Commonly Seen Differentials to be Grouped
Notice that we have simple formulas for some recurring differentials:
- \( d(x/y) = (ydx -xdy)/y^2 \)
- \( d(xy) = xdy + ydx \)
- \( d(x^2 + y^2) = 2(xdx + ydy) \)
- \( d(\tan^{-1}(x/y) ) = (ydx -xdy)/(x^2+y^2) \)
- \( d(\ln(x/y)) = (ydx -xdy)/xy \)
where the roles of \(x\) and \(y\) can be interchanged. So, we can rewrite and solve some ODEs by grouping terms into these forms. Particularly, if the expression \( ydx -xdy \) appears, it can be divided by \(x^2\), \(xy\), \(y^2\), \(x^2 + y^2\) respectively to produce other new differentials (1,4,5 above).
Example
A small example is to solve
\begin{equation}
x^2y dy -(x dy -ydx) = 0 \tag{1}
\end{equation}
We can utilize (1) in the previous list (with \(x,y\) swapped):
\begin{equation}
d(\frac{y}{x}) = \frac{xdy -ydx}{x^2} \tag{2}
\end{equation}
Dividing both sides of (1) by \(x^2\) and using (2), we arrive at the solution:
\begin{align}
y dy -\frac{x dy -ydx}{x^2} &= 0 \\
d(\frac{1}{2}y^2) -d(\frac{y}{x}) &= 0 \\
\frac{1}{2}y^2 -\frac{y}{x} &= C \tag{3}
\end{align}
where we have integrated the two differentials.
Exercise
Find the general solution of
\begin{equation}
ydx -xdy = x^3ydx
\end{equation}
Answer
Dividing both sides by \(xy\) leads to
\begin{align}
\frac{ydx -xdy}{xy} &= x^2dx \\
d(\ln(\frac{x}{y})) &= d(\frac{1}{3}x^3) \\
\ln(\frac{x}{y}) &= \frac{1}{3}x^3 + C \\
\frac{x}{y} &= e^{\frac{1}{3}x^3 + C} \\
y &= x e^{-(\frac{1}{3}x^3 + C)} = Ax e^{-\frac{x^3}{3}}
\end{align}
where we set \( A = e^{-C} \).
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