Differential Equations problems

[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.

 

[HallsofIvy]

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.

Good. I'll wait for it.

 

[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 got to 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 I am 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...

****... going to go kill myself...

 

[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 haven't been able to integrate u'= {c_1 \over (x^2-x^4)}... Assuming I did everything right ealier, and I am supposed to integrate that..

 

[camilus]

and I am pretty dumb, I think I am able to integrate that using integration by partial fractions.. going to try right now.

 

[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...

 

[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. I am too lazy to type the work right now, I think its right so I rather work on another problem.

 

[camilus]

Im almost done with #4 but I am 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!

 

[camilus]

maybe integration by parts?

 

[Defennder]

we get y'_c = c_2\sqrt{\lambda}cos(\sqrt{\lambda}t),

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.

 

[Defennder]

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

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?

I think the general solution to #6 is
y(x) = c_1cos x + c_2sin x + xsin x

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.

 

[Defennder]

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!

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.

 

[camilus]

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

and I' ll post my work to number 6 tomorrow, I am 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...

 

[Defennder]

it has nothing to do with the minus sign, what did you get for you integrating factor, mu?

My integrating factor is the same as yours.

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

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

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...

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.

 

[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 haven't been able to integrate u'= {c_1 \over (x^2-x^4)}... Assuming I did everything right ealier, and I am supposed to integrate that..

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

 

[Defennder]

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

 

[camilus]

My integrating factor is the same as yours.

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.

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?

Yes. question 6 is undetermined coefficients.

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

please don't lol. I really need help, and the help is greatly appreciated.

 

[Defennder]

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)

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

please don't lol. I really need help, and the help is greatly appreciated.

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

 

[camilus]

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.

yeah, there you go. I used 1-x^2, that's 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!

 

[camilus]

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?

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)}

 

[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.

 

[camilus]

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.

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

 

[camilus]

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

we got:

Yeah, you forgot to cancel a \pi there in your answer when you plug in L=\pi. You should just have \lambda_n=\frac{n^2}{4}, where n is odd, i.e. n=2k+1,\ k=0,1,2,\ldots. Once you've finished solving and substituted back for x, the easiest thing to do if you're not sure is to check if your solutions satisfy the ODE and the boundary conditions.

I had wrote that first, \lambda_n=\frac{n^2}{4}, but then I changed it when I noticed we needed a multiple of pi in the eigenfunction. Because the eigenvalues in my notes had a pi..

Although I though I think you're right, and its a pi^2 and not a pi^4 in lambda_n. The pi will come from the initial conditions in this case, and cancel out with pi that's supposed to be there leaving just \lambda_n=\frac{n^2}{4}.

So (hopefully) we're done. The eigenvalues are \lambda_n=\frac{n^2}{4}, and the eigenfunctions are y_n = c_2sin({n \over 2}x) where n is odd.

HURRAY! lol. Now that we (again, hopefully) knocked this one off, and we have an idea of what these eigenvalues/eigenfunctions problems are asking us to find, let's tackle number two. The differential equation is more complicated, but thankfully, we don't need a derivative, both boundaries are y(x)s, no y'(x)s..