Is This ODE Solvable by Substitution?

[find_the_fun]

Solve the ODE by making an appropriate substitution. \(\displaystyle (x-y)dx+xdy=0\)

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

For homoegeneous equations it's recommended to make the substitution y=ux so then the question becomes \(\displaystyle (x-ux)dx+x(xdu+udx)=0\)
\(\displaystyle (x-ux)dx+x^2 du + xudx =0\)
which simplifies to \(\displaystyle x dx +x^2 du =0\) 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 \(\displaystyle (x-ux)dx+x(xdu+udx)=0\)
\(\displaystyle (x-ux)dx+x^2 du + xudx =0\)
which simplifies to \(\displaystyle x dx +x^2 du =0\) 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. \(\displaystyle (x-y)dx+xdy=0\)

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

For homoegeneous equations it's recommended to make the substitution y=ux so then the question becomes \(\displaystyle (x-ux)dx+x(xdu+udx)=0\)
\(\displaystyle (x-ux)dx+x^2 du + xudx =0\)
which simplifies to \(\displaystyle x dx +x^2 du =0\) 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:

\(\displaystyle x\d{y}{x}=y-x\)

Divide through by $x$:

\(\displaystyle \d{y}{x}=\frac{y}{x}-1\)

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

\(\displaystyle v+x\d{v}{x}=v-1\)

Subtract through by $v$:

\(\displaystyle x\d{v}{x}=-1\)

Divide through by $x$:

\(\displaystyle \d{v}{x}=-\frac{1}{x}\)

This is equivalent to what is given by chisigma above.

 

[blue_raver22]

The second type is exact equations: if M(x,y)dx+N(x,y)dy=0 where M_y=N_x, then the equation can be solved by finding a function F(x,y) such that F_x=M and F_y=N. In this case, the equation is not exact as M_y=1 and N_x=1, so there is no function F(x,y) that satisfies the condition.

The third type is linear equations: if the equation can be written in the form dy/dx+p(x)y=q(x), then the substitution y=v(x)u(x) can be made to solve the equation. However, in this case, the equation does not fit this form as there is no p(x) term.

In conclusion, none of the three types of equations discussed in the textbook apply to this specific equation. It is possible that there may be another substitution or method that can be used to solve this equation, but without more context or information, it is difficult to determine the most appropriate approach.