Mathematical modeling / English

Exact Differential Equations - Example 1

Houston Math Prep · YouTube · 3:14

YouTube · Houston Math PrepOriginal video

IN THIS VIDEO

01Condition for an Exact Differential Equation
02Method for Solving Exact Equations: Step 1
03Method for Solving Exact Equations: Step 2

FROM THE LEARNING NOTES

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Condition for an Exact Differential Equation

SELECTED MOMENT

Boxing the answer and leaving it implicit

SUGGESTED PREPARATION

Partial derivatives
Integration of polynomial functions
Standard form of a first-order ODE
Differential equations in the form $M(x,y)dx + N(x,y)dy = 0$
Partial differentiation
Antidifferentiation with respect to one variable while treating the other as constant

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Reviewed learning material · Video analysis · English

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This video demonstrates the first steps of solving an exact differential equation. It begins by identifying the functions $M(x,y)$ and $N(x,y)$ from the equation (-3x^2 - y) $dx + (3y^2 - x)dy = 0$ . The instructor then verifies that the equation is exact by showing that the partial derivative of M with respect to y equals the partial derivative of N with respect to x, both being -1. Following this confirmation, the method for finding the potential function $F(x,y)$ is initiated. The instructor integrates M with respect to x, yielding -x^3 - xy + $c_1(y)$ , and then integrates N with respect to y, yielding $y^3$ - xy + $c_2(x)$ . The video concludes before these two results are combined to find the final general solution. This 44-second whiteboard segment completes the solution of the exact differential equation (-3x^2 - y) $dx + (3y^2 - x)dy = 0$ . The board already shows $M = -3x^2 - y$ , $N = 3y^2 - x$ , the exactness check ∂M/∂y = ∂N/∂ $x = -1$ , and the two antiderivatives $\int M dx = -x^3$ - xy + $c_1(y)$ and $\int N dy = y^3$ - xy + $c_2(x)$ . The presenter then combines the unique terms, explicitly noting that the shared term -xy is included only once, and writes the final implicit solution -x^ $3 + y^3$ - xy = C. Because the expression contains multiple y terms, the answer is left implicit rather than solved for y.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Problem Statement and Identifying M and N 0:35Checking for Exactness 1:22Integrating M with respect to x 1:59Integrating N with respect to y 2:30Review of exactness and antiderivatives 2:40Assembling the final implicit solution 2:57Boxing the answer and leaving it implicit

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The video presents the differential equation $(-3x^2 - y)dx + (3y^2 - x)dy = 0$ . The first step is to identify the components $M(x,y)$ and $N(x,y)$ , which are the coefficients of $dx$ and $dy$ respectively. Here, $M = -3x^2 - y$ and $N = 3y^2 - x$ .

To determine if the equation is solvable by the method for exact equations, we must check the condition for exactness. This requires verifying that the partial derivative of $M$ with respect to $y$ is equal to the partial derivative of $N$ with respect to $x$ . That is, $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

Calculating the partial derivatives: $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(-3x^2 - y) = 0 - 1 = -1$ . And $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(3y^2 - x) = 0 - 1 = -1$ . Since both partial derivatives equal $-1$ , the condition is met, and the differential equation is confirmed to be exact.

Because the equation is exact, there exists a potential function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$ . The general solution will be $F(x,y) = C$ . To find $F(x,y)$ , we begin by integrating $M$ with respect to $x$ : $F(x,y) = \int M dx = \int (-3x^2 - y) dx$ .

Performing the integration with respect to $x$ (treating $y$ as a constant), we get $\int -3x^2 dx = -x^3$ and $\int -y dx = -xy$ . The 'constant' of integration is not a simple number but an arbitrary function of the other variable, $y$ . Let's call this $c_1(y)$ . So, our first expression for the potential function is $F(x,y) = -x^3 - xy + c_1(y)$ .

Next, we integrate $N$ with respect to $y$ : $F(x,y) = \int N dy = \int (3y^2 - x) dy$ . This provides a second expression for the same potential function $F(x,y)$ .

Performing the integration with respect to $y$ (treating $x$ as a constant), we get $\int 3y^2 dy = y^3$ and $\int -x dy = -xy$ . The 'constant' of integration here is an arbitrary function of $x$ , which we'll call $c_2(x)$ . So, our second expression is $F(x,y) = y^3 - xy + c_2(x)$ .

The video ends here. The next logical step, not shown, would be to equate the two expressions for $F(x,y)$ to solve for the unknown functions $c_1(y)$ and $c_2(x)$ and thus find the complete potential function.

The board presents Example 1 for solving an exact differential equation. The equation is $(-3x^2-y)dx+(3y^2-x)dy=0$ , with $M=-3x^2-y$ and $N=3y^2-x$ . The exactness test has already been carried out: $\frac{\partial M}{\partial y}=-1$ and $\frac{\partial N}{\partial x}=-1$ , so the equation is labeled Exact.

