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

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

classification: this technique applies to first-order, separable differential equations

An equation that allows us to separate the variables is called a separable differential equation.

This is one of the simplest techniques when solving differential equations and sometimes it is introduced in first year calculus while studying integration. This technique is used when it is possible to isolate the variables, using algebra and differentials, on separate sides of the equal sign. This allows us to integrate both sides of the equation individually with only one variable on each side.

The idea with this technique is that the differential equation is in a form where we can isolate the two variables to each side of the equal sign. Then we can integrate each side separately. It's really that easy. The trick is to use algebra to get the equation into the right form.

Separable Equation Form

Some textbooks write separable equations as $$\displaystyle{ N(y)\frac{dy}{dx}=M(x) }$$. This notation tells you that $$N(y)$$ is some function that contains only $$y$$'s and maybe some constants. Similarly, $$M(x)$$ is some function that contains only $$x$$'s and maybe some constants.

A better way to write a separable equation is an equation of the form $$\displaystyle{ \frac{dy}{dx}=g(x)h(y) }$$. In this equation, it is easy to see what needs to done. We need to have two functions multipled together, with one having only x's and the other with only y's.

As we work this example, we will give you a heads up on something. Some instructors (most, from what I've seen), do a trick with constants that confuse a lot of people. Let's do an example to demonstrate the separation of variables technique.

Example Find y for $$y' = x$$.

For this separable equation, using the form $$\displaystyle{ \frac{dy}{dx}=g(x)h(y) }$$, we have $$g(x)=x$$ and $$h(y)=1$$

 $$\displaystyle{ y' = x ~~~ \to ~~~ \frac{dy}{dx} = x }$$ First, we rewrite the derivative so that it is more obvious what we need to do. $$\displaystyle{ \frac{dy}{dx} = x ~~~ \to ~~~ dy = x~dx}$$ Now we write the differential equation by 'moving' the $$dx$$ to the other side. It looks like we are multiplying $$dx$$ on both sides but that's not what is really happening. However, it helps me remember what to do by thinking of it this way. What we are doing is writing the equation in differential form. $$\begin{array}{rcl} \int{dy} & = & \int{x~dx} \\ y + c_1 & = & \frac{x^2}{2} + c_2 \end{array}$$ Next, we integrate both sides. On the left we integrate with respect to y, on the right with respect to x. $$\displaystyle{ y = \frac{x^2}{2} + c_2-c_1 }$$ Okay, so now we have two constants, one from each integration (don't forget these constants; they are extremely important when solving differential equations). We are going to move $$c_1$$ to the other side and combine it with $$c_2$$. $$\displaystyle{ y = \frac{x^2}{2} + C }$$ Let $$C = c_2 - c_1$$ to get our final answer.

Notes
1. This is a general solution. To get the particular solution, we need initial conditions to determine the value of C.
2. Many instructors skip the steps showing the two constants and combining them, i.e. $$C = c_2 - c_1$$ . They just jump to one constant. This can be confusing at first, if you don't know what is happening. Here is a video explaining this in more detail.

### Krista King Math - How can the constant C absorb all that other stuff? [3mins-54secs]

video by Krista King Math

And, that's it! That's all there is to it. Just separate the variables and integrate. There is nothing fancy going on and no tricks. But this is a basic technique that you need to understand how to do and practice thoroughly since, later on in your study of differential equations, you will learn other techniques to convert more complicated equations into this form.

Here is an in-depth video discussing first-order linear equations, separation of variables and steady-state and transient solutions. If you are first starting to learn differential equations, this may be a bit above you but you can still get a lot out of it and being exposed to a little more advanced techniques will help you learn them later.

### MIT OCW - Solving First-order Linear ODE's; Steady-state and Transient Solutions. [mins-secs]

video by MIT OCW

After you work through the practice problems we recommend learning about first-order, linear differential equations next.

### 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 separable differential equations and find the particular solution, if possible. Give your answers in exact terms.

Basic Problems

$$y'+2y=0; ~~~ y(0)=1$$

Problem Statement

Solve the initial value problem $$y'+2y=0; ~~~ y(0)=1$$

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

Problem Statement

Solve the initial value problem $$y'+2y=0; ~~~ y(0)=1$$

Solution

$$\begin{array}{rcl} y' + 2y & = & 0 \\ \displaystyle{\frac{dy}{dx}} & = & -2y \\ \displaystyle{\frac{dy}{y}} & = & -2dx \\ \displaystyle{\int{\frac{dy}{y}}} & = & \int{-2dx} \\ \ln\abs{y} & = & -2x + C \\ y & = & Ae^{-2x} \end{array}$$
Using the initial conditions $$y(0)=1$$,
$$1 = Ae^0 ~~~ \to ~~~ A = 1$$

