Two eqns: 3y' + 3xy = e^-3(x^2)/6; x^2 y'' - 3xy' + 3y = 0

M

|m|

New member
Joined
Apr 28, 2006
Messages
17
1.) 3y' + 3xy = e^-3(x^2)/6

My result y=C*e^-3(x^2)/6 + x*e^-3(x^2)/6 but it's not correct.

2.) x^2 y'' - 3xy' + 3y = 0

I don't know how to solve this one :(

Does anyone know where can I find a website with solved differential equations step by step and good explanation?

How can I solve diff. equations with program Mathematica. I think I have to use DSolve but then I don't know the correct form (especialy how to type y' or y'')
 
1) Divide the whole equation by 3, and it looks like you'll have to take an integrating factor.

2) Looks like a Euler-Cauchy equation. Go here to tackle them. Should not be very hard, since the characteristic equation is simple.
 
I ran the second one through Maple and got:

\(\displaystyle \L\\y=C_{1}x+C_{2}x^{3}\)


I don't know about Mathematica, but the Maple command for DE is:

dsolve(diff(y(x),x,x)*x^2-3x*diff(y(x),x)+3y(x)=0)
 
I've solved the second problem, tanx morson. The first one is still a mystery for me...
 
You have: 3y' + 3xy = e^-3(x^2)/6

Divide through by 3: y' + xy = 1/18*e^{-3x^2}, assuming you meant "e^-3(x^2)/6" to be 1/6 * e^{-3x^2}. If you meant e^{-3x^2/6}, the equation is much, much easier.

Integrating factor: e^{1/2*x^2}
 
Well, are you able to solve it? Where are you having trouble? Do you understand the concept of integrating factors?
 
morson said:
Where are you having trouble? Do you understand the concept of integrating factors?
Click to expand...

No, I don't understand it. This is the first time I'm solving differential equations so I have a lot to learn :oops:

I'm reading that page that you gave me link for and I found expression sec(x). What does it mean?
 
Your first equation is of the form: y' + P(x)y = Q(x), whose solution is discussed here.

"sec(x)" is the reciprocal of cos(x), ie: 1/cos(x).

Also, just curious as to why you're doing second-order equations when you're only just starting.
 
I have to solve 5 equation (2 linear, 2 second-order, 1 third-order). If I don't solve them I can't enter for math exam.
 
I solved the first one to. Result y = ((x/3)+C)/(e^{(x^2)/2}
 
That's correct. Good.

You could write it as:

\(\displaystyle \L\\y=\left(\frac{x}{3}+C\right)e^{\frac{-x^{2}}{2}}\)

No big deal. Same thing.
 
You must log in or register to reply here.
Top