1. ## 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

2. 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?

3. 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 $\displaystyle y = \sum_{n=0}^\infty a_n x^n$
And set this equal to the Taylor expansion of $\displaystyle x\ln(x)$ to find the coefficients $\displaystyle a_n$

4. 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?

5. Sorry I'm not a student of math department. So can you explain ur ideas deeply with some expressions? Pls...

6. So, the first question is $\displaystyle x^2 y'' + xy' + y = 0$?

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

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

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

You need the initial conditions to find $\displaystyle a_0$ and $\displaystyle a_1$.

7. No it's x²y'' + xy' - y = xlnx

8. snowtea; i used euler. But at the end there is xlnx. So what will i do at the end???

9. If it is =0, the result is y(x)=c1(first c)x+c2(second c)/x

10. 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.

11. euler is what you had written y,y',y'' ...
but the main point is xlnx here. im a rookie, pls forgive me.

The way an Euler differential equation is transformed into a linear ODE is by substituting $\displaystyle x = e^t$ or $\displaystyle t = \ln(x)$.

So substitue this for x on the right hand side.

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

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

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

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

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

$\displaystyle y = Ax + \frac{B}{x} + Cx\ln(x)$

13. ImageShack&#174; - Online Photo and Video Hosting
See what i did under the line for =0. I think it's true.

14. 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).

15. Lemme try it ...

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