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. 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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