# help me with differential equation

• Oct 24th 2008, 02:00 PM
Black Kawairothlite
help me with differential equation
hey all this is my first post and im sorry that i begin asking bt well ill try to help too whenever i can...

consider the equation

$xy''-(x+N)y'+Ny=0$ where N>=0

show that $y_1=e^x$ is a solution and the second solution can be in the form

$y_2(x)=e^x \int x^n e^{-x} dx$
• Oct 24th 2008, 07:15 PM
mr fantastic
Quote:

Originally Posted by Black Kawairothlite
hey all this is my first post and im sorry that i begin asking bt well ill try to help too whenever i can...

consider the equation xy''-(x+N)y'+Ny=0 where N>=0

show that y1= $e^x$ is a solution and the second solution can be in the form

$y2(x)=e^x \int x^n e^-x dx$

I assume you can calculate $y_1'$ and $y_1''$ from $y_1 = e^x$. Substitute the results into the left hand side of the differential equation and show that the result simplifies to zero.

Try finding a solution of the form $y_2(x) = y_1(x) u(x) = e^x u(x)$. Calculate $y_2'$ and $y_2''$ and substitute the results into the left hand side of the differential equation. Simplify and solve the resulting differential equation for u(x).
• Oct 26th 2008, 07:39 PM
Black Kawairothlite
thanks for your answer Mr. Fantastic...ill show ya what i did cuz i have a mistake and i don't know where it is...so let's see:

first of all i took the first solution and i evaluated it in the equation and yes it was a solution (i had already done it before but i didn't do it well)

so then i did grade reduction to get the second solution which in the exercise said it had to be:

$y_2= e^x\int x^Ne^{-x}dx$

well so this is what i got:

$y_2= e^xv(x)$ where v(x) is my unknown solution

$y_2'=e^xv(x)+e^xv'(x)$

$y_2'=e^xv(x)+2e^xv'(x)+e^xv''(x)$

then replacing on the diff eq. (which was $xy''-(x+N)y'+Ny=0$ with N>0)

$x(e^xv(x)+2e^xv'(x)+e^xv''(x))-(x+N)(e^xv(x)+e^xv'(x))+N(e^xv(x))=0$

some sums and products later...

$v''xe^x-v'e^x(x+N)=0$

then i suppouse that $u=v'(x)$ so the new equation is:

$u'xe^x-ue^x(x+N)=0$ which is a separable diff. equation

here is the problem i think...

$u'=u((x+N)/x)$

$du/dx=u((x+N)/x)$

$du/u=((x+N)/x)dx$

(while i was writing this i noticed my mistake, i was not ok because i put $du/u=dx/((x+N)/x))$ let's c if i'm right...)

$ln u=x+ln x^N$

$u=e^xx^N$

remember that u=v' so:

$v=\int e^xx^Ndx$ and finally:

$y_2=e^x\int e^xx^Ndx$ but im still missing the minus where do i get it (omg it worked pretty good)

and the other point i care is that i gotta prove that this second solution is a polynomial of N grade...

• Oct 26th 2008, 08:15 PM
mr fantastic
Quote:

Originally Posted by Black Kawairothlite
thanks for your answer Mr. Fantastic...ill show ya what i did cuz i have a mistake and i don't know where it is...so let's see:

first of all i took the first solution and i evaluated it in the equation and yes it was a solution (i had already done it before but i didn't do it well)

so then i did grade reduction to get the second solution which in the exercise said it had to be:

$y_2= e^x\int x^Ne^{-x}dx$

well so this is what i got:

$y_2= e^xv(x)$ where v(x) is my unknown solution

$y_2'=e^xv(x)+e^xv'(x)$

$y_2'=e^xv(x)+2e^xv'(x)+e^xv''(x)$

then replacing on the diff eq. (which was $xy''-(x+N)y'+Ny=0$ with N>0)

$x(e^xv(x)+2e^xv'(x)+e^xv''(x))-(x+N)(e^xv(x)+e^xv'(x))+N(e^xv(x))=0$

some sums and products later...

$v''xe^x-v'e^x(x+N)=0$ Mr F says: I think you'll find this should be ${\color{red}v'' x e^x + v' e^x (x-N) = 0}$. And you can divide out the common factor of e^x.

then i suppouse that $u=v'(x)$ so the new equation is:

$u'xe^x-ue^x(x+N)=0$ which is a separable diff. equation

here is the problem i think...

