# desperate seconds

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• Dec 17th 2010, 04:36 PM
freakymath
desperate seconds
I have 3 questions and NO TIME.

I'll be sucking on one of the exam desks just a few hrs later.

Pls help me coz late studies may bring me just a few points...

1) x²y'' + xy' - y = xlnx maybe about euler

2) y'' – 2y' + y = x-½ ex

3) ty''+(1-2t)y'-2y=0 y(0)=1 , y'(0)=2 Laplace

Pls help me(Crying)
• Dec 17th 2010, 05:03 PM
pickslides
Quote:

Originally Posted by freakymath
2) y'' – 2y' + y = x-½ ex

This one can't be too tricky, have you tried to solve the non homogenous version of this D.E?

• Dec 17th 2010, 05:09 PM
snowtea
Quote:

Originally Posted by freakymath
I have 3 questions and NO TIME!
1) x²y'' + xy' - y = xlnx maybe about euler

For this one, you can use the power series substitution $y = \sum_{n=0}^\infty a_n x^n$
And set this equal to the Taylor expansion of $x\ln(x)$ to find the coefficients $a_n$
• Dec 17th 2010, 05:10 PM
freakymath
I solved it. What about the first one: I solved it too but the other side of the equation is =0 NOT =xlnx. What can i do about that problem?
• Dec 17th 2010, 05:13 PM
freakymath
Sorry I'm not a student of math department. So can you explain ur ideas deeply with some expressions? Pls...
• Dec 17th 2010, 05:13 PM
snowtea
So, the first question is $x^2 y'' + xy' + y = 0$?

Well, then let
$y = \sum_{n=0}^\infty a_nx^n$
$y' = \sum_{n=0}^\infty a_nnx^{n-1}$
$y'' = \sum_{n=0}^\infty a_nn(n-1)x^{n-2}$

$x^2 y'' + xy' + y = \sum_{n=0}^\infty((n+1)(n+2)a_{n+2} + (n+1)a_{n+1} + a_n)x^n = 0$

This means:
$(n+1)(n+2)a_{n+2} + (n+1)a_{n+1} + a_n = 0$
Now you have the recurrence
$a_{n+2} = -\frac{a_{n+1}}{n+2} - \frac{a_n}{(n+1)(n+2)}$

You need the initial conditions to find $a_0$ and $a_1$.
• Dec 17th 2010, 05:16 PM
freakymath
No it's x²y'' + xy' - y = xlnx
• Dec 17th 2010, 05:22 PM
freakymath
snowtea; i used euler. But at the end there is xlnx. So what will i do at the end???
• Dec 17th 2010, 05:24 PM
freakymath
If it is =0, the result is y(x)=c1(first c)x+c2(second c)/x
• Dec 17th 2010, 05:29 PM
snowtea
I'm not sure what you mean by Euler. Do you mean Euler's method to numerically estimate the function? If not, I do not know enough to help you on problem 1, sorry.
• Dec 17th 2010, 05:33 PM
freakymath
euler is what you had written y,y',y'' ...
but the main point is xlnx here. im a rookie, pls forgive me.
• Dec 17th 2010, 05:40 PM
snowtea
The way an Euler differential equation is transformed into a linear ODE is by substituting $x = e^t$ or $t = \ln(x)$.

So substitue this for x on the right hand side.

$x\ln(x) = te^t$

The new differential equation is (where all the derivatives are now w.r.t. t):

$(y'' - y') + y' - y = te^t$
$y'' - y = te^t$

Solve this differential equation to get something of the form:
$y = Ae^{t} + Be^{-t} + Cte^t$

Now substitute in $t = \ln(x)$ to get the final solution.

$y = Ax + \frac{B}{x} + Cx\ln(x)$
• Dec 17th 2010, 05:50 PM
freakymath
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See what i did under the line for =0. I think it's true.
• Dec 17th 2010, 06:02 PM
snowtea
Yes, for x^2y'' + xy' + y =0, you are correct.
For x^2y'' + xy' + y = xlnx, just substitute x=e^t on the left hand side also (see what I posted above).
• Dec 17th 2010, 06:06 PM
freakymath
Lemme try it ...
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