Two antiderivative lines are visible: $\int M\,dx=-x^3-xy+c_1(y)$ and $\int N\,dy=y^3-xy+c_2(x)$ . Green boxes highlight the non-arbitrary parts of these expressions, indicating the terms that will be merged into the final answer.

The presenter now constructs the solution from those boxed terms. First comes $-x^3$ from the $x$ -integration, then $+y^3$ from the $y$ -integration. The term $-xy$ appears in both antiderivatives, and the narration explicitly says it should be included only once, not multiple times.

The completed relation is written as $-x^3+y^3-xy=C$ . The constant $C$ absorbs the arbitrary functions from the intermediate antiderivatives, yielding a single implicit equation.

Because the expression contains multiple powers and products of $y$ , the presenter chooses not to solve explicitly for $y$ . The final boxed answer is therefore left in implicit form: $-x^3+y^3-xy=C$ .

Knowledge cards

Condition for an Exact Differential Equation

A first-order ordinary differential equation of the form $M(x,y)dx + N(x,y)dy = 0$ is called an exact differential equation if there exists a function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M(x,y)$ and $\frac{\partial F}{\partial y} = N(x,y)$ . A necessary and sufficient condition for the equation to be exact (assuming $M$ and $N$ have continuous first partial derivatives) is that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ . This condition is checked first before attempting to solve the equation using the method for exact equations.

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Method for Solving Exact Equations: Step 1

Once an equation is confirmed to be exact, the goal is to find the potential function $F(x,y)$ . The first step is to integrate $M(x,y)$ with respect to $x$ . Because this is a partial integration, the 'constant' of integration is not a simple constant $C$ , but an arbitrary function of the other variable, $y$ . This is denoted as $c_1(y)$ . The result is an initial expression for the potential function: $F(x,y) = \int M(x,y) dx + c_1(y)$ .

F(x,y) = \int M(x,y) dx + c_1(y)

Method for Solving Exact Equations: Step 2

The second step in finding the potential function $F(x,y)$ is to integrate $N(x,y)$ with respect to $y$ . Similar to the first step, the 'constant' of integration is an arbitrary function of the other variable, $x$ , denoted as $c_2(x)$ . This gives a second expression for the potential function: $F(x,y) = \int N(x,y) dy + c_2(x)$ . This expression must be consistent with the one found in Step 1. By comparing the two expressions, one can determine the functions $c_1(y)$ and $c_2(x)$ and thus find the complete potential function $F(x,y)$ . The general solution to the differential equation is then $F(x,y) = C$ .

F(x,y) = \int N(x,y) dy + c_2(x)

The 'Constant' of Integration in Partial Integration

When performing an indefinite integral of a function of multiple variables with respect to one variable, all other variables are treated as constants. Consequently, the 'constant' of integration is not a single numerical constant, but rather any function that depends only on the variables being treated as constants. For example, when integrating $f(x,y)$ with respect to $x$ , the result is $\int f(x,y) dx + g(y)$ , where $g(y)$ is an arbitrary function of $y$ . This is a crucial concept in solving exact differential equations, as it introduces the unknown functions $c_1(y)$ and $c_2(x)$ that must be resolved in later steps.

Exact equation setup

The example begins with the differential equation $(-3x^2-y)dx+(3y^2-x)dy=0$ . The coefficient of $dx$ is $M=-3x^2-y$ , and the coefficient of $dy$ is $N=3y^2-x$ .

(-3x^2 - y)dx + (3y^2 - x)dy = 0

Exactness check

Exactness is verified by comparing mixed partial derivatives. The board shows $\frac{\partial M}{\partial y}=-1$ and $\frac{\partial N}{\partial x}=-1$ , so the equality condition holds and the equation is marked Exact.

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Antiderivatives used to build the solution

Integrating $M$ with respect to $x$ gives $-x^3-xy+c_1(y)$ , while integrating $N$ with respect to $y$ gives $y^3-xy+c_2(x)$ . These two lines supply the terms of the final implicit solution.

\int M\,dx = -x^3 - xy + c_1(y), \quad \int N\,dy = y^3 - xy + c_2(x)

Combine unique terms only once

When assembling the answer, the presenter takes $-x^3$ from the first antiderivative, $+y^3$ from the second, and the shared term $-xy$ only once. The narration explicitly warns against including that common term multiple times.

-x^3 + y^3 - xy = C

Final implicit solution

The solution is left as an implicit equation because there are multiple $y$ terms. The final boxed answer is $-x^3+y^3-xy=C$ , and the presenter does not solve directly for $y$ .

-x^3 + y^3 - xy = C

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 15

M

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes $M = -3x^2 - y$ .
Audio
Observation
He says, "So here our M is this expression here, so M is equal to negative 3x squared minus y."

Symbol

Meaning

The coefficient of dx in the differential equation.

Domain

A function of x and y.

N

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes $N = 3y^2 - x$ .
Audio
Observation
He says, "And then our expression N, the dy term here, so we have 3y squared minus x for N."

Symbol

Meaning

The coefficient of dy in the differential equation.

Domain

A function of x and y.

