Homework Help: Differential Equations problems

1. Jul 14, 2008

camilus

1. Find the eigenvalues and eigenfunctions for the given boundary-value problem.

$${d \over dx}[x{dy \over dx}] + {\lambda \over x}{y} = 0,$$ subject to $$y(1)=0, y'(e^\pi)=0$$

2. Find the eigenvalues and eigenfunctions for the given boundary-value problem.

$${d \over dx}[{1 \over 3x^2+1}{dy \over dx}] + {\lambda (3x^2+1)}{y} = 0,$$ subject to $$y(0)=0, y(\pi)=0$$

hint: let $$t=x^3+x$$
3. Find the general solution of the Cauchy-Euler equation Assume x>0.

$$x^3{d^3y \over dx^3} - 4x^2{d^2y \over dx^2} + 8x{dy \over dx} - 8y = 4ln(x)$$
4. Use teh variation of parameters to find the general solution of the given differential equation.

$${d^2y \over dx^2} +y = {1 \over 1+sin x}$$

5. Given that $$y_1(x)=x$$ is a solution of the DE $$(1-x^2)y'' - 2xy' + 2y = 0, -1<x<1,$$ use the reduction of order to find a second solution (Do not use the formula.)
6. Use the method of undetermined coefficients to find the general solution of the give DE.

$$y''+y = x sinx$$

I need a lot of help here! In a little bit, I'll post my work so far, which should lead you in the right direction.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 14, 2008

HallsofIvy

Good. I'll wait for it.

3. Jul 14, 2008

camilus

1.$${d \over dx}[x{dy \over dx}] + {\lambda \over x}{y} = 0,$$ subject to $$y(1)=0, y'(e^\pi)=0$$

differentiate that first term.

$$x{d^2y \over dx^2} +{dy \over dx} +{\lambda \over x}{y} = 0,$$

multiply through by x to make it a Cauchy-Euler equation (making the powers of x and derivatives of y the same).

$$x^2{d^2y \over dx^2} +x{dy \over dx} +{\lambda}{y} = 0,$$ (1)

now using the substitution $$x=e^t$$ and $$t=ln x$$, and taking derivatives of t, you can substitute $$x^2y"= {d^2y \over dt^2} - {dy \over dt}$$ and $$xy'= {dy \over dt}$$

When substituted, the $${dy \over dt}$$ cancels out leaving

$${d^2y \over dt^2} + \lambda{y} = 0$$ (2)

now getting the auxiliary/characteristic equation of (2) gives us
$$m^2+\lambda=0$$ or $$m=+/- i \sqrt{\lambda}$$

that gives us the complimentary solution $$y_c = c_1cos(sqrt{\lambda}t) + c_2 sin (sqrt{\lambda}t)$$

now since we're working with t instead of x, we need to modify our boundaries using t=ln(x). That gives us $$y(0)=0, y'(\pi)=0$$

plugging in the first boundary y(0)=0, gives us $$c_1 = 0$$

now this is where I'm stuck I think.

we gotta get the first derivative of $$y_c$$, and we could ignore the cosine term because c1=0.

we get $$y'_c = c_2\sqrt{\lambda}cos(\sqrt{\lambda}t)$$, but when I plug in y(pi)=0, which I am not even sure if its right, I get stuck. All I know is I get to this line $$0=c_2\sqrt{\lambda}cos(\sqrt{\lambda}\pi)$$ and Im clueless. I know cos(n*pi)=-1 for n=1,3,5... and =1 for n=2,4,6...

now from my notes it says the eigenvalues are $$\lambda_n = {n^2\pi^2 \over L^2}$$ where L is the second boundary y(L)=y1, and the eigenfunctions are supposed to be $$y_n(x) = c_2 sin ({n\pi \over L}{x})$$for n=1,2,3,4...

****... gonna go kill myself...

4. Jul 14, 2008

camilus

number 5.... watch this, this is sad...

