Posted by xX_Supaman_Xx on Saturday, December 29, 2007 at 6:20am.
What is wrong is this: In step 3, you cannot treat x as a constant in the dy integration (on the left), nor can your treat y as a constant in the dy integration (on the right). You need to totally separate y and x variables on opposite sides somehow, or find some other trick.
This is a first order linear differential equation that can be written as
dy/dx - Py = Q
where P and Q are functions of x.
Equations of this type can be solved by the "integrating factor" method. I suggest you familiarize yourself with the method. It goes like this:
Compute the function
rho(x) = exp(integral of P dx)
The solution will be
rho*y = [Integral of rho*Q(x)dx] + C
I should have written the standard form as dy/dx + P(x)y = Q(x)
I got the sign wrong on the P.
In your case, Q = -P = (1/x) + 1
The form of the solution which I wrote should be correct.
For the first step, I get
rho(x) = 1/(x*e^x)
The final step would be to solve
y(x) = (x*e^x)*(Integral)[(1/x^2)*e^-x + (1/x)*e^-x] dx + C
Yes, I worked it through mostly the same way and got
y = c x e^x - 1
By the way, having gotten interested, I also solved this by brute force trial and error.
I assumed an exponential type solution:
y = Ae^x + Bxe^x+ C
then
dy/dx = Ae^x + Bxe^x + Be^x
plug that in
Ae^x + Bxe^x + Be^x - Ae^x/x - Be^x - C/x = 1/x + 1 + Ae^x + Bxe^x+ C
combine like terms
-Ae^x/x - C/x = 1/x + 1+ C
then
-Ae^x -C = 1 + x + Cx
well now, A = 0 because we have no other term in e^x
C(x+1)= -(x+1)
so
C = -1
B can be any constant
so
y = B e^x - 1
same answer
Whoops, correct last line
y = B xe^x - 1
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