Homogeneous, linear, first-order, ordinary differential equation mistake

[kalish1]

I am trying to solve a homogeneous, first-order, linear, ordinary differential equation but am running into what I am sure is the wrong answer. However I can't identify what is wrong with my working?!

$$\frac{dy}{dx}=\frac{-x+y}{x+y}=\frac{1-\frac{x}{y}}{1+\frac{x}{y}}.$$ Let $z=x/y$, so that $y=x/z$ and $$\frac{dy}{dx}=\frac{z-x\frac{dz}{dx}}{z^2}=\frac{1-z}{1+z}.$$ Then $z-x\frac{dz}{dx}=\frac{z^2-z^3}{1+z} \implies -x\frac{dz}{dx}=-\frac{z^3+z}{z+1} \implies \frac{dx}{x}=\frac{dz}{z^2+1} \implies \arctan(z)=\ln(|x|)+C \implies z=\tan(\ln(|x|)+C)$

Thus $$y = \frac{x}{\tan(\ln(|x|)+C)}$$.

However, my book: *A Modern Introduction to Differential Equations 2nd edition by Henry Ricardo* says "this first-order equation is homogeneous and can be solved implicitly".

They pursued the method of letting $z=y/x$ and obtained the solution as
$$2\arctan(y/x)+\ln(x^2+y^2)-C=0.$$

Why are the two different substitutions, which should both be suitable, giving me two different answers?

 

[alyafey22]

Re: Homogenous, linear, first-order, ordinary differential equation mistake

They pursued the method of letting $z=y/x$ and obtained the solution as

Using this substitution is better because differentiation becomes easier

$$y=zx \,\,\to \,\, \frac{dy}{dx}= z+x \frac{dz}{dx}$$

 

[MarkFL]

Re: Homogenous, linear, first-order, ordinary differential equation mistake

I am with you up to this point (where I have negated both sides):

$$x\frac{dz}{dx}=\frac{z^3+z}{z+1}$$

However, when you separated variables, you should get:

$$\frac{z+1}{z^3+z}\,dz=\frac{1}{x}\,dx$$

Using partial fractions, we may write:

$$\left(\frac{1}{z^2+1}-\frac{z}{z^2+1}+\frac{1}{z} \right)\,dz=\frac{1}{x}\,dx$$

This will lead you to a solution that differs from that given by your textbook by only a constant, which can then be "absorbed" by the constant of integration if you desire to get it in the same form.