$\frac{\partial M}{\partial y}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes ∂ $M/y = -1$ .
Audio
Observation
He explains taking the partial derivative of M with respect to y.

Symbol

$\frac{\partial M}{\partial y}$

Meaning

The partial derivative of M with respect to y.

Domain

A real number or function.

$\frac{\partial N}{\partial x}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes ∂N/∂ $x = -1$ .
Audio
Observation
He explains taking the partial derivative of N with respect to x.

Symbol

$\frac{\partial N}{\partial x}$

Meaning

The partial derivative of N with respect to x.

Domain

A real number or function.

$c_1(y)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes + $c_1(y)$ at the end of the integral of M dx.
Audio
Observation
He says, "it's possible we could have some constant terms which are really in terms of y".

Symbol

$c_1(y)$

Meaning

An arbitrary function of y arising as the 'constant' of integration when integrating with respect to x.

Domain

A function of y.

$c_2(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes + $c_2(x)$ at the end of the integral of N dy.
Audio
Observation
He says, "plus possibly some constant type terms involving x".

Symbol

$c_2(x)$

Meaning

An arbitrary function of x arising as the 'constant' of integration when integrating with respect to y.

Domain

A function of x.

M

Clear evidence

Shown in the video

Evidence

Formula
Observation
$M = -3x^2 - y$

Symbol

Meaning

Coefficient of dx in the differential equation; treated as a function of x and y.

Domain

Function of x and y

N

Clear evidence

Shown in the video

Evidence

Formula
Observation
$N = 3y^2 - x$

Symbol

Meaning

Coefficient of dy in the differential equation; treated as a function of x and y.

Domain

Function of x and y

∂ $M/y$

Clear evidence

Shown in the video

Evidence

Formula
Observation
∂ $M/y = -1$

Symbol

∂ $M/y$

Meaning

Partial derivative of M with respect to y.

Domain

Real number

∂ $N/x$

Clear evidence

Shown in the video

Evidence

Formula
Observation
∂ $N/x = -1$

Symbol

∂ $N/x$

Meaning

Partial derivative of N with respect to x.

Domain

Real number

$\int M dx$

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int M dx = -x^3$ - xy + $c_1(y)$

Symbol

$\int M dx$

Meaning

Antiderivative of M with respect to x, including an arbitrary function of y.

Domain

Expression involving x, y, and $c_1(y)$

$c_1(y)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
$c_1(y)$

Symbol

$c_1(y)$

Meaning

Arbitrary function of y arising from integration with respect to x.

Domain

Function of y

Knowledge points · 9

Condition for Exactness

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor states, "the first thing we want to do is we want to check exactness. And the way we do that is we make sure that the partial of M with respect to y is equal to the partial of N with respect to x."
Formula
Observation
He writes and calculates ∂M/∂ $y = -1$ and ∂N/∂ $x = -1$ , concluding they are equal.

Definition

Explanation

A differential equation of the form $M(x,y)dx + N(x,y)dy = 0$ is exact if the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x.

Formula

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Conditions

The equation must be in the form $M(x,y)dx + N(x,y)dy = 0$ .
M and N must have continuous first-order partial derivatives.

Prerequisites

M
N
$\frac{\partial M}{\partial y}$
$\frac{\partial N}{\partial x}$

Step 1 of Solving an Exact Equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor says, "So we can use the steps to solve it, finding the potential function. We need to integrate M in terms of x."
Formula
Observation
He writes and calculates $\int M dx = -x^3$ - xy + $c_1(y)$ .

Method

Explanation

To find the potential function $F(x,y)$ , the first step is to integrate the function $M(x,y)$ with respect to x. The 'constant' of integration is an arbitrary function of y, denoted $c_1(y)$ .

Formula

F(x,y) = \int M(x,y) dx = ... + c_1(y)

Conditions

The differential equation has been verified as exact.

Prerequisites

Condition for Exactness
M
$c_1(y)$

Step 2 of Solving an Exact Equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor says, "So now integrating N dy, which means we're treating x as a constant."
Formula
Observation
He writes and calculates $\int N dy = y^3$ - xy + $c_2(x)$ .

Method

Explanation

The second step is to integrate the function $N(x,y)$ with respect to y. The 'constant' of integration is an arbitrary function of x, denoted $c_2(x)$ . This result must be consistent with the result from Step 1.

Formula

F(x,y) = \int N(x,y) dy = ... + c_2(x)

Conditions

The differential equation has been verified as exact.

Prerequisites

Condition for Exactness
N
$c_2(x)$

Form of the exact differential equation example

Clear evidence

Shown in the video

Evidence

Formula
Observation
(-3x^2 - y) $dx + (3y^2 - x)dy = 0$
Formula
Observation
$M = -3x^2 - y$ , $N = 3y^2 - x$

Definition

Explanation

The worked example starts from the differential equation (-3x^2 - y) $dx + (3y^2 - x)dy = 0$ , with M identified as the coefficient of dx and N as the coefficient of dy.