Given that $$y_1(x)=x$$ is a solution of the DE $$(1-x^2)y'' - 2xy' + 2y = 0, -1<x<1,$$ use the reduction of order to find a second solution (Do not use the formula.)

attempted solution:

if its given that y1=x, the method of reduction of order says that y2=u(x)y1(x).

so $$y_2=ux$$ and $$y'_2= u + u'x$$ and $$y''_2=u''x+2u'$$

and putting the DE in standard form.

$$y'' - {2x \over 1-x^2}y' + {2 \over 1-x^2}y = 0$$

plugging in $$y_2, y'_2, and y''_2$$ you get

$$u''x+2u' - {2x \over 1-x^2}(u + u'x) + {2 \over 1-x^2}(ux) = 0$$

expanding cancels out the u term, leaving $$xu''+(2-{2x^2 \over 1-x^2})u' = 0$$, and divide by x, $$u''+({2 \over x}-{2x \over 1-x^2})u' = 0$$

Now is where the reduction of order method gets its name. you let w=u', so w'=u", reducing the second-order equation into an eaiser first-order DE.

so $$w'+( {2 \over x}-{2x \over 1-x^2})w = 0$$ is the result after the substitution. But getting the integrating factor is what got me.

$$\mu(x)=e^{\int ({2 \over x}-{2x \over 1-x^2})dx}$$, which I would like someone to check this, I get $$\mu(x)=e^{ln[x^2(1-x^2)]}=x^2(1-x^2) = x^2-x^4$$

Now if this mu is correct, than multiplying both sides of the DE by the mu will result by definition as

$${d \over dx}{[(x^2-x^4)w] = 0*[x^2(1-x^2)] = 0$$

But this is were I run into problems. Integrating this equation gives me

$$(x^2-x^4)w= c_1$$, and $$w= {c_1 \over (x^2-x^4)}$$.

Substituting u' for w again, will give us u after integration. But its this integration that got me stuck.

I havent been able to integrate $$u'= {c_1 \over (x^2-x^4)}$$... Assuming I did everything right ealier, and Im supposed to integrate that..

5. Jul 14, 2008

camilus

and im pretty dumb, I think Im able to integrate that using integration by partial fractions.. gonna try right now.

6. Jul 14, 2008

camilus

Well I knocked off #5 I believe, I integrated by partial fractions and got

$$u' = c_1\left[ \int {{-1 \over x^2} + {-1/2 \over x+1} + {1/2 \over x-1}} \right]dx$$

so
$$u = c_1\left[{1 \over x} + {1 \over 2}{ln(x-1)} - {1 \over 2}{ln(x+1)}\right] +c_2$$

and for convinience, we'll let c1=1 and c2=0 rendering

$$u = {1 \over x} + {1 \over 2}{ln(x-1)} - {1 \over 2}{ln(x+1)}$$

and since from the beginning we assumed that $$y_2(x) = y_1(x)u(x)$$ and y1=x, then $$y_2(x) = x \left [ {1 \over x} + {1 \over 2}{ln(x-1)} - {1 \over 2}{ln(x+1)} \right ]$$

sweet, 1 down, 5 to go....

7. Jul 14, 2008

camilus

I think the general solution to #6 is

$$y(x) = c_1cos x + c_2sin x + xsin x$$

let me know what you get. Im too lazy to type the work right now, I think its right so I rather work on another problem.

8. Jul 14, 2008

camilus

Im almost done with #4 but Im stuck at the integration again. This time its serious.

for u2, I get $$u_2=\int {cosx \over 1+sinx}dx$$, which simple substitution u=1+sinx can integrate.

I get stuck on u1, where I get $$u_1=\int -{sinx \over 1+sinx}dx$$...

Now this is one hell of an integral to compute, I can tell just by putting it into the Wolfram Mathematica Online Integrator. It says that

$$\int -{sinx \over 1+sinx}dx = - {{[ cos ({x \over 2}) + sin ({x \over 2}) ] [ xcos({x \over 2}) + (x-2)sin( {x \over 2} ) ]} \over 1+ sin x}$$

how in the world would I do an integral by hand and get this crazy function!!

9. Jul 14, 2008

camilus

maybe integration by parts?

10. Jul 14, 2008

Defennder

I don't think this is right. Remember that y'(e^pi) in the boundary conditions refers to dy/dx. Now you're working in dy/du. The chain rule for dy/dx = dy/du du/dx.

