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# Exact Equations – General Form, Solutions, and Examples

**Exact equations **are unique differential equations that satisfy certain conditions leading to a simpler way to find their corresponding solutions. Knowing how to simplify and solve exact equations is important when you want to have a solid foundation on differential equations. This leads to you being confident when dealing with more advanced topics and when taking STEM-related courses that utilize differential equations.

** Exact equations are first-order differential equations that appear in the form of **$\boldsymbol{P(x, y) \phantom{x} +Q(x, y) \phantom{x} \phantom{x}dy = 0}$

**$\boldsymbol{P}$**

*. We can solve exact equations by utilizing the partial derivatives of***$\boldsymbol{Q}$**

*and*

*.*Since this deals with differential equations, familiarity with the topic is a must. In this article, weâ€™ll define the conditions for a differential equation to be exact, weâ€™ll show you how to test equations for their exactness, and weâ€™ll break down the process of finding the solutions for different types of exact equations.

**What Is an Exact Equation?**

An exact equation is a differential equation that has the general form shown below.

\begin{aligned}P(x, y) \phantom{x}dx + Q(x, y)\phantom{x}dy&= 0 \end{aligned}

For this differential equation to be an exact differential equation, then the function, $f$, must satisfy the following conditions:

\begin{aligned}\dfrac{\partial f}{\partial x} &=P(x, y)\\ \dfrac{\partial f}{\partial y} &= Q(x, y)\end{aligned}

Keep in mind that the function, $f$, must have continuous partial derivatives with respect to $x$ and $y$. When the differential equation is an exact equation, its general solution has the implicit form, $f(x, y)=C$, where $C$ is a constant. Why donâ€™t we check this by proving that by finding the expression for $df$? By applying the chain rule, we have the resulting expression:

\begin{aligned} df &= \dfrac{\partial f}{\partial x} \phantom{x}dx +\dfrac{\partial f}{\partial y} \phantom{x}dy \end{aligned}

Since $f(x, y) = C$, the left-hand side of the equationÂ is equal to $0$. If we let $P(x, y) = \dfrac{\partial f}{\partial x}$ and $Q(x, y)=\dfrac{\partial f}{\partial y}$, we can rewrite the equation proving the general form of the exact equation to remain true.

\begin{aligned} 0 &= \dfrac{\partial f}{\partial x} \phantom{x}dx +\dfrac{\partial f}{\partial y} \phantom{x}dy\\0 &= P(x, y) \phantom{x}dx + Q(x, y) \phantom{x}dy\\\\&\Rightarrow \boldsymbol{P(x, y)\phantom{x}dx + Q(x, y)\phantom{x}dy=0} \end{aligned}

Letâ€™s now go back to the general form of the exact equation and rewrite them in the following forms:

\begin{aligned} P(x, y) + Q(x, y) \phantom{x}\dfrac{dy}{dx} &= 0 \phantom{xxx}(1)\\P(x, y)\phantom{x}\dfrac{dx}{dy}Â + Q(x, y) &= 0\phantom{xxx}(2) \end{aligned}

We use the first form (Equation $(1)$) when $x$ is the independent variable and $y$ is the dependent variable. The second form is best used when $y$ is the independent variable and $x$ is the dependent variable.

\begin{aligned} 0 &= \dfrac{\partial f}{\partial x} \phantom{x}dx +\dfrac{\partial f}{\partial y} \phantom{x}dy\\0 &= P(x, y) \phantom{x}dx + Q(x, y) \phantom{x}dy \end{aligned}