Formula

(-3x^2 - y)dx + (3y^2 - x)dy = 0

Conditions

M is the coefficient of dx
N is the coefficient of dy

Exactness condition for the example

Clear evidence

Shown in the video

Evidence

Formula
Observation
∂M/∂ $y = -1$
Formula
Observation
∂N/∂ $x = -1$
Formula
Observation
Exact

Method

Explanation

The video checks exactness by comparing the mixed partials ∂M/∂y and ∂N/∂x. Both are shown equal to -1, and the board labels the equation Exact.

Formula

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Conditions

M and N are differentiable enough for the displayed partial derivatives to exist
Equality of the mixed partials is used to justify exactness

Prerequisites

Form of the exact differential equation example

Integrating M with respect to x

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int M dx = -x^3$ - xy + $c_1(y)$

Method

Explanation

To recover part of the potential function, the video integrates $M = -3x^2 - y$ with respect to x, treating y as constant and adding an arbitrary function $c_1(y)$ .

Formula

\int M\,dx = -x^3 - xy + c_1(y)

Conditions

Integration is with respect to x
The constant of integration may depend on y

Prerequisites

Form of the exact differential equation example

Integrating N with respect to y

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int N dy = y^3$ - xy + $c_2(x)$

Method

Explanation

The video also integrates $N = 3y^2 - x$ with respect to y, treating x as constant and adding an arbitrary function $c_2(x)$ .

Formula

\int N\,dy = y^3 - xy + c_2(x)

Conditions

Integration is with respect to y
The constant of integration may depend on x

Prerequisites

Form of the exact differential equation example

Combining the two antiderivatives into one implicit solution

Clear evidence

Shown in the video

Evidence

Audio
Observation
Then includes all of my terms located here, includes all of my terms located here.
Audio
Observation
We only need to include that once, not multiple times.
Formula
Observation
-x^ $3 + y^3$ - xy = C

Method

Explanation

The final solution is assembled by taking the union of the nonconstant terms from the two antiderivatives: -x^3 from $\int M dx$ , $y^3$ from $\int N dy$ , and the shared term -xy included only once. The result is written as an implicit equation equal to a constant.

Formula

-x^3 + y^3 - xy = C

Conditions

Shared terms appearing in both antiderivatives are counted once
The arbitrary functions are absorbed into a single constant C in the final implicit form

Prerequisites

Integrating M with respect to x
Integrating N with respect to y

Leaving the answer in implicit form

Clear evidence

Shown in the video

Evidence

Audio
Observation
since we have multiple y terms here, I think we'll probably leave this and not solve this directly for y.
Formula
Observation
-x^ $3 + y^3$ - xy = C

Method

Explanation

Because the solution contains multiple powers and products of y, the presenter leaves the answer implicit rather than solving explicitly for y.

Formula

-x^3 + y^3 - xy = C

Conditions

The equation contains multiple y terms
An explicit solution for y is not attempted

Prerequisites

Combining the two antiderivatives into one implicit solution

Claims and conditions · 1

Equality of mixed partials implies the equation is exact in this example

Clear evidence

Shown in the video

Evidence

Formula
Observation
∂M/∂ $y = -1$ , ∂N/∂ $x = -1$ , Exact

Uncertainties

The video does not verbally state the full theorem hypothesis beyond displaying the equality and writing Exact.

Proposition

Statement

For the displayed equation, because ∂M/∂y = ∂N/∂ $x = -1$ , the equation is labeled Exact.

Hypotheses

$M = -3x^2 - y$
$N = 3y^2 - x$
∂M/∂y and ∂N/∂x are computed and compared

Quantifiers

For the specific M and N shown on the board.

Derivations and proofs · 4

Derivation: Checking for Exactness

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor calculates ∂M/∂y and ∂N/∂x on the board.
Audio
Observation
He verbally explains the differentiation process for each term.

Proof

Steps

Expression
$M = -3x^2 - y$
Explanation
Identify M from the given differential equation.
Justification
Given in the problem statement.
Shown in the video
Expression
$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(-3x^2 - y)$
Explanation
Set up the partial derivative of M with respect to y.
Justification
Definition of the exactness condition.
Shown in the video
Expression
$\frac{\partial M}{\partial y} = 0 - 1 = -1$
Explanation
Calculate the partial derivative. The term -3x^2 is treated as a constant with respect to y, so its derivative is 0. The derivative of -y with respect to y is -1.
Justification
Rules of partial differentiation.
Shown in the video
Expression
$N = 3y^2 - x$
Explanation
Identify N from the given differential equation.
Justification
Given in the problem statement.
Shown in the video
Expression
$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(3y^2 - x)$
Explanation
Set up the partial derivative of N with respect to x.
Justification
Definition of the exactness condition.
Shown in the video
Expression
$\frac{\partial N}{\partial x} = 0 - 1 = -1$
Explanation
Calculate the partial derivative. The term $3y^2$ is treated as a constant with respect to x, so its derivative is 0. The derivative of -x with respect to x is -1.
Justification
Rules of partial differentiation.
Shown in the video
Expression
$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = -1$
Explanation
Compare the two partial derivatives.
Justification
Direct comparison of the calculated results.
Shown in the video

Conclusion

Since the partial derivatives are equal, the differential equation is exact.

