(2x+y+1)y' = 1

Although this intially appeared to be a simple problem I've tried every method I know how to do on this problem and I just can't get a solution.

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- Feb 2nd 2010, 10:10 PMjrw391Solving a nonhomogenous D.E.
(2x+y+1)y' = 1

Although this intially appeared to be a simple problem I've tried every method I know how to do on this problem and I just can't get a solution. - Feb 2nd 2010, 11:39 PMCalculus26
dy/dx = 1/(2x+y+1)

Invert:

dx/dy = 2x + 2y +1

dx/dy -2x = 2y +1

use integrating factor e^(-2y)

d[xe^(-2y)] = [(2y +1)e^(-2y)]dy

x e^(-2y) = -e^(-2y)(y+1) + C

x(y) = -(y + 1) + C e^(2y) - Feb 2nd 2010, 11:58 PMCalculus26
alternatively

(2x+y+1)y' = 1

can be written

(2x+y+1)dy = dx

-dx + (2x+y+1)dy = 0

can be made exact with the integrating factor e^(-2y)

-e^(-2y)dx + e^(-2y)(2x+y+1)dy = 0

dF/dx = -e^(-2y)

F = -xe^(-2y) + f(y)

dF/dy = 2xe^(-2y) + f ' (y) = e^(-2y)(2x+y+1)

f ' (y) = (y+1)e^(-2y)

f(y) = -e^(-2y)(y+1)

F = -xe^(-2y) -e^(-2y)(y+1) = C

-x = Ce^(2y) +(y+1)

x = -(y+1) + Ce^(2y)

as in the previous post - Feb 3rd 2010, 06:15 AMjrw391
Thanks! My problem was when I integrated the I.F. I kept on integrating with respect to x instead of y. So I was getting e^(-2x) instead of e^(-2y).