# Thread: help me with differential equation

1. ## 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

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

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

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

2. 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=$\displaystyle e^x$ is a solution and the second solution can be in the form

$\displaystyle y2(x)=e^x \int x^n e^-x dx$
I assume you can calculate $\displaystyle y_1'$ and $\displaystyle y_1''$ from $\displaystyle y_1 = e^x$. 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 $\displaystyle y_2(x) = y_1(x) u(x) = e^x u(x)$. Calculate $\displaystyle y_2'$ and $\displaystyle y_2''$ and substitute the results into the left hand side of the differential equation. Simplify and solve the resulting differential equation for u(x).

3. 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:

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

well so this is what i got:

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

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

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

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

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

some sums and products later...

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

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

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

here is the problem i think...

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

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

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

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

$\displaystyle ln u=x+ln x^N$

$\displaystyle u=e^xx^N$

remember that u=v' so:

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

$\displaystyle y_2=e^x\int e^xx^Ndx$ 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...

4. 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:

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

well so this is what i got:

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

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

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

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

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

some sums and products later...

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

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

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

here is the problem i think...

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

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

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

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

$\displaystyle ln u=x+ln x^N$

$\displaystyle u=e^xx^N$

remember that u=v' so:

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

$\displaystyle y_2=e^x\int e^xx^Ndx$ 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

5. 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:

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

6. 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:

$\displaystyle u'=u((x+N)/x)$
Not so. From what I said it follows that

$\displaystyle \frac{du}{dx} = u \, \frac{(N - x)}{x}$.

7. 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)

8. Originally Posted by Black Kawairothlite
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)

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.

9. well just for the record im gonna show the second part of the problem i hope this could help someone in the future 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:

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

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

$\displaystyle u=x^N$

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

$\displaystyle dv=e^{-x}dx$

$\displaystyle v=-e^{-x}$

then:

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

again and again: (i gonna do it directly)

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

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

where $\displaystyle M,N>=0$

then the second solution...

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

and finally

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