Derivation: Integrating M with respect to x

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes and solves the integral $\int M dx$ .
Audio
Observation
He explains the integration of each term with respect to x.

Proof

Steps

Expression
$\int M dx = \int (-3x^2 - y) dx$
Explanation
Set up the integral of M with respect to x.
Justification
Step 1 of the method for solving exact equations.
Shown in the video
Expression
$\int M dx = \int -3x^2 dx - \int y dx$
Explanation
Split the integral into two parts using linearity.
Justification
Linearity of the integral.
Derived from the video
Expression
$\int M dx = -x^3 - xy + c_1(y)$
Explanation
Integrate each term. The integral of -3x^2 is -x^3. The integral of -y with respect to x is -xy. Since y was treated as a constant, the constant of integration is an arbitrary function of y, $c_1(y)$ .
Justification
Power rule for integration and rules for integrating with respect to one variable.
Shown in the video

Conclusion

The partial solution for the potential function is $F(x,y) = -x^3$ - xy + $c_1(y)$ .

Derivation: Integrating N with respect to y

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes and solves the integral $\int N dy$ .
Audio
Observation
He explains the integration of each term with respect to y.

Proof

Steps

Expression
$\int N dy = \int (3y^2 - x) dy$
Explanation
Set up the integral of N with respect to y.
Justification
Step 2 of the method for solving exact equations.
Shown in the video
Expression
$\int N dy = \int 3y^2 dy - \int x dy$
Explanation
Split the integral into two parts using linearity.
Justification
Linearity of the integral.
Derived from the video
Expression
$\int N dy = y^3 - xy + c_2(x)$
Explanation
Integrate each term. The integral of $3y^2$ is $y^3$ . The integral of -x with respect to y is -xy. Since x was treated as a constant, the constant of integration is an arbitrary function of x, $c_2(x)$ .
Justification
Power rule for integration and rules for integrating with respect to one variable.
Shown in the video

Conclusion

The partial solution for the potential function is $F(x,y) = y^3$ - xy + $c_2(x)$ .

Derivation of the implicit solution for the exact equation example

Clear evidence

Shown in the video

Evidence

Formula
Observation
(-3x^2 - y) $dx + (3y^2 - x)dy = 0$
Formula
Observation
∂ $M/y = -1$ , ∂N/∂ $x = -1$ , Exact
Formula
Observation
$\int M dx = -x^3$ - xy + $c_1(y)$
Formula
Observation
$\int N dy = y^3$ - xy + $c_2(x)$
Formula
Observation
-x^ $3 + y^3$ - xy = C

Proof

Steps

Expression
$M = -3x^2 - y,\quad N = 3y^2 - x$
Explanation
Identify the coefficients of dx and dy from the given differential equation.
Justification
Direct reading from the board.
Shown in the video
Expression
$\frac{\partial M}{\partial y} = -1,\quad \frac{\partial N}{\partial x} = -1$
Explanation
Compute the mixed partial derivatives needed for the exactness test.
Justification
Differentiation of the displayed M and N.
Shown in the video
Expression
$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
Explanation
Compare the two partial derivatives and conclude the equation is exact.
Justification
Both displayed values are -1, and the board writes Exact.
Shown in the video
Expression
$\int M\,dx = -x^3 - xy + c_1(y)$
Explanation
Integrate M with respect to x while treating y as constant.
Justification
Antidifferentiation rule applied to -3x^2 - y.
Shown in the video
Expression
$\int N\,dy = y^3 - xy + c_2(x)$
Explanation
Integrate N with respect to y while treating x as constant.
Justification
Antidifferentiation rule applied to $3y^2 - x$ .
Shown in the video
Expression
$-x^3 + y^3 - xy = C$
Explanation
Combine the distinct terms from the two antiderivatives, counting the shared term -xy only once, and absorb the arbitrary functions into a single constant.
Justification
Audio explanation that the solution includes all terms from both lines but only needs the common term once.
Shown in the video

Conclusion

The exact differential equation has implicit solution -x^ $3 + y^3$ - xy = C.

Worked examples · 2

Example 1: Solving an Exact Differential Equation

Clear evidence

Shown in the video

Evidence

Formula
Observation
The entire video is dedicated to solving this single example.
Audio
Observation
The instructor narrates the process from start to finish.

Uncertainties

The final step of combining the results to find the general solution $F(x,y)=C$ is not shown in this clip.

Problem

Solve the differential equation (-3x^2 - y) $dx + (3y^2 - x)dy = 0$ .

Given

The differential equation (-3x^2 - y) $dx + (3y^2 - x)dy = 0$ .

Goal

Find the general solution by determining the potential function $F(x,y)$ .

