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)
This one can't be too tricky, have you tried to solve the non homogenous version of this D.E?
Originally Posted by freakymath
It may be worth reading this thread http://www.mathhelpforum.com/math-he...ial-38182.html
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?
Sorry I'm not a student of math department. So can you explain ur ideas deeply with some expressions? Pls...
So, the first question is ?
Well, then let
Now you have the recurrence
You need the initial conditions to find and .
No it's x²y'' + xy' - y = xlnx
snowtea; i used euler. But at the end there is xlnx. So what will i do at the end???
If it is =0, the result is y(x)=c1(first c)x+c2(second c)/x
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.
euler is what you had written y,y',y'' ...
but the main point is xlnx here. im a rookie, pls forgive me.
Ok, I see. Did not know about this type of differential equation was also Euler's.
The way an Euler differential equation is transformed into a linear ODE is by substituting or .
So substitue this for x on the right hand side.
The new differential equation is (where all the derivatives are now w.r.t. t):
Solve this differential equation to get something of the form:
Now substitute in to get the final solution.
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See what i did under the line for =0. I think it's true.
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).