Posted by xX_Supaman_Xx on .
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.
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
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
-Ae^x -C = 1 + x + Cx
well now, A = 0 because we have no other term in e^x
C = -1
B can be any constant
y = B e^x - 1
Whoops, correct last line
y = B xe^x - 1