Steps

Expression
$M = -3x^2 - y, \quad N = 3y^2 - x$
Explanation
Identify M and N from the standard form of the equation.
Justification
Definition of M and N.
Shown in the video
Expression
$\frac{\partial M}{\partial y} = -1, \quad \frac{\partial N}{\partial x} = -1$
Explanation
Calculate the partial derivatives to check for exactness.
Justification
Condition for an exact differential equation.
Shown in the video
Expression
$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
Explanation
Conclude that the equation is exact because the partial derivatives are equal.
Justification
Definition of an exact equation.
Shown in the video
Expression
$F(x,y) = \int M dx = \int (-3x^2 - y) dx = -x^3 - xy + c_1(y)$
Explanation
Integrate M with respect to x to find part of the potential function $F(x,y)$ .
Justification
Method for solving exact equations.
Shown in the video
Expression
$F(x,y) = \int N dy = \int (3y^2 - x) dy = y^3 - xy + c_2(x)$
Explanation
Integrate N with respect to y to find another part of the potential function $F(x,y)$ .
Justification
Method for solving exact equations.
Shown in the video

Answer

The clip ends before the final answer is stated. The two expressions for $F(x,y)$ are derived but not yet combined.

Verification

Verification would involve differentiating the final $F(x,y)$ to ensure ∂F/∂ $x = M$ and ∂F/∂ $y = N$ . This step is not performed in the clip.

Solving Exact Equations - Example 1

Clear evidence

Shown in the video

Evidence

Formula
Observation
Solving Exact Equations - Example 1
Formula
Observation
(-3x^2 - y) $dx + (3y^2 - x)dy = 0$
Formula
Observation
-x^ $3 + y^3$ - xy = C

Problem

Solve the differential equation (-3x^2 - y) $dx + (3y^2 - x)dy = 0$ .

Given

Equation: (-3x^2 - y) $dx + (3y^2 - x)dy = 0$
$M = -3x^2 - y$
$N = 3y^2 - x$

Goal

Find the implicit solution of the exact differential equation.

Steps

Expression
$\frac{\partial M}{\partial y} = -1,\quad \frac{\partial N}{\partial x} = -1$
Explanation
Check exactness by comparing mixed partials.
Justification
Exactness test shown on the board.
Shown in the video
Expression
$\int M\,dx = -x^3 - xy + c_1(y)$
Explanation
Integrate the dx-coefficient with respect to x.
Justification
Standard step in solving an exact equation.
Shown in the video
Expression
$\int N\,dy = y^3 - xy + c_2(x)$
Explanation
Integrate the dy-coefficient with respect to y.
Justification
Standard step in solving an exact equation.
Shown in the video
Expression
$-x^3 + y^3 - xy = C$
Explanation
Collect all unique terms from the two antiderivatives and write the result equal to a constant.
Justification
Audio states that shared terms are included once and the final expression is set equal to some constant.
Shown in the video

Answer

-x^ $3 + y^3$ - xy = C

Verification

The video verifies exactness by showing ∂M/∂y = ∂N/∂ $x = -1$ before presenting the final implicit solution.

Visual events · 5

Writing the Initial Equation

Clear evidence

Shown in the video

Evidence

Animation
Observation
The differential equation is written on the screen.

Objects

Differential equation (-3x^2 - y) $dx + (3y^2 - x)dy = 0$

Changes

The equation appears on the screen.

Invariants

The equation itself does not change.

Interpretation

This establishes the problem to be solved.

Verifying Exactness

Clear evidence

Shown in the video

Evidence

Animation
Observation
The instructor writes the partial derivatives ∂M/∂y and ∂N/∂x, calculates them to be -1, and writes the word 'Exact'.

Objects

Expressions for ∂M/∂y and ∂N/∂x
The word 'Exact'

Changes

The partial derivatives are calculated and shown to be equal.
The conclusion 'Exact' is written.

Invariants

The original equation remains on screen.

Interpretation

This visual sequence demonstrates the test for whether the differential equation is exact.

Finding the Potential Function

Clear evidence

Shown in the video

Evidence

Animation
Observation
The instructor writes and solves the integrals $\int M dx$ and $\int N dy$ .

Objects

Integral expressions $\int M dx$ and $\int N dy$
Results of the integrations

Changes

The integral of M with respect to x is calculated.
The integral of N with respect to y is calculated.

Invariants

The original equation and the exactness check remain on screen.

Interpretation

This shows the two parallel processes of integrating M and N to build the potential function $F(x,y)$ .

Highlighting the terms to be combined

Clear evidence

Shown in the video

Evidence

Animation
Observation
Green boxes are drawn around -x^3 - xy in the first integral line and around $y^3$ - xy in the second integral line.

Objects

Integral expression $\int M dx = -x^3$ - xy + $c_1(y)$
Integral expression $\int N dy = y^3$ - xy + $c_2(x)$
Green rectangular boxes

Changes

Boxes appear around the non-arbitrary-function parts of each antiderivative
The shared term -xy is visibly present in both boxed expressions

Invariants

The arbitrary functions $c_1(y)$ and $c_2(x)$ remain outside the boxes
The earlier exactness work remains visible above

Interpretation

The visual emphasis identifies which terms from the two antiderivatives will be merged into the final implicit solution.

