Posted by xX_Supaman_Xx on Saturday, December 29, 2007 at 6:20am.
Solve the following differential equation, i.e. solve for y:
dy/dx  y/x = 1/x + y + 1
I have this so far,
1. dy/dx= (y + 1 + xy + x)/x
2. x dy = (y + 1 + xy + x)dx
3. Integrate both sides...
xy + c = xy + x + .5x^2y + .5x^2 + k
where c and k are constants
4. c = x + .5x^2 (y + 1) + k
5. .5x^2 (y + 1) = c  k x
6. y = 2(C x)x^2  1
where C = c  k
but I have been told this is wrong. so can soneone please direct me in the error of my method and show me the correct working out.

Mathematics  drwls, Saturday, December 29, 2007 at 7:18am
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

Mathematics  drwls (correction), Saturday, December 29, 2007 at 7:26am
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.

Mathematics  drwls, Saturday, December 29, 2007 at 1:33pm
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

Mathematics  Damon, Saturday, December 29, 2007 at 5:07pm
Yes, I worked it through mostly the same way and got
y = c x e^x  1

Mathematics  Damon, Saturday, December 29, 2007 at 5:55pm
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

Mathematics  Damon, Saturday, December 29, 2007 at 5:57pm
Whoops, correct last line
y = B xe^x  1
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