EDIT: This is in reply to qn 1, by the way.

11. Jul 14, 2008

Defennder

That's odd. By partial fractions I got $$\frac{1}{x^2} + \frac{1}{2(1-x)} + \frac{1}{2(1+x)}$$. Where did your minus sign come from?

Actually it's a lot more complicated than that. This appears to be a straightforward application of the variation of parameters. The particular solution for y(x) is in the form $$y_p(x) = u(x)y_1(x) + v(x)y_2(x)$$, where y1 and y2 are individual linearly independent solutions to the homogenous DE.

Last edited: Jul 14, 2008
12. Jul 14, 2008

Defennder

This is for qn 4:

To integrate that you might want to use something known as t-formulae or tangent half-formulae. Or also known by its fancy name Weierstrass substitution.

EDIT: I just tried it out and it seems to work.

13. Jul 14, 2008

camilus

it has nothing to do with the minus sign, what did you get for you integrating factor, mu? EDIT: Im gonna make it BOLD on my work so you see what Im talking about.

and I' ll post my work to number 6 tomorrow, im too lazy to write that all out right now. But the problem says use the method of undetermined coefficients.

you get the homogeneous case out the way, and for the particular solution you let y_p=(Ax+B)cos x (Cx+D)sin x. After the two differentiations and plugging in and cancelling terms, you're left with -2Asinx + 2Bcosx. Since these terms are duplicated in the homogeneous case, you multiply thru by x.

you you got that -2Axsinx+2Bxcos=xsinx

you match up coefficients, making B=0 and A=-1/2...

Last edited: Jul 14, 2008
14. Jul 14, 2008

Defennder

My integrating factor is the same as yours.

Apparently I didn't see that, so yeah, I suppose you should use that instead of variation of parameters.

This is with reference to qn 6?

EDIT: Woohoo! 999 posts. Great number. Maybe I should stop posting for a while just to retain that post number.

15. Jul 14, 2008

camilus

strange, it didnt let me edit.. so I had 2 quote myself..

16. Jul 14, 2008

Defennder

Yes, and I did say that I got the same integrating factor as you did.

17. Jul 15, 2008

camilus

Okay, then this is my work.
$$\mu(x)=e^{\int ({2 \over x}-{2x \over 1-x^2})dx} = e^{ln x^2}*e^{ln(1-x^2)}dx}= x^2(1-x^2) = x^2(x+1)(x-1)$$

doing partials fractions on that is where I got my answer.

Yes. question 6 is undetermined coefficients.

please dont lol. I really need help, and the help is greatly appreciated.

18. Jul 15, 2008

Defennder

Well for one thing $$x^2(1-x^2)$$ is not the same as $$x^2(x^2-1)$$.

Oh well I guess I'm past that anyway. Maybe next time when I'm at 1,111 posts.

19. Jul 15, 2008

camilus

yeah, there you go. I used 1-x^2, thats how the problem appears on my paper.

any more help?? especially with the first 2. I could probably get the rest on my own but those first two eigeinvalues/eigenfunctions are killing me!

20. Jul 15, 2008

camilus

HAHA I made the same mistake. I used 1-x^2, but when I split it I put (x-1)(x+1) instead of (1-x)(1+x)... so your right in the above quote. Although the work is the same, its just a matter of factoring out the minus sign you're mentioning.

EDIT: nevermind, neither of us is right.

its supposed to be $$\frac{1}{2(1+x)} - \frac{1}{x^2} - \frac{1}{2(1-x)}$$

21. Jul 15, 2008

Defennder

You used x^2-1 and not 1-x^2. And as for the first two questions, I did reply earlier for qn 1. I haven't looked at 2 yet.

22. Jul 15, 2008

camilus

I did this, instead of y(x) boundaries, transformed the equation using t=ln(x), so the boundaries instead of y(1)=0 and y'(e^pi)=0 become y(0)=0 and y'(pi)=0

23. Jul 15, 2008

camilus

Well, with help from a friend on another forum, I think we solved #1.

we got: