The Ultimate Crash Course for STEM Majors exact differential equations free preview

Testing an Exact Differential Equation Using Partial Derivatives

This page is nothing more than a free preview from The Ultimate Crash Course for STEM Majors. This differential equations example demonstrates how to identify an ODE, formulate the exactness condition, calculate the required partial derivatives, and verify that an equation is exact.

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The above equation with dependent variable, \(y\) and independent variable \(x\) says that the ODE is first-order-linear-nonhomogeneous. We can put it into exact form from different books as well. We are in [21].

\(\mathcal{F}_1\) Identify the ODE “first-order-linear-nonhomogeneous” with potential exactness.

Next, identify the method needed to solve the ODE. In this case, we are told what to do. However, that may not be the case on exams. You should take account of ODE-formation to be prepped for exam alteration of structure—that is, the form will tell you what to use—usually.

\[ M(x,y)=\frac{y}{x}+6x, \qquad N(x,y)=(\color{red}{\ln x}-2). \]

\(\mathcal{F}_2\) Formulation of the theorem(s)/definition(s)/equation(s)

The equation is exact iff \(M_y=N_y\).

\[ M_y = \frac{\partial}{\partial y} \left( \frac{y}{x}+6x \right) = \frac{\partial}{\partial y}\frac{y}{x} + \frac{\partial}{\partial y}6x = \frac{1}{x}\frac{\partial}{\partial y}y + 6x\frac{\partial}{\partial y}(1) = \frac{1}{x}\frac{d}{dy}y + 6x\frac{d}{dx}(1) \]

\[ = \frac{1}{x}\frac{dy}{dy} + 6x(0) = \frac{1}{x}(1)+(0) = \color{#00a651}{\frac{1}{x}}. \]

\[ N_y = \frac{\partial}{\partial x}(\ln x-2) = \frac{\partial}{\partial x}\ln x – \frac{\partial}{\partial x}2 = \frac{d}{dx}\ln x-(0) = \frac{1}{x}\frac{d}{dx}x = \frac{1}{x}\frac{dx}{dx} = \color{#00a651}{\frac{1}{x}}. \]

Thus,

\[ M_y=\frac{1}{x}=N_x. \]

We can now see by the theorem that the equation is indeed exact.

So, how do we solve it? Well, this is essentially the same as a conservative function in multivariable calculus techniques. But we are not in calculus-MV, so we must stick with the book!!!

Let \(\psi_x=M\) and \(\psi_y=N\)

\[ \psi_x=M(x,y)=\frac{y}{x}+6x, \qquad \psi_y=N(x,y)=\ln x-2. \]

Now, we have a relation of antiderivative nature. Let us take the antiderivatives.

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