Writing and boxing the final answer

Clear evidence

Shown in the video

Evidence

Animation
Observation
The presenter writes -x^3, then + $y^3$ , then - xy, then = C beneath the boxed expressions.
Animation
Observation
A green box is drawn around the completed final equation.

Objects

New line below the integrals
Final equation -x^ $3 + y^3$ - xy = C
Green box around the final equation

Changes

Terms are added sequentially from left to right
The completed solution is enclosed in a box

Invariants

The upper board content remains unchanged while the final line is written

Interpretation

The animation shows the assembly of the final implicit solution from the previously identified terms.

Misconceptions · 3

Constant of Integration in Partial Integration

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor explicitly states that the constant of integration is a function of the other variable: "it's possible we could have some constant terms which are really in terms of y".
Formula
Observation
He writes $c_1(y)$ and $c_2(x)$ instead of a simple constant C.

Misconception

When integrating a multivariable function with respect to one variable, the constant of integration is just a number.

Clarification

The constant of integration is an arbitrary function of the other variables. When integrating with respect to x, the 'constant' is a function of y, $c_1(y)$ . When integrating with respect to y, the 'constant' is a function of x, $c_2(x)$ .

Do not duplicate a term that appears in both antiderivatives

Clear evidence

Shown in the video

Evidence

Audio
Observation
and then they both have a minus xy. We only need to include that once, not multiple times.

Misconception

When combining the results of $\int M dx$ and $\int N dy$ , one might incorrectly write the shared term -xy twice.

Clarification

The video explicitly says the common term -xy should be included only once in the final solution.

Implicit form is acceptable when solving for y is cumbersome

Clear evidence

Shown in the video

Evidence

Audio
Observation
since we have multiple y terms here, I think we'll probably leave this and not solve this directly for y.

Misconception

One might assume the final answer must always be solved explicitly for y.

Clarification

Because the expression contains multiple y terms, the presenter leaves the answer in implicit form instead of solving directly for y.

Concept relations · 5

Condition for Exactness → Step 1 of Solving an Exact Equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
After confirming the equation is exact, the instructor says, "So we can use the steps to solve it..."

Prerequisite

Explanation

Verifying that the equation is exact is a necessary first step before applying the method of finding a potential function by integration.

Step 1 of Solving an Exact Equation → Step 2 of Solving an Exact Equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor moves directly from writing the result of $\int M dx$ to starting the calculation for $\int N dy$ .

Contains

Explanation

The overall method for solving an exact equation contains these two sequential integration steps, whose results must be consistent to form the final potential function.

Exactness condition for the example → Integrating M with respect to x

Clear evidence

Shown in the video

Evidence

Formula
Observation
∂M/∂ $y = -1$ , ∂N/∂ $x = -1$ , Exact, followed by $\int M dx$ and $\int N dy$

Prerequisite

Explanation

The video first verifies exactness and only then proceeds to integrate M and N to build the solution.

Integrating N with respect to y → Combining the two antiderivatives into one implicit solution

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int M dx = -x^3$ - xy + $c_1(y)$
Formula
Observation
$\int N dy = y^3$ - xy + $c_2(x)$
Audio
Observation
Then includes all of my terms located here, includes all of my terms located here.

Application

Explanation

The second antiderivative supplies the $y^3$ term and confirms the shared -xy term used in forming the final implicit solution.

Combining the two antiderivatives into one implicit solution → Leaving the answer in implicit form

Clear evidence

Shown in the video

Evidence

Audio
Observation
I think we'll probably leave this and not solve this directly for y.
Formula
Observation
-x^ $3 + y^3$ - xy = C

Contains

Explanation

The combined expression is presented as the final answer in implicit form rather than being rearranged to isolate y.

Find an answer · 7

How do you check if a differential equation is exact?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor explains the process of checking for exactness.

Knowledge points

Condition for Exactness

Why is the constant of integration a function of y when integrating M with respect to x?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor explains why the constant of integration is a function of y.

Knowledge points

Constant of Integration in Partial Integration
Step 1 of Solving an Exact Equation

What are the steps to solve an exact differential equation once you've confirmed it's exact?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor outlines the steps to solve the equation after confirming it is exact.

Knowledge points

Step 1 of Solving an Exact Equation
Step 2 of Solving an Exact Equation

How do you check whether (-3x^2 - y) $dx + (3y^2 - x)dy = 0$ is exact?

Clear evidence

Shown in the video

Evidence

Formula
Observation
∂M/∂ $y = -1$ , ∂N/∂ $x = -1$ , Exact

Knowledge points

Form of the exact differential equation example
Exactness condition for the example

Why does integrating M with respect to x produce $c_1(y)$ instead of just a constant?

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int M dx = -x^3$ - xy + $c_1(y)$

Knowledge points

Integrating M with respect to x

Why is the term -xy included only once in the final solution?

Clear evidence

Shown in the video

Evidence

Audio
Observation
We only need to include that once, not multiple times.

