Equations Soluble for y (or x)
First-order ODEs soluble for \( y \), such that
\begin{equation}
y = f(x, y’) \tag{1}
\end{equation}
can be reduced to first-degree first-order equations in \( y’ \) by taking a derivative on both sides with respect to \( x \) via Chain Rule for multiple variables, yielding
\begin{align}
y’ = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y’}\frac{dy’}{dx} \tag{2}
\end{align}
Denote the solution of (2) by \( g(x,y’) = 0 \), it can be combined with (1) to eliminate \( y’ \) and obtain the general solution. An extra singular solution may also be generated alongside. Similarly, if an ODE is soluble for \( x \), such that \( x = h(y, y’) \), then we can use (try to show it!)
\begin{align}
\frac{1}{y’} = \frac{\partial h}{\partial y} + \frac{\partial h}{\partial y’}\frac{dy’}{dy} \tag{3}
\end{align}
Example
To illustrate the idea, we will solve
\begin{equation}
y = x^3(y’)^2 -2xy’ \tag{4}
\end{equation}
By (2), we have
\begin{align}
y’ &= (3x^2(y’)^2 -2y’) + (2x^3y’ -2x)\frac{dy’}{dx} \\
0 &= (3x^2(y’)^2 -3y’) + 2x(x^2y’ -1)\frac{dy’}{dx} \\
0 &= (3y’ + 2x\frac{dy’}{dx})(x^2y’ -1) \tag{5}
\end{align}
After factorization, the general solution will be retrieved by solving the ODE indicated by the first factor:
\begin{align}
3y’ + 2x\frac{dy’}{dx} &= 0 \\
x\frac{dy’}{dx} &= -\frac{3}{2}y’ \\
\int\frac{dy’}{y’} &= -\int\frac{3}{2x} dx \\
\ln |y’| &= -\frac{3}{2}\ln |x| + C \\
y’ &= \pm e^Cx^{-3/2} = Ax^{-3/2} \tag{6}
\end{align}
as we set \( A = \pm e^C \). Subsequently, substituting (6) into (4) gives
\begin{align}
y &= x^3(Ax^{-3/2})^2 -2xAx^{-3/2} \\
y &= A^2 -2Ax^{-1/2} \tag{7}
\end{align}
Meanwhile, the second factor in (5) represents the singular solution, \(y’ = 1/x^2\), and plugging it into (4) produces (or by a direct integration)
\begin{align}
y &= x^3(\frac{1}{x^2})^2 -2x(\frac{1}{x^2}) \\
&= \frac{1}{x} -\frac{2}{x} = -\frac{1}{x} \tag{8}
\end{align}
Exercise
Solve the ODE
\begin{equation}
y^2 -yy’ -x(y’)^2 = 0 \tag{9}
\end{equation}
Answer
Rearranging easily shows that the ODE is soluble in \( x \):
\begin{align}
x = (\frac{y}{y’})^2 -\frac{y}{y’} = (\frac{y}{p})^2 -\frac{y}{p}
\end{align}
For notational clarity we have written \( p = y’ \). By (3), we have
\begin{align}
\frac{1}{p} &= (2\frac{y}{p^2} -\frac{1}{p}) + (-2\frac{y^2}{p^3} + \frac{y}{p^2})\frac{dp}{dy} \\
0 &= (2\frac{y}{p^2} -\frac{2}{p}) + (-2\frac{y^2}{p^3} + \frac{y}{p^2})\frac{dp}{dy} \\
0 &= (2\frac{y}{p} -2) + (-2\frac{y^2}{p^2} + \frac{y}{p})\frac{dp}{dy}
\end{align}
This is a homogeneous ODE and we can let \( v = p/y \) (see this tutorial):
\begin{align}
(2\frac{1}{v} -2) + (-2\frac{1}{v^2} + \frac{1}{v})(v+y\frac{dv}{dy}) &= 0 \\
(2v -2v^2) + (-2 + v)(v+y\frac{dv}{dy}) &= 0 \\
v+y\frac{dv}{dy} &= \frac{2v^2 -2v}{v -2} \\
y\frac{dv}{dy} &= \frac{2v^2 -2v -v(v-2)}{v -2} = \frac{v^2}{v -2} \\
\int \frac{(v-2)dv}{v^2} &= \int \frac{dy}{y} \\
\int (\frac{1}{v} -\frac{2}{v^2}) dv &= \ln |y| + C \\
\ln |v| + \frac{2}{v} &= \ln |y| + C \\
ve^{2/v} &= Ay \\
(\frac{p}{y})e^{2y/p} &= Ay
\end{align}
where we set \( A = \pm e^C \) as usual. Note that the ODE can also be written as a quadratic equation in \( p \), so by high-school algebra:
\begin{align}
0 &= y^2 -yp -xp^2 \\
\Rightarrow p &= \frac{-y \pm \sqrt{y^2 + 4xy^2}}{2x} \\
\frac{p}{y} &= \frac{-1 \pm \sqrt{1 + 4x}}{2x} = \frac{2}{\pm \sqrt{4x+1} + 1}
\end{align}
So
\begin{align}
\frac{2}{\pm \sqrt{4x+1} + 1}e^{\pm \sqrt{4x+1} + 1} &= Ay \\
\Rightarrow y &= \frac{Be^{\pm \sqrt{4x+1}}}{\sqrt{4x+1} \pm 1}
\end{align}
where \( B= 2e/A \).
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