Free preview of The Ultimate Crash Course for STEM Majors exact differential equations lesson

Solving an Exact Differential Equation Using Antiderivatives and the Potential Function

This lesson is nothing more than a free preview of The Ultimate Crash Course for STEM Majors. This worked differential equations example continues the exact-equation method by taking antiderivatives, comparing the resulting potential functions, handling the logarithmic absolute value, and writing the final implicit solution.

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\[ \psi_x=\frac{y}{x}+6x, \qquad \psi_y=\ln x-2 \quad\Rightarrow\quad \psi=\int \frac{y}{x}+6x\,\partial x, \qquad \psi=\int \ln x-2\,\partial y \]

\[ \Rightarrow\quad \psi=\int \frac{y}{x}+6x\,dx, \qquad \psi=\int \ln x-2\,dy \]

\[ \Rightarrow\quad \psi=\int \frac{y}{x}\,dx+\int 6x\,dx, \qquad \psi=\int \ln x\,dy-\int 2\,dy \]

\[ \Rightarrow\quad \psi=y\int\frac{1}{x}\,dx+6\int x\,dx, \qquad \psi=\ln x\int dy-2\int dy \]

\[ \Rightarrow\quad \psi=y\ln|x|+6\left[\frac{1}{2}x^2\right]+g_1(y), \qquad \psi=\ln x[y]-2[y]+g_2(x) \]

We now have two equation that are the same but with mix and match variable.

\[ \psi = y\ln|x| + \color{#00a651}{3x^2} + \color{red}{g_1(y)}, \qquad \psi = y\ln x – \color{red}{2y} + g_2(x) \]

\[ \Rightarrow\quad \psi = y\ln|x| + \color{#00a651}{3x^2} + \color{red}{g_1(y)}, \qquad \psi = y\ln x + \color{red}{(-2y)} + g_2(x) \]

\[ \Rightarrow\quad \psi = y\ln|x| + \color{#00a651}{3x^2} + \color{red}{g_1(y)}, \qquad \psi = y\ln x + \color{red}{(-2y)} + \color{#00a651}{g_2(x)} \]

\[ h(x,y)=y\ln x. \]

Thus, you can omit the absolute value since the original statement states \(x>0\), we know that it is positive so we can drop the absolute value. (Don’t forget the scalar constant.)

\[ \therefore\quad \psi = y\ln x + \color{#00a651}{3x^2} – \color{red}{2y} + C. \]

NOTE I have written the answer as \(\psi=y\ln x+3x^2-2y+C\), however, the books answer is

\[ y\ln x+3x^2-2y=c. \]

[RP] Do you know why the book has equal to \(c\) and mine is not? Is this acceptable?

\(\mathcal{F}_3\) Finalize the answer stated correctly.

In a 3D calculus situation, it is common to add the constant and equate to a function.

\[ y\ln x+3x^2-2y=c \quad vs \quad \psi=y\ln x+3x^2-2y+C \quad \text{iff} \quad \psi=0, \qquad c=-C. \]

The former is the same relation—that is, one is state as a function that equals zero whereas the other is stated as a function which equals zero.

\[ \psi=y\ln x+3x^2-2y+C=0, \qquad y\ln x+3x^2-2y-c=0. \]

Continue Learning Exact Differential Equations

This lesson is nothing more than a free preview of The Ultimate Crash Course for STEM Majors. Explore the complete series for additional worked lessons covering exact differential equations, ordinary differential equations, calculus, mathematics, physics, engineering, and other STEM subjects.

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