You CAN Ace Differential Equations

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

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17calculus > differential equations > exact equations

 How To Solve Integrating Factors Practice

Exact differential equations are first-order differential equations of the form $$\displaystyle{ M(x,y)~dx + N(x,y)~dy = 0 }$$ where $$M_y = N_x$$. The requirement that $$M_y = N_x$$ means the differential equation is what we call exact.

classification: this technique applies to first-order differential equations

$$M(x,y)~dx + N(x,y)~dy = 0$$ and $$M_y = N_x$$

Here are some equivalent ways of writing this differential equation.

 $$\displaystyle{ M(x,y)~dx + N(x,y)~dy = 0 }$$ $$\displaystyle{ M(x,y) + N(x,y)y' = 0 }$$ $$\displaystyle{ M(x,y) + N(x,y)\frac{dy}{dx} = 0 }$$

How To Solve

Lets jump right into a video to see how we solve these. Here is a great video clip explaining the technique and showing a couple of examples.

### Dr Chris Tisdell - exact equations [1hr-6mins-16secs]

video by Dr Chris Tisdell

In the video above, he does a good job of emphasizing that, before you start actually solving the differential equation, you need to make sure it is exact by calculating $$M_y$$ and $$N_x$$ and checking that they are equal. Otherwise, this technique will take you down a path that leads nowhere. Most instructors will take off points if you don't show work that you did this. So check with your instructor. But even if they don't take off points, it could save you a lot of time on homework and exams if you do this and it takes only a few seconds. So it is a good investment of your time.

The main difference with these types of problems is that you need to remember the constant of integration is actually a function of integration. This occurs because we are taking the partial integral of one variable while holding the other variable constant. This concept is covered in detail on the multi-variable partial integration page.

Here is another video that is very similar to the first one, with a few different examples. But watching this one also, will help you understand this technique better.

### Dr Chris Tisdell - more exact equations [23mins-3secs]

video by Dr Chris Tisdell

Integrating Factors

If we have a first-order equation of the form $$M(x,y)~dx + N(x,y)~dy = 0$$ but $$M_y \neq N_x$$, i.e. the equation is not exact (also called inexact), we may be able to convert the equation to exact using integrating factors. You first came across integrating factors when you studied linear, first-order equations. This technique also uses an integrating factor but it is calculated differently.

There are actually two possible integrating factors that convert the differential equation to an exact equation.

 $$\displaystyle{ \mu_1 = \exp \int{ \frac{M_y - N_x}{N} dx } }$$ $$\displaystyle{ \mu_2 = \exp \int{ \frac{M_y - N_x}{-M} dy } }$$

The trick comes in when you are asked to evaluate these integrals. The evaluation is not always possible and can get quite messy. Also, one integral may yield an integrating factor that is quite complicated while the other one may be much easier to use. Regardless, these integrating factors can, at times, be quite useful. Let's look at an example in this next video.

In this first video, he starts an in-depth example where he calculates an integrating factor for a specific inexact equation. He finishes the example in the next video.

### Khan Academy - Integrating Factors (1) [10mins-16secs]

Here is the completion of the example he starts in the previous video. He calculated an integrating factor in the above video and he finishes the problem, which is now exact, in this video.

### Khan Academy - Integrating Factors (2) [8mins-25secs]

Okay, time for some practice problems.

### Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, solve these differential equations. Make sure to check that the equation is exact before attempting to solve. If the equation is not exact, calculate an integrating factor and use it make the equation exact. Give your answers in exact form.

