two questions on differential equations

[fudge]

Any thoughts on how to to any of these!! (I'm sorry if i'm insulting you but y' = dy/dx, etc!)

1)Find a continous solution with continous first derivative of the system:

y'' + 2y' + 2y = sin x + f(x)

subject to y(-pi/2)=y(pi)=0, where

f(x)= 0 (x<or= 0)
=x^2 (x>0)

2)You may ignore the first bits (i am including them just incase they're relavent for the last bit)

Verify y = x+1 is a soln of

(x^2-1)y'' + (x+1)y' - y = 0 *

Writing y=(x+1)u show that u'=0

hence show the gen soln of * is

y=K[0.25(x+1)ln((x-1)/(x+1))-0.5] + K'(x+1)

where K and K' are arbitary constants.

[HallsofIvy]

You have the equation y'' + 2y' + 2y = sin x + f(x)
subject to y(-pi/2)=y(pi)=0.

I take it that you know how to solve non-homogeneous linear differential equations with constant coefficients and that the problem is just that f(x)= 0 if x<or= 0 and f(x)=x^2 if x>0.

Think of this first as the equation y'' + 2y' + 2y = sin x. Find the general solution to that equation. That will involve 2 unknown constants. Call that solution Y1(x).

Now, find the general solution to the non-homogeneous equation y''+ 2y'+ 2y= sin x+ x2. That will also involve 2 unknown constants. Call that solution Y2(x).

Use the fact that Y1(-pi/2)= 0, Y2(pi)= 0, Y1(0)= Y2=0 (the solution must be continuous at x=0) and Y1'(0)= Y2'(0) (the derivative of the solution must be continuous at x= 0) to get 4 equations to solve for the 4 unknown constants.

For the second question "Verify y = x+1 is a soln of (x^2-1)y'' + (x+1)y' - y = 0", you should be able to do that part. Find the first and second derivatives of x+1, put them into the equation and see what happens!
"Writing y=(x+1)u show that u'=0."
You can't prove that- it's not true. Unless I've made a stupid mistake (it's rather late!) y'= (x+1)u'+ u and y''= (x+1)u''+ 2u so the equation reduces to (x-1)u"+ 3u'= 0.

[fudge]

Thanks for the advice, I've tried it out and it works fine!

I made a very careless typing error for the second part:

"Writing y=(x+1)u show that u'=0"

I actually meant: u'=du/dx !!

Sorry!

Is partial diffrentiation the way to go inorder to show:
du'/dx + [(3x-1)/(x^2-1)]u' = 0 ?

[HallsofIvy]

"Writing y=(x+1)u show that u'=0"

I actually meant: u'=du/dx !!

Surely, you don't mean "show that u'= du/dx"??? Isn't that how it's defined?

Is partial diffrentiation the way to go inorder to show:
du'/dx + [(3x-1)/(x^2-1)]u' = 0 ?
Partial differentiation has nothing to do with it. y is a function of the single variable x, u is a function of the single variable x. The product rule is all you need.

[fudge]

sorry, i am being very careless. I'll write it out again!

Verify y = x+1 is a soln of

(x^2-1)y'' + (x+1)y' - y = 0 *

Writing y=(x+1)u show that u' = du/dx satisfies:

du'/dx + [(3x-1)/(x^2-1)]u' = 0

Hence show the gen soln of * is

y=K[0.25(x+1)ln((x-1)/(x+1))-0.5] + K'(x+1)

where K and K' are arbitary constants.


Partial differentiation has nothing to do with it. y is a function of the single variable x, u is a function of the single variable x. The product rule is all you need. [/B]

Yes ofcourse, I'll try this (I am really concerned... i miss basic steps [:(])

Thank you Halls