1. ## Non-linear DE #2

(1-x)y'' + xy' - y = 0

i have a midterm tomorrow and i'm having a lot of trouble with these types of DEs

2. Originally Posted by haneek
(1-x)y'' + xy' - y = 0

i have a midterm tomorrow and i'm having a lot of trouble with these types of DEs
once again can you "guess" as solution? It is a simple solution.

3. guess what?

4. Originally Posted by haneek
guess what?
Did you read what was posted at your other thread! This is the exact same type of problem. If you can guess one solution you can use reduction of order to find the complete soluton set.

5. yes i did! is one of the solutions (x+1)?

6. Originally Posted by haneek
yes i did! is one of the solutions (x+1)?
No, remember that you can check if it is by plugging it into the ode.,

7. ok so i continued solving the problem and i choose that one of the solutions is x and i plugged it and it works. then i got the derivates of y = u(x) x
and i reached this step

u''[x+x^2] + u'[x^2-2x+2] = 0

but now i'm stuck i dont know how to continue to find y2

8. Originally Posted by haneek
ok so i continued solving the problem and i choose that one of the solutions is x and i plugged it and it works. then i got the derivates of y = u(x) x
and i reached this step

u''[x+x^2] + u'[x^2-2x+2] = 0

but now i'm stuck i dont know how to continue to find y2
Im also curious how one would tackle this although I get u''[x-x^2] + u'[x^2-2x+2] = 0

Would this be a correct attempt -

$\displaystyle \dispalystyle \int\frac{u''(x)}{u'(x)}dx=\int\frac{x^2-2x+2}{x-x^2}dx$

9. By the way, you titled this thread "non-linear DE #2". You should understand that this is NOT a non-linear differential equation- it is a perfectly good linear equation.

10. Originally Posted by HallsofIvy
By the way, you titled this thread "non-linear DE #2". You should understand that this is NOT a non-linear differential equation- it is a perfectly good linear equation.
I realise its not non-linear but I am not the OP. I am just also interested in the solution

Would that be a correct approach?

11. Originally Posted by bugatti79
Im also curious how one would tackle this although I get u''[x-x^2] + u'[x^2-2x+2] = 0

Would this be a correct attempt -

$\displaystyle \dispalystyle \int\frac{u''(x)}{u'(x)}dx=\int\frac{x^2-2x+2}{x-x^2}dx$
Yes it just takes some work. Here we go!

$\displaystyle \ln|u'(x)|=x+\ln|x-1|-2\ln|x|+\ln(c) \iff u'(x)=c\left( \frac{1}{x}e^{x}-\frac{1}{x^2}e^{x}\right)$

Now if we integrate by parts with

$\displaystyle u=e^{x} \implies du=e^x \text{ and } dv=\frac{1}{x^2} \implies v= -\frac{1}{x}$

This gives

$\displaystyle \frac{1}{c}u(x)=\int \frac{e^x}{x}dx-\left[ -\frac{e^x}{x}+\int \frac{e^x}{x}dx\right]=\frac{e^x}{x}+d$

So pick c=1 d=0 and we get the 2nd solution is

$\displaystyle y=xu(x)=x\left( \frac{e^x}{x}\right)=e^x$

$\displaystyle u=c_1x+c_2e^x$

12. Originally Posted by TheEmptySet
Yes it just takes some work. Here we go!

$\displaystyle \ln|u'(x)|=x+\ln|x-1|-2\ln|x|+\ln(c) \iff u'(x)=c\left( \frac{1}{x}e^{x}-\frac{1}{x^2}e^{x}\right)$

Now if we integrate by parts with

$\displaystyle u=e^{x} \implies du=e^x \text{ and } dv=\frac{1}{x^2} \implies v= -\frac{1}{x}$

This gives

$\displaystyle \frac{1}{c}u(x)=\int \frac{e^x}{x}dx-\left[ -\frac{e^x}{x}+\int \frac{e^x}{x}dx\right]=\frac{e^x}{x}+d$

So pick c=1 d=0 and we get the 2nd solution is

$\displaystyle y=xu(x)=x\left( \frac{e^x}{x}\right)=e^x$

$\displaystyle u=c_1x+c_2e^x$
Hmmm, the same here, why/how c=1, d=0? :-)