Basic Problems

$$2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0$$

Problem Statement

$$2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0$$

$$y = x^2y^2 + x^4 + y^4 + C$$

Problem Statement

$$2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0$$

Solution

First, check for exactness. Rewriting the equation in differential form, we have $$(2xy^2 + 4x^3)dx + (2x^2y +4y^3)dy = 0$$
$$M(x,y) = 2xy^2 + 4x^3$$ $$\to$$ $$M_y = (2x)(2y) + 0 = 4xy$$
$$N(x,y) = 2x^2y+4y^3$$ $$\to$$ $$N_x = 2y(2x)+ 0 = 4xy$$
Since $$M_y = N_x$$, the differential equation is exact and so we can solve it.
$$\psi_x = M(x,y) = 2xy^2+4x^3$$ $$\to$$ $$\psi = 2y^2 (x^2/2) + (4x^4)/4 + h(y) =$$ $$x^2y^2 + x^4 + h(y)$$
$$\psi_y = N(x,y) = 2x^2y+4y^3$$ $$\to$$ $$\psi = 2x^2 (y^2/2) + (4y^4)/4 + g(x) =$$ $$x^2y^2 + y^4 + g(x)$$
Equating these results, we get $$\psi = x^2y^2 + x^4 + y^4 + C$$

$$y = x^2y^2 + x^4 + y^4 + C$$

$$y+2e^x + (1+e^{-x})y' = 0$$

Problem Statement

$$y+2e^x + (1+e^{-x})y' = 0$$

$$y = ye^x +y + e^{2x} + C$$

Problem Statement

$$y+2e^x + (1+e^{-x})y' = 0$$

Solution

First we check for exactness. Rewriting the differential equation, we have $$(y+2e^x)dx + (1+e^{-x})dy = 0$$.
$$M(x,y) = y+2e^x$$ $$\to$$ $$M_y = 1$$
$$N(x,y) = 1+e^{-x}$$ $$\to$$ $$N_x = -e^{-x}$$
Since $$M_y \neq N_x$$, this equation is not exact. We need to find an integrating factor.
$$\begin{array}{rcl} \displaystyle{\frac{d\mu}{dx}} & = & \displaystyle{\frac{M_y - N_x}{N}\mu} \\ & = & \displaystyle{\frac{1+e^{-x}}{1+e^{-x}}\mu} \\ & = & \mu \\ \displaystyle{\frac{d\mu}{\mu}} & = & dx \\ ln|\mu| & = & x + k ~~ \text{(let k=0)}\\ \mu & = & e^x \end{array}$$
So our integrating factor is $$\mu=e^x$$. Multiplying the differential equation by this factor, we get
$$e^x[t+2e^x] + e^x[1+e^{-x}]y' = 0$$ $$\to$$ $$(ye^x+2e^{2x})dx + (e^x+1)dy = 0$$
Let's check that we now have an exact equation.
$$M(x,y) = ye^x + 2e^{2x}$$ $$\to$$ $$M_y = e^x$$
$$N(x,y) = e^x+1$$ $$\to$$ $$N_x = e^x$$
Since we now have an exact equation, we can solve it.
$$\psi_x = M(x,y) = ye^x + 2e^{2x}$$ $$\to$$ $$\psi = ye^x + e^{2x} + h(y)$$
$$\psi_y = N(x,y) = e^x+1$$ $$\to$$ $$\psi = ye^x + y + g(x)$$
Equating we get $$\psi = ye^x +y + e^{2x} + C$$