Here are some examples of exact differential equations in three possible forms:

\begin{aligned}6x^2y^2 \phantom{x}dx + 4x^3y \phantom{x}dy = 0\end{aligned} | \begin{aligned}6x^2y^2+4x^3y \phantom{x}\dfrac{dy}{dx} = 0\end{aligned} | \begin{aligned}6x^2y^2 \phantom{x}\dfrac{dx}{dy}+ 4x^3y =0\end{aligned} |

\begin{aligned}(x^3 +y^3) \phantom{x}dx + 3xy^2 \phantom{x}dy = 0\end{aligned} | \begin{aligned}(x^3 +y^3)+ 3xy^2 \phantom{x}\dfrac{dy}{dx} =0\end{aligned} | \begin{aligned}(x^3 +y^3) \phantom{x}\dfrac{dx}{dy}+ 3xy^2 = 0\end{aligned} |

\begin{aligned}4x \cos y \phantom{x}dx -2y^2 \sin y \phantom{x}dy = 0\end{aligned} | \begin{aligned}4x \cos y â€“ 2y^2 \sin y \phantom{x}\dfrac{dy}{dx} = 0\end{aligned} | \begin{aligned}4x \cos y \phantom{x}\dfrac{dx}{dy} â€“ 2y^2 \sin y = 0\end{aligned} |

From the definition of exact equations and the examples weâ€™ve shown you, letâ€™s establish a rule for testing exact differential equations:

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Suppose that there exist continuous partial derivatives for $P(x, y)$ and $Q(x, y)$, we can use the condition shown below, \begin{aligned} \dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}\end{aligned} to confirm whether the differential equation, $P(x, y)\phantom{x}dx +Q(x, y) \phantom{x}dy$, is an exact equation. |

Letâ€™s take a look at the differential equation, $6x^2y^2 \phantom{x}dx +4x^3y \phantom{x}dy = 0$. Find the partial derivatives of $6x^2y^2$ and $4x^3y$ with respect to $y$ and $x$, respectively, to confirm that the equation is indeed an exact equation.

\begin{aligned}\dfrac{\partial }{\partial y}(6x^2y^2) &= (6x^2)(2y)\\&= 12x^2y\\\dfrac{\partial }{\partial x}(4x^3y) &= (12x^2)(y)\\&= 12x^2y\\\\\dfrac{\partial }{\partial y}(6x^2y^2) &= \dfrac{\partial }{\partial x}(4x^3y)\end{aligned}

Since both expressions are equal, the differential exact equation satisfies the test for exact equations. Now that weâ€™ve established the core definition and conditions for exact equations, itâ€™s time for us to break down the steps in simplifying and solving different types of exact equations.

**How To Solve Exact Equations? **

There can be different approaches when solving exact equations â€“ but to help you have a more systematic approach, weâ€™ve prepared a guideline for you.

1. First, manipulate the equation so that it is in the exact equationâ€™s general form:

\begin{aligned}P(x, y) \phantom{x} dx + Q(x,y ) \phantom{x} dy\end{aligned}

2. Identify the functions representing $P(x,y)$ and $Q(x, y)$ then test the equation for exactness as we have discussed earlier.

\begin{aligned} \dfrac{\partial P}{\partial y}= \dfrac{\partial Q}{\partial x}\end{aligned}

3. When the equation satisfies the condition for exact equations, assess the equation and see whether itâ€™s better to integrate $P(x, y)$ with respect to $x$ or $Q(x, y)$ with respect to $y$.

*For the next few steps, we assume that you chose to integrate *$P(x, y)$*.*

4. After integrating $P(x, y)$ with respect to $x$, we can rewrite the function shown below.

\begin{aligned} f(x, y) &= \int P(x, y) \phantom{x}dx + g(y)\end{aligned}

Since weâ€™re integrating with respect to $x$, weâ€™re treating the function in terms of $y$ as a constant, hence we have $g(y)$.

5. Once weâ€™ve simplified the expression for $f(x, y)$ from the previous step, find $g(y)$ by taking the partial derivative of $f(x, y)$ with respect to $y$. \begin{aligned} f(x, y) &= \int P(x, y) \phantom{x}dx +g(y)\\\dfrac{\partial f}{\partial y} &= \dfrac{\partial}{\partial y}\left(\int P(x, y) \phantom{x}dx \right ) + g^{\prime}(y)\end{aligned}

6. Weâ€™ve established before that $Q(x, y) = \dfrac{\partial f}{\partial y}$, so use this to solve for $g^{\prime}(y)$ and find the expression for $f(x, y)$.

