MHB Is This ODE Solvable by Substitution?

[find_the_fun]

Solve the ODE by making an appropriate substitution. [math](x-y)dx+xdy=0[/math]

So the textbook presents "three different kinds of first-order differential equations that are solvable by means of a substitution" so I guess my first step is to determine which of the 3 it is.

The first is homogeneous equation: if [math]f(tx, ty)=t^\alpha (x, y)[/math] I'm unclear how to test this on the question because I don't know what f is. Is it [math]f(x, y)=(x-y)+x[/math] where you just drop the dx and dy? If it is, then this works and [math]f(tx, ty)=(tx-ty)+tx=t(f(x,y))[/math]

For homoegeneous equations it's recommended to make the substitution y=ux so then the question becomes $$(x-ux)dx+x(xdu+udx)=0$$
$$(x-ux)dx+x^2 du + xudx =0$$
which simplifies to [math]x dx +x^2 du =0[/math] I don't know what to do next.

 

[chisigma]

... for homoegeneous equations it's recommended to make the substitution y=ux so then the question becomes $$(x-ux)dx+x(xdu+udx)=0$$
$$(x-ux)dx+x^2 du + xudx =0$$
which simplifies to [math]x dx +x^2 du =0[/math] I don't know what to do next.

... first You can further simplify writing...

$\displaystyle d x = - x\ d u \implies d u = - \frac{d x}{x}\ (1)$

... and now You can integrate both terms of (1)

Kind regards

$\chi$ $\sigma$

 

[Ackbach]

Solve the ODE by making an appropriate substitution. [math](x-y)dx+xdy=0[/math]

So the textbook presents "three different kinds of first-order differential equations that are solvable by means of a substitution" so I guess my first step is to determine which of the 3 it is.

The first is homogeneous equation: if [math]f(tx, ty)=t^\alpha (x, y)[/math] I'm unclear how to test this on the question because I don't know what f is. Is it [math]f(x, y)=(x-y)+x[/math] where you just drop the dx and dy? If it is, then this works and [math]f(tx, ty)=(tx-ty)+tx=t(f(x,y))[/math]

For homoegeneous equations it's recommended to make the substitution y=ux so then the question becomes $$(x-ux)dx+x(xdu+udx)=0$$
$$(x-ux)dx+x^2 du + xudx =0$$
which simplifies to [math]x dx +x^2 du =0[/math] I don't know what to do next.

Homogeneous DE's are of the form $f(x,y) \, dx+g(x,y) \, dy=0$, where $f$ and $g$ are homogeneous functions of the same type. That is,
\begin{align*}
f(\lambda x,\lambda y)&=\lambda^n f(x,y),\quad\text{and} \\
g(\lambda x,\lambda y)&=\lambda^n g(x,y).
\end{align*}

So in answer to your question about how to identify homogeneous DE's, you have a separate function for each coefficient of a differential.

Then you do a substitution as you've mentioned, and as chisigma has hinted at, a nice thing happens when you continue.

 

[MarkFL]

If I were to solve the give ODE, I would first arrange as:

$$x\d{y}{x}=y-x$$

Divide through by $x$:

$$\d{y}{x}=\frac{y}{x}-1$$

Use the substitution $$v=\frac{y}{x}\implies \d{y}{x}=v+x\d{v}{x}$$ and we now have:

$$v+x\d{v}{x}=v-1$$

Subtract through by $v$:

$$x\d{v}{x}=-1$$

Divide through by $x$:

$$\d{v}{x}=-\frac{1}{x}$$

This is equivalent to what is given by chisigma above.