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The instructor in our Differential Equations class gave us the following to solve: $$ \frac{dy}{dx} = \frac{x - y}{xy} $$

It was an item under separable differential equations. I have gotten as far as $ \frac{dy}{dx} = \frac{1}{y} - \frac{1}{x} $ which to me doesn't really seem much. I don't even know if it really is a separable equation.

I tried treating it as a homogeneous equation, multiplying both sides with $y$ to get (Do note that I just did the following for what it's worth)... $$ y\frac{dy}{dx} = 1 - \frac{y}{x} $$ $$ vx (v + x \frac{dv}{dx}) = 1 - v $$ $$ v^2x + vx^2 \frac{dv}{dx} = 1 - v $$ $$ vx^2 \frac{dv}{dx} = 1 - v - v^2x$$

I am unsure how to proceed at this point.

What should I first do to solve the given differential equation?

Cesareo
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Mark Johnson
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3 Answers3

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We write the differential equation as

\begin{align*} xyy^\prime=x-y\tag{1} \end{align*}

and follow the receipt I.237 in the german book Differentialgleichungen, Lösungsmethoden und Lösungen I by E. Kamke.

We consider $y=y(x)$ as the independent variable and use the substitution \begin{align*} v=v(y)=\frac{1}{y-x(y)}=\left(y-x(y)\right)^{-1}\tag{2} \end{align*}

We obtain from (2) \begin{align*} v&=\frac{1}{y-x}\qquad\to\qquad x=y-\frac{1}{v}\\ v^{\prime}&=(-1)(y-x)^{-2}\left(1-x^{\prime}\right)=\left(\frac{1}{y^{\prime}}-1\right)v^2 \end{align*} From (1) we get by taking $v$: \begin{align*} \frac{1}{y^{\prime}}=\frac{xy}{x-y}=\left(y-\frac{1}{v}\right)y(-v)=y-y^2v\tag{3} \end{align*}

Putting (2) and (3) together we get \begin{align*} v^{\prime}=\left(y-y^2v-1\right)v^2 \end{align*} respectively \begin{align*} \color{blue}{v^{\prime}+y^2v^3-(y-1)v^2=0}\tag{4} \end{align*}

and observe (4) is an instance of an Abel equation of the first kind.

Markus Scheuer
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I don't think it is a homogeneous ordinary differential equation.

In order to be homogeneous, one should get the ODE in the form $\frac{dy}{dx}=F\left(\frac{y}{x}\right)$,

i.e the derivative of $y$ with respect to $x$ is some function of $\frac{y}{x}$.

If yo try that for your equation, you get the form $\frac{dy}{dx}=\frac{1-y/x}{y}$ which is clearly not a function of $y/x$ due to the presence of $y$ in the denominator.

gammanorm
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This equation is not homogeneous for $(x,y)$.

One simple test for homogeneity $\bigl(y' = F(\frac xy)\bigr)$ is if $y' = f(x,y) = f(tx,ty)$ for any real $t$. Here, this is not at all the case.

As a matter of fact, I don't think this is an ordinary differential equation whose family of solutions is in the form of common functions. If $y'$ were $1-y\over xy$ we could substitute $xy = r$ to get $$r' = \frac rx +\frac 1{\frac rx} - 1$$ which is homogeneous for $(x, r)$ but as posed I have a feeling that this DE does not have a "nice" solution.

Certainly not a dog
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