• Apr 24th 2008, 12:50 AM
cyy
hi given (2x - 4y + 5 ) * ( dy/dx ) + x - 2y + 3 = 0

I'v let u = x - 2y; dy/dx = 1-2*(dy/dx); dy/dx = 1/2 * ( 1 - du/dx )

sub into origin eqn:
( 2u + 5 ) [ 1/2 ( 1-du/dx ) ] + u + 3 = 0

( 2u + 5 ) [ 1/2 - 1/2 * du/dx ] + u + 3 = 0
( 2u + 5 ) / 2 - ( ( 2u + 5 ) / 2 ) * du/dx = -u - 3
( 2u + 5 ) du/dx = 4u + 11 ....

so how i should carry on from here? thx u!!!
• Apr 24th 2008, 12:55 AM
Isomorphism
Quote:

Originally Posted by cyy
hi given (2x - 4y + 5 ) * ( dy/dx ) + x - 2y + 3 = 0

I'v let u = x - 2y; dy/dx = 1-2*(dy/dx); dy/dx = 1/2 * ( 1 - du/dx )

sub into origin eqn:
( 2u + 5 ) [ 1/2 ( 1-du/dx ) ] + u + 3 = 0

( 2u + 5 ) [ 1/2 - 1/2 * du/dx ] + u + 3 = 0
( 2u + 5 ) / 2 - ( ( 2u + 5 ) / 2 ) * du/dx = -u - 3
( 2u + 5 ) du/dx = 4u + 11 ....

so how i should carry on from here? thx u!!!

So integrate now :)

$\int \frac{2u + 5}{4u + 11} \, du = \int dx$

$\frac12\int \frac{4u + 11 - 1}{4u + 11} \, du = \int dx$

$\frac12\int \left(1 - \frac{1}{4u + 11}\right) \, du = \int dx$

$\frac{u}2 - \frac18 \ln|4u+11| = x + C$

Resubstituting back:

$\frac{x - 2y}2 - \frac18 \ln|4x - 8y+11| = x + C$