3
$\begingroup$

Using the substitution $p=x+y$, find the general solution of $\frac{dy}{dx}=(3x+3y+4)/(x+y+1)$ Here are my steps:

Since $p=x+y$, $\frac{3x+3y+4}{x+y+1}=\frac{3p+4}{p+1}=\frac{1}{p+1}+3$

Therefore, integrate both sides $y=\ln(p+1)+3p+c$

$y=\ln(x+y+1)+3(x+y)+c$

But the answer in my book is $x+y-\frac{1}{4}\ln(4x+4y+5)=4x+c$ Is that correct?

2 Answers 2

2

Since $p(x)=x+y(x)$ therefore $y(x)=p(x)-x$. Thus $ dy/dx=y'(x)=p'(x)-1. $ So the new equation is $ \frac{dp}{dx}=p'(x)=\frac{1}{p(x)+1}+4=\frac{4p(x)+5}{p(x)+1}=\frac{4p+5}{p+1}. $ This equation is separable. Using the usual method $ \frac{p+1}{4p+5}dp=1 dx, $ integrating $ \frac{p}{4}-\frac{1}{16}\log(4p+5)+C=x $ Substituting $p(x)=x+y(x)$ we obtain the solution that is the same as in your book.

3

For the left hand side of your differential equation, you should substitute $ \frac{dy}{dx} = \frac{dp}{dx} - 1\,, $ since $ p = x + y $, and then advance with your solution by separation of variable method.

  • 0
    @Vic: I already told you how to advance with the problem in my answer. It always happens when you solve a differential equation you get different forms of closed form solutions. But you can bring one form to the other.2012-07-19