Some Notable Product Differentials
As for first-order exact equations, some higher-order ODEs can also be found to be exact by grouping terms, usually as the derivative of some product involving the dependent variable or its derivatives. Some frequently recurring examples are:
\begin{align}
\frac{d}{dx}(y\frac{dy}{dx}) &= (\frac{dy}{dx})^2 + y\frac{d^2y}{dx^2} \tag{1} \\
\frac{d}{dx}[(\frac{dy}{dx})^2] &= 2\frac{dy}{dx}\frac{d^2y}{dx^2} \tag{2} \\
\frac{d}{dx}(y\frac{d^2y}{dx^2}) &= \frac{dy}{dx}\frac{d^2y}{dx^2} + y\frac{d^3y}{dx^3} \tag{3}
\end{align}
Example
Solve
\begin{align}
\frac{d^2y}{dx^2} + y\frac{dy}{dx} &= 0 \tag{4}
\end{align}
It is not hard to notice that \( y\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2}y^2) \), and thus the ODE is exact as
\begin{align}
\frac{d}{dx}(\frac{dy}{dx}) + \frac{d}{dx}(\frac{1}{2}y^2) &= 0 \\
\frac{dy}{dx} + \frac{1}{2}y^2 &= c_1 \tag{5}
\end{align}
This is a simple separable ODE and can be solved as
\begin{align}
\frac{dy}{dx} &= c_1 -\frac{1}{2}y^2 \\
\int \frac{dy}{c_1 -\frac{1}{2}y^2} &= \int dx \\
\frac{\sqrt{2}}{\sqrt{c_1}} \int \frac{d(\frac{y}{\sqrt{2c_1}})}{1 -(\frac{y}{\sqrt{2c_1}})^2} &= \int dx \\
\frac{\sqrt{2}}{\sqrt{c_1}} \tanh^{-1}(\frac{y}{\sqrt{2c_1}}) &= x + c_2 \\
y &= \sqrt{2c_1}\tanh(\frac{\sqrt{c_1}}{\sqrt{2}}(x + c_2)) \tag{6}
\end{align}
Exercise
Find the general solution family to the ODE
\begin{align}
y\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 + y\frac{dy}{dx} = x \tag{7}
\end{align}
Answer
Using the facts of (1) and \( y\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2}y^2) \) allows us to write
\begin{align}
\frac{d}{dx}(y\frac{dy}{dx}) + \frac{d}{dx}(\frac{1}{2}y^2) &= x \\
y\frac{dy}{dx} + \frac{1}{2}y^2 &= \int x dx = \frac{1}{2}x^2 + c_1
\end{align}
Then letting \( w = y^2 \) and using \( y\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2}y^2) = \frac{1}{2}\frac{dw}{dx}\) again produces
\begin{align}
\frac{1}{2}\frac{dw}{dx} + \frac{1}{2}w &= \frac{1}{2}x^2 + c_1 \\
\frac{dw}{dx} + w &= x^2 + 2c_1 \\
\frac{d}{dx}(e^{x}w) = e^{x}(\frac{dw}{dx} + w) &= x^2e^x + 2c_1 e^x \\
e^{x}w &= \int x^2e^x dx + \int 2c_1 e^x dx \\
e^{x}w &= [x^2e^x -\int e^xd(x^2)] + 2c_1 e^x + c_2 \\
e^{x}w &= [x^2e^x -\int 2xe^xdx] + 2c_1 e^x + c_2 \\
e^{x}w &= [x^2e^x -2xe^x + 2 \int e^x dx] + 2c_1 e^x + c_2 \\
e^{x}w &= e^x (x^2 -2x + 2) + 2c_1 e^x + c_2 \\
w &= x^2 -2x + 2 + 2c_1 + c_2e^{-x} \\
y &= \pm \sqrt{2c_1 + c_2e^{-x} + x^2 -2x + 2}
\end{align}
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