7. When given an initial condition for the exact differential equation, use the implicit form, $f(x, y) =C$, to solve for the constant.

The best way to master solving exact equations is by working on different types of differential equations. Why donâ€™t we begin by finding the solution for the differential equation, $4xy \phantom{x}dx + (2x^2 +8y^2)\phantom{x} dy=0$?

When given a differential equation, we first confirm the exactness of the differential equation. From our equation, we have $P = 4xy$ and $Q = 2x^2 + 8y^2$. Below are their respective partial derivatives:

Â

\begin{aligned}\boldsymbol{\dfrac{\partial P}{\partial y}}\end{aligned} | \begin{aligned}\boldsymbol{\dfrac{\partial Q}{\partial x}}\end{aligned} |

\begin{aligned}\dfrac{\partial P}{\partial \partial y} = \dfrac{\partial}{\partial y} (4xy) = 4x\end{aligned} | \begin{aligned}\dfrac{\partial Q}{\partial \partial x} = \dfrac{\partial}{\partial x} (2x^2 + 8y^2) = 4x\end{aligned} |

Since $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$, we can confirm that the differential equation is an exact equation. This means that we have the function, $f(x, y)$, defined by the two differential equations shown below.

\begin{aligned} \dfrac{\partial f}{\partial x} &= 4xy\\\dfrac{\partial f}{\partial y} &= 2x^2 + 8y^2 \end{aligned}

Since $P(x, y)$ is a monomial, it is much easier to integrate $4xy$ with respect to $x$. We have $\int 4xy \phantom{x}dx = 2x^2y$, so we can now rewrite $f(x, y)$ as a sum of the antiderivative of $4x$ and $g(y)$.

\begin{aligned}f(x, y)&= \int P(x) \phantom{x}dx + g(y)\\&= \int 4x \phantom{x} dx + g(y) \\&= 2x^2y + g(y)\end{aligned}

Recall that $\dfrac{\partial f}{\partial y} = Q(x, y) = (2x^2 + 8y^2)$ and we know that $f(x, y) = 2x^2y + g(y)$. Using these equations, take the partial derivative of $f(x, y)$ with respect to $y$ then equate this to $(2x ^2 + 8y^2)$.

\begin{aligned}\dfrac{\partial}{\partial y}f(x, y)&= \dfrac{\partial}{\partial y}\left[2x^2y + g(y) \right] \\&= 2x^2 + g^{\prime}(y)\\\\2x^2 + g^{\prime}(y) &= 2x^2 + 8y\\g^{\prime}(y) &= 8y\end{aligned}

Integrate $8y$ to find the expression for $g(y)$. There is no initial condition provided for this differential equation, account for the constant, $C$, when integrating $8y$.

\begin{aligned}\int g(y) \phantom{x}dy &= \int 8y^2 \phantom{x}dy\\&= \dfrac{8}{3}y^3 + C\end{aligned}

Complete the expression for $f(x, y)$ using the result and write the solution of the exact equation in the form of $f(x, y) = C$ to account for the arbitrary constant.

\begin{aligned}f(x, y) &= 2x^2y + \dfrac{8}{3}y^3 \\2x^2y + \dfrac{8}{3}y^3 &= C \end{aligned}

This means that the general solution of our exact differential equation is equal to $2x^2y + \dfrac{8}{3}y^3 = C$.

This example shows us how we can find the general solution of a simple exact differential equation. Donâ€™t worry, we have prepared more examples for you to work on! When youâ€™re ready, head over to the section below to try out more examples and eventually master solving this type of differential equation!

*Example 1*

Find the general solution for the differential equation, $(8xÂ + 4y^2) \phantom{x}dx + y(8x â€“ 6y) \phantom{x}dy = 0$.