$u'=u((x+N)/x)$

$du/dx=u((x+N)/x)$

$du/u=((x+N)/x)dx$

(while i was writing this i noticed my mistake, i was not ok because i put $du/u=dx/((x+N)/x))$ let's c if i'm right...)

$ln u=x+ln x^N$

$u=e^xx^N$

remember that u=v' so:

$v=\int e^xx^Ndx$ and finally:

$y_2=e^x\int e^xx^Ndx$ but im still missing the minus where do i get it (omg it worked pretty good)

and the other point i care is that i gotta prove that this second solution is a polynomial of N grade...

Good job showing all your work (Yes)
• Oct 26th 2008, 08:19 PM
Black Kawairothlite
thank you M. Fantastic i'm learning LaTex code here lol and btw i already did what u said you can c that when i put this:

$u'=u((x+N)/x)$
• Oct 26th 2008, 09:21 PM
mr fantastic
Quote:

Originally Posted by Black Kawairothlite
thank you M. Fantastic i'm learning LaTex code here lol and btw i already did what u said you can c that when i put this:

$u'=u((x+N)/x)$

Not so. From what I said it follows that

$\frac{du}{dx} = u \, \frac{(N - x)}{x}$.
• Oct 26th 2008, 09:32 PM
Black Kawairothlite
(Clapping) Wonderful i got the answer!! thank you Mr. Fantastic you're fantastic lol btw what do u think about the other thing? ( i mean about proving that the second solution is a polynomial of N grade)

(Rofl) (Clapping)
• Oct 27th 2008, 02:51 AM
mr fantastic
Quote:

Originally Posted by Black Kawairothlite
(Clapping) Wonderful i got the answer!! thank you Mr. Fantastic you're fantastic lol btw what do u think about the other thing? ( i mean about proving that the second solution is a polynomial of N grade)

(Rofl) (Clapping)

I'd suggest evaluating the integral (use integration by parts N times - try substituting some concrete values for N first to get the feel of things. N= 1, 2, 3). Then multiply the result by e^x.
• Oct 27th 2008, 07:50 PM
Black Kawairothlite
well just for the record im gonna show the second part of the problem i hope this could help someone in the future :D and if i'm wrong correct me plz...

well with the changes that MR. Fantastic adviced me i got the answer of the solution which was:

$y_2(x)=e^x \int x^n e^{-x} dx$

and as he said i did the integration by parts N times:

$u=x^N$

$du= Nx^{N-1}dx$

$dv=e^{-x}dx$

$v=-e^{-x}$

then:

$\int e^{-x}x^Ndx= -x^Ne^{-x} + \int e^{-x}Nx^{N-1}dx$

again and again: (i gonna do it directly)

$\int e^{-x}x^Ndx= -x^Ne^{-x} - Nx^{N-1}e^{-x} + \int e^{-x}(N^2-N)x^{N-2}dx$

$\int e^{-x}x^Ndx= -x^Ne^{-x} - (N^2-N)x^{N-2}e^{-x}$ $+ \int e^{-x}x^{N-3}(N^3-3N^2-2N)dx+ ... -$ $\frac{{d^M(-x^N)}}{{dx^M}}e^{-x}$ $+\int \frac{{d^{M+1}(-x^N)}}{{dx^{M+1}}}e^{-x}dx$

where $M,N>=0$

then the second solution...

$y_2= e^x(-x^Ne^{-x} - (N^2-N)x^{N-2}e^{-x}$ $+ \int e^{-x}x^{N-3}(N^3-3N^2-2N)dx+ ... -$ $\frac{{d^M(-x^N)}}{{dx^M}}e^{-x}$ $+\int \frac{{d^{M+1}(-x^N)}}{{dx^{M+1}}}e^{-x}dx)$

and finally

$y_2=-x^N - (N^2-N)x^{N-2}$ $+ e^x\int e^{-x}x^{N-3}(N^3-3N^2-2N)dx+...-$ $\frac{{d^M(-x^N)}}{{dx^M}}$ $+e^x\int \frac{{d^{M+1}(-x^N)}}{{dx^{M+1}}}e^{-x}dx$

as you can see is a polynomial of N grade

(yesterday)
i've checked it again and found some errors but right now i'm too tired to solve them so i hope you help me with that :)

(today)
ok i think it's done