2nd order ODE reduction of order method

[pat666]

Homework Statement

$$x^2 y''+2xy'-12y=0,y_1=x^3$$
$$y''+(2y')/x-12y/x^2 =0$$

Homework Equations

The Attempt at a Solution

$$y=uy_1=ux y'=u' x+u y''=u'' x+2u'$$
subbing that in instead of y,y',y''
$$u'' x+2u'+2(u'x+u)/x+12(ux)/x^2 =0$$

now my book says to reduce the order v=u' which is easy enough except for the fact that I have "u" in there with no derivative. What do I sub here (integral of v)? STUMPED!

thanks for any help

 

[lurflurf]

x is not a solution, x^3 is.

 

[pat666]

Yes, that's one mistake so far:
$$y_1=uy_1=u*x^3 y'=u'*x^3+u y''=u''x^3+2u'$$
before going on can you tell me if that's correct??

thanks

 

[lurflurf]

I do not follow what you wrote, the subscripts are unclear and it all runs together.
The product rule is
(uv)'=u'v+uv'
(uv)''=u''v+2u'v'+uv''

 

[pat666]

sorry, Ill fix it:
$$y_1=u*x^3--------------y'=u'*x^3+u----------y''=u''x^3+2u'$$

 

[lurflurf]

(x^3)'=3 x^2
(x^3)''=6 x
You might be differentiating x
x'=1
x''=0

 

[pat666]

sorry, I don't know what you mean. I was following how the textbook does it which doesn't seem to be normal differentiation?

 

[lurflurf]

It is normal differentiation.
let
y=u y₁
y'=(uy₁)'=u'y₁+u y₁'
y''=(uy₁)''=u''y₁+2u'y₁'+uy₁''
in this case
y₁=x^3
y₁'=3x^2
y₁''=6x
so
y=x^3 u
y'=x^3 u'+3x^2 u
y''=x^3 u''+6x^2 u'+6x u
which we substitute into the differential equation

 

[pat666]

Again your right, I was treating u as a constant and not what it actually is.
so what I have now is:
$$x^3*u''+6x^2u'+6x*u+2(x^3 u'+3x^2 u)/x-12(x^3 u)/x^2=0$$
now I can let v=u' or do I neaten things up first?

 

[lurflurf]

You can let v=u' if you want, you will notice the u's cancel out as expected.

 

[pat666]

Hi again, just neatened it up, in actually comes out quite neatly:
$$x^3u''+8x^2u' =0$$
then subbing v=u', separation of variables + integration gives
$$ln(v)=-8ln(x)$$ can you give me some more help here please.

 

[lurflurf]

Just reverse the substitutions
v=A x^-8
u'=v
y=x^3 u
to find y
y will have the form A y1+B y2
where A and B are arbitrary constants

 

[HallsofIvy]

Hi again, just neatened it up, in actually comes out quite neatly:
$$x^3u''+8x^2u' =0$$
then subbing v=u', separation of variables + integration gives
$$ln(v)=-8ln(x)$$ can you give me some more help here please.

Well, $-8ln(x)= ln(x^{-8})$, of course, so you have $ln(v)= ln(x^{-8})$. Can you solve that for v? And then integrate to find u.

 

[pat666]

Certainty can:
$$v=1/x^8$$
since v=u' so ' = integral of v
so $$u=\int 1/x^8$$
$$u=-1/(7x^7)$$
Do I need a constant of integration here or not because I am looking for a particular solution?

Thanks

also not sure what the next step is?

 

[pat666]

Hey got a little further:
I have y=x^3u so u=y/x^3
and u=-A/(7x^7 )
so y/x^3=-A/(7x^7 )
But I don't have A and B??

Thanks

 

[HallsofIvy]

Actually there are two constants of integration and you (and I) completely forgot the first one- which is the one you need.

ln(v)= -8ln(x)+ c gives $v= Cx^{-8}$ ($C= e^c$.)

Then $u'= v= Cx^{-8}$ so $u(x)= -\frac{C}{7}x^{-7}+ C_2$

And then $y(x)= u(x)x^3= -\frac{C}{7}x^{-4}+ C_2x^3$ is the general solution. Notice that the second constant of integration gives the original solution, $x^3$. Of course, you could have ignored both constants of integration to just get that $x^{-4}$ is a second independent solution and, already knowing that $x^3$is a solution, write the general solution as
$$y(x)= Ax^{-4}+ Bx^3$$.

 

[pat666]

Thanks heaps for all your help HallsofIvy and lurflurf! mucho appreciato!