Strange equation

May 2010
3
0
Hello

I am new to this forum. I think that you are doing a very good job here helping people solve their problems !!!!

In few days i have one of my final course exam (I will graduate) and i have some questions that concerns me

One of them is this :

\(\displaystyle y'''' * y'''= 1 \)

I had an idea to put y'''=u , y''''=u' but it didn't worked

How can i solve this ???

Thanks in advance
 

shawsend

MHF Hall of Honor
Aug 2008
903
379
That's the same as vdv=dx isn't it? I mean isn't y'''' the derivative of y'''?
Same dif if is was yy'=1 or y'y'' and so forth.
 
Last edited:
May 2010
3
0
So how is this equation can be solved ??? (Happy)
 

TheEmptySet

MHF Hall of Honor
Feb 2008
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2,029
Yuma, AZ, USA
So how is this equation can be solved ??? (Happy)

Your orginal idea will work

let \(\displaystyle u=y''' \implies u'=y^{(4)}\) this gives the equation

\(\displaystyle \frac{du}{dx}u=1 \iff udu=dx \iff u^2=2x+C \iff u=\sqrt{2x+C}\)

Now \(\displaystyle u=y'''\) is just integrate the above three times to recover the function y.