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February 1, 2015

February 1, 2015

Posted by **xX_Supaman_Xx** on Saturday, December 29, 2007 at 6:20am.

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 7:18amWhat 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 7:26amI 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 1:33pmFor 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 5:07pmYes, I worked it through mostly the same way and got

y = c x e^x - 1

- Mathematics -
**Damon**, Saturday, December 29, 2007 at 5:55pmBy 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 5:57pmWhoops, correct last line

y = B xe^x - 1

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