$$y = ye^x +y + e^{2x} + C$$

$$2xy~dx + (x^2+3y^2)~dy = 0$$

Problem Statement

$$2xy~dx + (x^2+3y^2)~dy = 0$$

Solution

### 585 solution video

video by Dr Chris Tisdell

$$2xy~dx + x^2~dy = 0$$

Problem Statement

$$2xy~dx + x^2~dy = 0$$

Solution

### 587 solution video

video by Dr Chris Tisdell

$$2x + 3 + (2y-2)y' = 0$$

Problem Statement

$$2x + 3 + (2y-2)y' = 0$$

$$x^2 + 3x + y^2 - 2y = C$$

Problem Statement

$$2x + 3 + (2y-2)y' = 0$$

Solution

### 595 solution video

$$x^2 + 3x + y^2 - 2y = C$$

$$2xy~dx + (x^2-1)~dy = 0$$

Problem Statement

$$2xy~dx + (x^2-1)~dy = 0$$

$$x^2y - y = C$$

Problem Statement

$$2xy~dx + (x^2-1)~dy = 0$$

Solution

### 597 solution video

video by MIP4U

$$x^2y - y = C$$

$$[\cos(x)\sin(x)-xy^2]~dx + y(1-x^2)~dy = 0$$

Problem Statement

$$[\cos(x)\sin(x)-xy^2]~dx + y(1-x^2)~dy = 0$$

$$(1/2)[y^2(1-x^2) + sin^2(x)] = C$$

Problem Statement

$$[\cos(x)\sin(x)-xy^2]~dx + y(1-x^2)~dy = 0$$

Solution

### 598 solution video

video by MIP4U

$$(1/2)[y^2(1-x^2) + sin^2(x)] = C$$

$$(y\cos(x)+2xe^y) + (\sin(x)+x^2e^y-1)y' = 0$$

Problem Statement

$$(y\cos(x)+2xe^y) + (\sin(x)+x^2e^y-1)y' = 0$$

$$y\sin(x) + x^2e^y - y = C$$

Problem Statement

$$(y\cos(x)+2xe^y) + (\sin(x)+x^2e^y-1)y' = 0$$

Solution

### 593 solution video

video by PatrickJMT

$$y\sin(x) + x^2e^y - y = C$$

$$(2xy+1) + (x^2+3y^2)y' = 0$$

Problem Statement

$$(2xy+1) + (x^2+3y^2)y' = 0$$

Solution

### 1845 solution video

video by Dr Chris Tisdell

Find the explicit solution for the exact differential equation $$(e^{x+y}-2x)dx + (e^{x+y}+2y)dy = 0$$, $$y(0)=0$$.

Problem Statement

Find the explicit solution for the exact differential equation $$(e^{x+y}-2x)dx + (e^{x+y}+2y)dy = 0$$, $$y(0)=0$$.

$$1 = e^{x+y} - x^2 + y^2$$

Problem Statement

Find the explicit solution for the exact differential equation $$(e^{x+y}-2x)dx + (e^{x+y}+2y)dy = 0$$, $$y(0)=0$$.

Solution

### 2086 solution video

video by Krista King Math

$$1 = e^{x+y} - x^2 + y^2$$

$$xy^3dx + (x^2y^2+1)dy = 0$$

Problem Statement

Solve $$xy^3dx + (x^2y^2+1)dy = 0$$ using exact integrating factors, if necessary.

$$\displaystyle{ \frac{1}{2}x^2y^2 + \ln|y| = C }$$

Problem Statement

Solve $$xy^3dx + (x^2y^2+1)dy = 0$$ using exact integrating factors, if necessary.

Solution

He makes several mistakes in his solution and answer. First, when calculating the integrating factor, he needs $$dy$$ in the integral. Second, since we can't tell if $$y$$ is always positive, he needs absolute value signs on his natural log term.

### 2206 solution video

video by Engineer4Free

$$\displaystyle{ \frac{1}{2}x^2y^2 + \ln|y| = C }$$

Intermediate Problems

$$(3x+y+1)~dx + (3y+x+1)~dy = 0$$

Problem Statement

$$(3x+y+1)~dx + (3y+x+1)~dy = 0$$

Solution

### 586 solution video

video by Dr Chris Tisdell

$$(2x+y+1)~dx + (2y+x+1)~dy = 0$$

Problem Statement

$$(2x+y+1)~dx + (2y+x+1)~dy = 0$$

Solution

### 588 solution video

video by Dr Chris Tisdell

$$(3x^2-2xy+2)~dx + (6y^2-x^2+3)~dy = 0$$

Problem Statement

$$(3x^2-2xy+2)~dx + (6y^2-x^2+3)~dy = 0$$

Solution

### 596 solution video

$$(2xy^{-1}+2ye^{2x}-1) + (e^{2x}+y^2-x^2y^{-2})\frac{dy}{dx} = 0$$
$$(2xy^{-1}+2ye^{2x}-1) + (e^{2x}+y^2-x^2y^{-2})\frac{dy}{dx} = 0$$