__Solution__

First, lets identify the expressions for $P(x, y)$ and $Q(x, y)$ from the given differential equation.

\begin{aligned}P(x, y) &= 8x + 4y^2\\ Q(x, y) &= y(8x â€“ 6y)\\&= 8xy â€“ 6y^2\end{aligned}

Confirm the exactness of the differential equation by comparing the expressions of $\dfrac{\partial P}{\partial y}$ and $\dfrac{\partial Q}{\partial x}$.

\begin{aligned}\boldsymbol{\dfrac{\partial P}{\partial y}}\end{aligned} | \begin{aligned}\boldsymbol{\dfrac{\partial Q}{\partial x}}\end{aligned} |

\begin{aligned}\dfrac{\partial P}{\partial \partial y} = \dfrac{\partial}{\partial y} (8x + 4y^2) = 8y\end{aligned} | \begin{aligned}\dfrac{\partial Q}{\partial \partial x} = \dfrac{\partial}{\partial x} (8xy â€“ 6y^2) = 8y\end{aligned} |

Since the two partial derivatives are equal, we can confirm that the differential equation is exact. We can integrate either expression but letâ€™s stick with the default option unless working with $Q$ is much more convenient. Integrate $P(x, y) = 8x + 4y^2$ with respect to $x$ first.

\begin{aligned}\int P(x, y) \phantom{x}dx &= \int (8x + 4y^2) \phantom{x}dx\\&= 4x^2 + 4xy^2\end{aligned}

We can now establish the equation, $f(x, y)= \int P(x, y) \phantom{x}dx + g(y)$, where $f(x, y)$ is defined by these two differential equations:

\begin{aligned} \dfrac{\partial f}{\partial x} &= 8xÂ + 4y^2\\\dfrac{\partial f}{\partial y} &= y(8x â€“ 6y) \end{aligned}

After writing the equation of $f(x, y)$ in terms of $\int P(x, y) \phantom{x}dx$ and $g(y)$, take the partial derivative of both sides of the equation with respect to $y$.

\begin{aligned} f(x, y) &= \int P(x, y) \phantom{x}dx + g(y)\\&= (4x^2 + 4xy^2)Â + g(y)\\\\\dfrac{\partial}{\partial y} f(x, y) &= \dfrac{\partial}{\partial y}(4x^2 + 4xy^2)Â + g^{\prime}(y)\end{aligned}Â

Use the fact that $\dfrac{\partial f}{\partial y} = Q(x, y) = 8xy â€“ 6y^2$ to find the expression for $g(y)$. Equate $8xy â€“ 6y^2$ with our previous expression for $\dfrac{\partial f}{\partial y}$ then solve for $g(y)$ by integrating $g^{\prime}(y)$.

\begin{aligned}\dfrac{\partial}{\partial y}(4x^2 + 4xy^2)Â + g^{\prime}(y) &= 8xy – 6y^2\\ [0 + 4x(2y)] + g^{\prime}(y) &= 8xy – 6y^2\\8xy + g^{\prime}(y) &= 8xy – 6y^2\\g^{\prime}(y) &= -6y^2\\\\g(y) &= \int (-6y^2) \phantom{x}dy\\&= -\dfrac{6y^3}{3}\\&= -2y^3 \end{aligned}

Rewrite our current expression for $f(x, y)$ and replace $g^{\prime}{(y)}$ with $-2y^3$. Since weâ€™re looking for a general solution again, weâ€™ll have to account for the arbitrary constant, $C$.

\begin{aligned}f(x, y) &= C\\ \int P(x, y) \phantom{x}dx + g(y) &= C\\(4x^2 + 4xy^2) + (-2y^3) &= C\\ 4x^2 + 4xy^2 â€“ 2y^3 &= C\end{aligned}

This means that the general solution of our exact equation is equal to $4x^2 + 4xy^2 â€“ 2y^3 = C$ when written in implicit form.

*Example 2*

Find the particular solution for the differential equation, $(4xy â€“ 12x^2) + (6y + 2x^2 +5) \phantom{x}\dfrac{dy}{dx} = 0$, given that the equation satisfies the initial condition, $y(0) = 4$.

