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Ordinary Differential Equations MA26600 Spring 2026

Jakayla Robbins

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Handout Lesson 8, Exact Differential Equations and More Substitutions

Textbook Section(s).

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This lesson is based on Section 1.6 of your textbook by Edwards, Penney, and Calvis.
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What is an Exact Differential Equation?

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We are going to do something a little different today. We are going to work backwards. In other words, I am going to tell you the solution, and we are going to find the differential equation associated with that solution.
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Suppose that you know the solution of your differential equation has the form:
\begin{equation*} F(x,y(x))=C \end{equation*}
What is a differential equation for which this is a solution? Let’s start by differentiating both sides of this equation with respect to \(x\text{.}\)
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Definition 47.

IF there is a a function \(F(x,y)\) that satisfies
\begin{equation*} \frac{\partial F}{\partial x}=M(x,y) \qquad {\text and } \qquad \frac{\partial F}{\partial y}=N(x,y) \end{equation*}
for all \((x,y)\) in a rectangle \(R\text{,}\)
\begin{gather} M(x,y)dx+N(x,y)dy=0\tag{✶} \end{gather}
is an exact differential equation and
\begin{equation*} F(x,y)=C \end{equation*}
is a solution to (✶).
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Moving forward, we will, of course, start with the equation and be asked to find the solution. So how do we know if a differential equation of the form in (✶) is exact?
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Example 49. Structure of exact differential equations.

Consider the differential equation
\begin{gather} (5x^4y)dx+ (x^5)dy = 0\tag{†} \end{gather}
  1. Is (†) an exact differential equation?
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  2. Is (†) an exact differential equation? Divide both sides of (†) by \(x^4\text{.}\) Is the resulting differential equation exact?
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  3. What are you supposed to learn from this example?
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Solving Exact Differential Equations.

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Given: An exact differential equation
\begin{gather} M(x,y)dx+N(x,y)dy=0 \tag{✶✶} \end{gather}
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We are seeking a function \(F(x,y)\) that satisfies
\begin{gather} \frac{\partial F}{\partial x}=M(x,y)\tag{#} \end{gather}
and
\begin{gather} \frac{\partial F}{\partial y}=N(x,y)\tag{##} \end{gather}
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If we have function \(F\) of the form
\begin{gather} F(x,y)=\int M(x,y) \, dx + g(y)\tag{✠} \end{gather}
then (#) is satisfied. We will use (##) to find \(g(y)\) and complete our solution, \(F(x,y)=C\text{,}\) for (✶✶).
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Example 50. Solving exact differential equations.

(Number 32 from Section 1.6 of your textbook by Edwards, et.al.) Solve.
\begin{equation*} (4x-y)dx+(6y-x)dy=0 \end{equation*}
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Example 51. At home: Solving exact differential equations.

(Number 36 from Section 1.6 of your textbook by Edwards, et.al.)
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Try the following problem at home.
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Verify that the differential equation is exact. Then solve it.
\begin{equation*} (1+ye^{xy})dx+(2y+xe^{xy})dy=0 \end{equation*}
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Answers: The equation is exact because \(M(x,y)\) and \(N(x,y)\) are continuous on \(\mathbb{R}^2\) and \(M_y=N_x=(1+xy)e^{xy}\text{.}\) The solutions can be written in the form \(x+e^{xy}+y^2=C\text{.}\)
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Reducible Second-order Differential Equations.

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Definition 52.

A second-order differential equation involves the second derivative, but no higher derivative, and can be written in the form:
\begin{equation*} F(x,y,y',y'')=0 \end{equation*}
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If the differential equation contains either no \(x\)’s or no \(y\)’s, then we can use the substitution:
\begin{equation*} p=y'=\frac{dy}{dx} \end{equation*}
to reduce the differential equation from a second-order differential equation to a first-order differential equation.
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Case 1: \(F(x,y',y'')=0\).

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If we have a second order differential equation that does not contain the dependent variable \(y\text{,}\) then the substitution \(p=y'=\frac{dy}{dx}\) gives
\begin{equation*} y''=\frac{dp}{dx} \end{equation*}
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Example 53. Reducible second-order differential equation with no \(y\) term.

(Number 43 from Section 1.6 of your textbook by Edwards, et.al.)
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Find a general solutions of the reducible second-order differential equation. You may assume \(x\text{,}\) \(y\text{,}\) and \(y'\) are positive if it is helpful.
\begin{equation*} xy''=y' \end{equation*}
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Case 2: \(F(y,y',y'')=0\).

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If we have a second order differential equation that does not contain the independent variable \(x\text{,}\) then the substitution \(p=y'=\frac{dy}{dx}\) gives
\begin{equation*} y''=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p\frac{dp}{dy} \end{equation*}
by the chain rule.
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Example 54. Reducible second-order differential equation with no \(x\) term.

(Number 51 from Section 1.6 of your textbook by Edwards, et.al.)
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Find a general solutions of the reducible second-order differential equation. You may assume \(x\text{,}\) \(y\text{,}\) and \(y'$\) are positive if it is helpful, and your solution may be implicit.
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\begin{equation*} y''=2y(y')^3 \end{equation*}
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A bit of homework help.

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Example 55. Hint for written homework problem 52 in section 1.6.

(Example 4 from Section 2.6 of the textbook by Nagle, Saff, and Snider)
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Consider the differential equation
\begin{gather} \frac{dy}{dx}=\frac{-2x+2y+6}{x-3y-5}\tag{‡} \end{gather}
Find constants \(h\) and \(k\) so that the substitutions
\begin{equation*} x=u+h \end{equation*}
and
\begin{equation*} y=v+k \end{equation*}
transform (‡) into the homogeneous equation
\begin{equation*} \frac{dv}{du}=\frac{2v-2u}{u-3v}\text{.} \end{equation*}
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