book!