__Solution__

Rewrite the equation so that it is of the form, $P(x,y) \phantom{x}dx + Q(x, y)\phantom{x} dy =0$.

\begin{aligned}(4xy – 12x^2)+ (6y + 2x^2 +5) \phantom{x}\dfrac{dy}{dx} = 0\\(4xy – 12x^2) dx+ (6y + 2x^2 +5) \phantom{x}dy = 0\end{aligned}

This means that $P(x, y) = 4xy – 12x^2$ and $Q(x, y) = (6y + 2x^2 +5)$, so letâ€™s take their partial derivatives with respect to $y$ and $x$, respectively.

\begin{aligned}\boldsymbol{\dfrac{\partial P}{\partial y}}\end{aligned} | \begin{aligned}\boldsymbol{\dfrac{\partial Q}{\partial x}}\end{aligned} |

\begin{aligned}\dfrac{\partial P}{\partial \partial y} = \dfrac{\partial}{\partial y} (4xy – 12x^2) = 4x\end{aligned} | \begin{aligned}\dfrac{\partial Q}{\partial \partial x} = \dfrac{\partial}{\partial x} (6y + 2x^2 +5) = 4x\end{aligned} |

We can say that the equation is exact since weâ€™ve shown that $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$. Weâ€™ll show you how itâ€™ll work if we choose to integrate $Q(x, y)$ with respect to $y$ this time.

\begin{aligned}\int Q(x, y)\phantom{x}dy &= \int (6y + 2x^2 +5) \phantom{x}dy \\&= 3y^2 + 2x^2y +5y\end{aligned}

Suppose that $f(x, y)$ is defined by the two differential equations: $\dfrac{\partial f}{\partial y} = 4xy â€“ 12x^2$ and $\dfrac{\partial f}{\partial x} = 6y + 2x^2 +5$. We can rewrite $f(x, y)$ using integral of $Q(x, y)$ and $g(x)$.

\begin{aligned}f(x, y) &= \int Q(x, y)\phantom{x}dy + g(x)\\&= (3y^2 + 2x^2y +5y ) +g(x) \end{aligned}

Solve for $g(x)$ by finding $\dfrac{\partial f}{\partial x}$ and equate the expression to $P(x, y)$.

\begin{aligned}\dfrac{\partial}{\partial x} f(x, y) &= \dfrac{\partial }{\partial x}[(3y^2 + 2x^2y +5y ) +g(x)]\\&= 4xy + g^{\prime}(x)\\\\(4xy – 12x^2) &= 4xy + g^{\prime}(x)\\g^{\prime}(x) &= -12x^2\\g(x) &= -\int 12x^2\phantom{x}dx\\&= -4x^3 \end{aligned}

This means that $f(x, y) = 3y^2 + 2x^2y +5y -4x^3$ and the general solution for the equation is $ 3y^2 + 2x^2y +5y -4x^3 =C$. Use $f(0) = 4$ to find the value of $C$.

\begin{aligned}3(4)^2 + 2(0)^2(4) +5(4) -4(0)^3 &= C\\C &= 68\end{aligned}

Hence, the particular solution of the exact equation is 3y^2 + 2x^2y +5y -4x^3 = 68$.

### Practice Questions

1. Find the general solution for the differential equation, $2(3x + xy) \phantom{x}dx + (4y^3 + x^2) \phantom{x}dy = 0$.

2. Find the general solution for the differential equation, $\dfrac{y^2}{2(1 + x^2)} \phantom{x} dx+ y\tan^{-1} x \phantom{x}dy =0$.

3. Solve the initial value problem, $(e^{4x}+2xy^2)\dfrac{dx}{dy} +(\cos y +2x^2y)=0$, given the initial condition, $y(0) = \pi$.

### Answer Key

1. $x^2y +3x^2 + y^4 = C$

2. $\dfrac{y^2}{2} \tan^{-1}x = C$

3. $\dfrac{1}{4}e^{4x} +x^2y^2 +\sin y =\dfrac{1}{4}$

*Images/mathematical drawings are created with GeoGebra.*