# Solving differential equations using a change of variable

• Mar 21st 2009, 09:44 AM
Erghhh
Solving differential equations using a change of variable
(x+y)dy/dx = x^2+xy+x+1 ; y=v-x

So i did the substitution and ended up with vdv/dx - v =xv+x+1
and, if correct, i've no idea how to solve.
• Mar 21st 2009, 09:51 AM
Quote:

Originally Posted by Erghhh
(x+y)dy/dx = x^2+xy+x+1 ; y=v-x

So i did the substitution and ended up with vdv/dx - v =xv+x+1
and, if correct, i've no idea how to solve.

(x+y)dy/dx = x^2+xy+x+1 ; y=v-x

vdv/dx - v =xv+x+1

vdv/dx = v+xv+x+1

vdv/dx = v(1+x) + (1+x)

vdv/dx = (v+1)(1+x)

$\frac{vdv}{v+1} =(x+1)dx$

Integrate both sides

$\int{\frac{vdv}{v+1}} =\int{(x+1)dx}$

Integrate this thing and enjoy the sunny day (Happy)
• Mar 21st 2009, 10:25 AM
Erghhh
So then i got (y+x)ln|x+y+1|=(1/2)(x^2) +x+c

The answer, according to my book, is y=ln|x+y+1| +(1/2)(x^2)+C

I'm unsure of how to get there..
• Mar 21st 2009, 10:45 AM
running-gag
Hi

A sign - is missing
(y+x)-ln|x+y+1|=(1/2)(x^2) +x+c

Then y=ln|x+y+1| +(1/2)(x^2)+c
• Mar 21st 2009, 10:46 AM
Erghhh
Can you explain where that minus sign has come from?
• Mar 21st 2009, 10:50 AM
running-gag
$\int{\frac{vdv}{v+1}} = \int{\frac{(v+1-1)dv}{v+1}} = \int \left(1-\frac{1}{v+1}\right)\:dv = v - \ln|v+1|$
• Mar 21st 2009, 10:51 AM
HallsofIvy
$\frac{v}{v+1}= 1- \frac{1}{v+1}$
Intgrate that.

running-gag got in ahead of me!