Equations Soluble for y’
For a higher-degree first-order ODE in the form of a polynomial in \( y’ \)
\begin{equation}
(y’)^n + a_{n-1}(x,y) (y’)^{n-1} + \cdots + a_1(x,y) y’ + a_0(x,y) = 0 \tag{1}
\end{equation}
where the coefficients \( a_k \) are now functions of \(x, y\), if it can be factorized into linear factors as
\begin{equation}
(y’ -f_1) (y’ -f_2) \cdots (y’ -f_n) = 0 \tag{2}
\end{equation}
where \( f_k = f_k(x,y) \) are also some functions of \(x, y\), then the general solution of (1) will be given by the product
\begin{equation}
g_1(x,y) g_2(x,y) \cdots g_n(x,y) = 0 \tag{3}
\end{equation}
where each of the forms \( g_k = 0 \) solves the corresponding first-degree equation \( y’ -f_k = 0 \) as suggested by the factors in (2). To see this, simply note that (3) implies that \( g_1(x,y) \) or \( g_2(x,y) \) or … or \( g_n(x,y) \) has to equal \( 0 \), which means that \( y’ -f_1 = 0 \) or \( y’ -f_2 = 0 \) or … or \( y’ -f_n = 0 \), translating to (2).
Example
The example is to solve the ODE
\begin{equation}
x^2 (y’)^2 -5 x yy’ + 6 y^2 = 0 \tag{4}
\end{equation}
It is not hard to check that it factorizes into two linear expressions:
\begin{align}
(xy’ -3y)(xy’ -2y) = 0 \tag{5}
\end{align}
(we can divide by \(x\) if needed to explicitly convert it into the form like (2)) The solution related to the first factor is
\begin{align}
x\frac{dy}{dx} -3y &= 0 \\
x^{-3}\frac{dy}{dx} -3x^{-4}y &= 0 \\
\frac{d}{dx}(x^{-3}y) &= 0 \\
x^{-3}y &= C \\
y &= Cx^3 \Rightarrow y -Cx^3 = 0 \tag{6}
\end{align}
Similarly, the solution for the second factor is \( y -Cx^2 = 0 \). Therefore the full solution is
\begin{align}
(y -Cx^3)(y -Cx^2) = 0 \tag{7}
\end{align}
Note that the arbitrary constants in these two solutions can be taken to be the same one, because first-order ODEs only require one integration constant (to be determined by one initial condition), and exactly only one factor in (3) needs to be satisfied.
Exercise
Solve
\begin{equation}
\sin(2x) (y’)^2 -2\cos(2x)yy’ -\sin(2x)y^2 = 0 \tag{8}
\end{equation}
Answer
Using the trigonometric identities \( \sin(2x) = 2\sin x\cos x\) and \( \cos(2x) = \cos^2 x -\sin^2 x \), we can express the ODE as
\begin{align}
(2\sin x\cos x) (y’)^2 -2(\cos^2 x -\sin^2 x )yy’ -(2\sin x\cos x)y^2 &= 0 \\
(\sin x\cos x) (y’)^2 -(\cos^2 x)yy’ + (\sin^2 x)yy’ -(\sin x\cos x)y^2 &= 0 \\
(\cos x) y’ ((\sin x)y’ -(\cos x)y) + (\sin x)y ((\sin x)y’ -(\cos x)y) &= 0 \\
((\cos x) y’ + (\sin x)y)((\sin x)y’ -(\cos x)y) &= 0
\end{align}
The solution to the linear ODE dictated by the first factor is computed as
\begin{align}
(\cos x) \frac{dy}{dx} + (\sin x)y &= 0 \\
\frac{dy}{dx} &= -(\tan x) y \\
\int \frac{dy}{y} &= -\int \frac{\sin x}{\cos x} dx \\
\ln |y| &= \int \frac{1}{\cos x} d(\cos x) \\
\ln |y| &= \ln|\cos x| + C \\
y &= \pm e^C \cos x = A \cos x
\end{align}
where we set \( A = \pm e^C \) as usual. Similarly the solution associated with the second factor is \( y = A \sin x \), and the full solution is thus
\begin{equation}
(y -A \cos x)(y -A\sin x) = 0
\end{equation}
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