Knowledge points

Combining the two antiderivatives into one implicit solution
Do not duplicate a term that appears in both antiderivatives

Why does the presenter leave the answer implicit instead of solving for y?

Clear evidence

Shown in the video

Evidence

Audio
Observation
since we have multiple y terms here, I think we'll probably leave this and not solve this directly for y.

Knowledge points

Leaving the answer in implicit form
Implicit form is acceptable when solving for y is cumbersome

Coverage and review notes

Covered · Introduction of the problem and identification of M and N.

Covered · Checking the condition for exactness by calculating partial derivatives.

Covered · First step of the solution method: integrating M with respect to x.

Covered · Second step of the solution method: integrating N with respect to y. The clip ends before the final solution is assembled.

Covered · Board already shows the example equation, exactness check, and both antiderivatives; the presenter begins explaining which terms will be included.

Covered · The presenter writes the final expression term by term and explains that the shared -xy term is included only once.

Covered · The completed implicit solution is boxed and the presenter states that it will not be solved directly for y.

Explore the knowledge in this video

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Exact differential equations ExplanationAt 0:35
Why this connection?
Reviewed English material v1: card "Condition for an Exact Differential Equation" at 0:35 — A first-order ordinary differential equation of the form $M(x,y)dx + N(x,y)dy = 0$ is called an exact differential equation if there exists a function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M(x,y)$ and $\frac{\partial F}{\partial y} = N(x,y)$ . A necessary and sufficient condition for the equation to be exact (assuming $M$ and $N$ have continuous first partial derivatives) is that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ . This condition is checked first before attempting to solve the equation using the method for exact equations.

Questions this video answers

Meet the concept

Understand why

Understand why

Understand why

Find a method

Meet the concept

Exact Differential Equations - Example 1

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Condition for an Exact Differential Equation

Method for Solving Exact Equations: Step 1

Method for Solving Exact Equations: Step 2

The 'Constant' of Integration in Partial Integration

Exact equation setup

Exactness check

Antiderivatives used to build the solution

Combine unique terms only once

Final implicit solution

Detailed learning notes

M

N

∂M∂y\frac{\partial M}{\partial y}∂y∂M​

∂N∂x\frac{\partial N}{\partial x}∂x∂N​

c1(y)c_1(y)c1​(y)

c2(x)c_2(x)c2​(x)

M

N

∂M/yM/yM/y

∂N/xN/xN/x

∫Mdx\int M dx∫Mdx

c1(y)c_1(y)c1​(y)

Condition for Exactness

Step 1 of Solving an Exact Equation

Step 2 of Solving an Exact Equation

Form of the exact differential equation example

Exactness condition for the example

Integrating M with respect to x

Integrating N with respect to y

Combining the two antiderivatives into one implicit solution

Leaving the answer in implicit form

Equality of mixed partials implies the equation is exact in this example

Derivation: Checking for Exactness

Derivation: Integrating M with respect to x

Derivation: Integrating N with respect to y

Derivation of the implicit solution for the exact equation example

Example 1: Solving an Exact Differential Equation

Solving Exact Equations - Example 1

Writing the Initial Equation

Verifying Exactness

Finding the Potential Function

Highlighting the terms to be combined

Writing and boxing the final answer

Constant of Integration in Partial Integration

Do not duplicate a term that appears in both antiderivatives

Implicit form is acceptable when solving for y is cumbersome

Condition for Exactness → Step 1 of Solving an Exact Equation

Step 1 of Solving an Exact Equation → Step 2 of Solving an Exact Equation

Exactness condition for the example → Integrating M with respect to x

Integrating N with respect to y → Combining the two antiderivatives into one implicit solution

Combining the two antiderivatives into one implicit solution → Leaving the answer in implicit form

How do you check if a differential equation is exact?

Why is the constant of integration a function of y when integrating M with respect to x?

What are the steps to solve an exact differential equation once you've confirmed it's exact?

How do you check whether (-3x^2 - y)dx+(3y2−x)dy=0dx + (3y^2 - x)dy = 0dx+(3y2−x)dy=0 is exact?

Why does integrating M with respect to x produce c1(y)c_1(y)c1​(y) instead of just a constant?

Why is the term -xy included only once in the final solution?

Why does the presenter leave the answer implicit instead of solving for y?

Explore the knowledge in this video

Questions this video answers

What is the final implicit solution for the example equation?+

Why is the term -xy included only once in the final solution?+

Why is the constant of integration a function of y when integrating M with respect to x?+

Why does the presenter leave the answer implicit instead of solving for y?+

How do you check if a differential equation is exact?+

What are the steps to solve an exact differential equation once you've confirmed it's exact?+

More in this subject

$\frac{\partial M}{\partial y}$

$\frac{\partial N}{\partial x}$

$c_1(y)$

$c_2(x)$

∂ $M/y$

∂ $N/x$

$\int M dx$

$c_1(y)$

How do you check whether (-3x^2 - y) $dx + (3y^2 - x)dy = 0$ is exact?

Why does integrating M with respect to x produce $c_1(y)$ instead of just a constant?