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

$$y'+ty=t; ~~~ y(0)=2$$

Problem Statement

Solve the initial value problem $$y'+ty=t; ~~~ y(0)=2$$

$$\displaystyle{ y = 1 + e^{-t^2/2} }$$

Problem Statement

Solve the initial value problem $$y'+ty=t; ~~~ y(0)=2$$

Solution

$$\begin{array}{rcl} y'+ty & = & t \\ & = & t-ty \\ & = & t(1-y) \\ \displaystyle{\frac{dy}{1-y}} & = & t~dt \\ \displaystyle{\int{ \frac{dy}{1-y} }} & = & \int{t~dt} \\ -\ln\abs{1-y} & = & \displaystyle{\frac{t^2}{2} + c} \\ \displaystyle{\frac{1}{1-y}} & = & Ae^{t^2/2} \\ 1-y & = & Be^{-t^2/2} \\ y & = & 1-Be^{-t^2/2} \end{array}$$
Using the initial condition $$y(0)=2$$,
$$2 = 1-B ~~~ \to ~~~ B = -1$$

$$\displaystyle{ y = 1 + e^{-t^2/2} }$$

$$y' = (x+1)y$$

Problem Statement

Find the general solution to $$y' = (x+1)y$$

$$\displaystyle{ y = A~\exp\left( \frac{x^2}{2} + x \right) }$$

Problem Statement

Find the general solution to $$y' = (x+1)y$$

Solution

$$\begin{array}{rcl} y' & = & (x+1)y \\ \displaystyle{\frac{dy}{y}} & = & (x+1)~dx \\ \displaystyle{\int{\frac{dy}{y}}} & = & \int{(x+1)~dx} \\ \ln\abs{y} & = & \displaystyle{\frac{x^2}{2} + x + C} \\ y & = & \displaystyle{A~\exp\left( \frac{x^2}{2}+x \right)} \end{array}$$

$$\displaystyle{ y = A~\exp\left( \frac{x^2}{2} + x \right) }$$

$$y' = y - y^2$$

Problem Statement

Find the general solution to $$y' = y - y^2$$

$$\displaystyle{ y = \frac{Ae^x}{1+Ae^x} }$$

Problem Statement

Find the general solution to $$y' = y - y^2$$

Solution

$$\displaystyle{y' = y-y^2 ~~~ \to ~~~ \frac{dy}{y(1-y)} = dx }$$
Use partial fraction expansion on the left side.
$$\begin{array}{rcl} \displaystyle{\frac{dy}{y(1-y)}} & = & dx \\ \displaystyle{\int{\frac{1}{y} + \frac{1}{1-y}dy}} & = & \int{1~dx} \\ \ln\abs{y} - \ln\abs{1-y} & = & x + C \\ \displaystyle{\ln \abs{\frac{y}{1-y}}} & = & \\ \displaystyle{\frac{y}{1-y}} & = & Ae^x \end{array}$$
Now we need to solve for $$y$$.
$$\begin{array}{rcl} \displaystyle{\frac{y}{1-y}} & = & Ae^x \\ y & = & Ae^x (1-y) \\ & = & Ae^x - yAe^x \\ y + yAe^x & = & Ae^x \\ y(1+Ae^x) & = & Ae^x \\ y & = & \displaystyle{\frac{Ae^x}{1+Ae^x}} \end{array}$$

$$\displaystyle{ y = \frac{Ae^x}{1+Ae^x} }$$

$$\displaystyle{ \frac{dy}{dx} = xy }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = xy }$$

Solution

### 559 solution video

video by Dr Chris Tisdell

$$\displaystyle{ \frac{dy}{dx} = y^2 (1+x^2) }$$; $$y(0) = 1$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = y^2 (1+x^2) }$$; $$y(0) = 1$$

Solution

### 560 solution video

video by Dr Chris Tisdell

$$\displaystyle{ \frac{dy}{dx} = \cos(x) }$$; $$y(0) = -1$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = \cos(x) }$$; $$y(0) = -1$$

Solution

### 562 solution video

video by PatrickJMT

$$\displaystyle{ \frac{dy}{dx} = x/y^2 }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = x/y^2 }$$

Solution

### 563 solution video

video by PatrickJMT

$$\displaystyle{ \frac{dy}{dx} \cdot \frac{y^3+y}{x^2+3x} = 1 }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} \cdot \frac{y^3+y}{x^2+3x} = 1 }$$

Solution

### 564 solution video

video by PatrickJMT

$$\displaystyle{ \frac{dy}{dx} = \frac{\cos(x)}{y-1} }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = \frac{\cos(x)}{y-1} }$$

Solution

### 565 solution video

video by PatrickJMT

$$\displaystyle{ \frac{dy}{dx} = e^{4x-y} }$$; $$y(0) = 5$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = e^{4x-y} }$$; $$y(0) = 5$$

Solution

### 566 solution video

video by PatrickJMT

$$\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y-1} }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y-1} }$$

Solution

### 568 solution video

video by Krista King Math

$$\displaystyle{ \frac{y+2}{x^2-x+2} \frac{dy}{dx} = \frac{x}{y} }$$; $$y(1) = 2$$

Problem Statement

$$\displaystyle{ \frac{y+2}{x^2-x+2} \frac{dy}{dx} = \frac{x}{y} }$$; $$y(1) = 2$$

Solution

### 567 solution video

video by PatrickJMT

$$\displaystyle{ y \frac{dy}{dx} = x^2 + \sech^2(x) }$$; $$y(0) = 2$$

Problem Statement

$$\displaystyle{ y \frac{dy}{dx} = x^2 + \sech^2(x) }$$; $$y(0) = 2$$

Solution

### 561 solution video

video by Dr Chris Tisdell

$$\displaystyle{ \frac{dy}{dx} = \frac{x^2+1}{x^2(3y^2+1)} }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = \frac{x^2+1}{x^2(3y^2+1)} }$$

Solution

### 569 solution video

video by Krista King Math

$$\displaystyle{ \frac{dy}{dx} = \frac{4-2x}{3y^2-5} }$$; $$y(1) = 3$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = \frac{4-2x}{3y^2-5} }$$; $$y(1) = 3$$

Solution

### 570 solution video

video by Krista King Math

$$\displaystyle{ \frac{dy}{dx} = 2 \sqrt{y} }$$; $$y(0) = 9$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = 2 \sqrt{y} }$$; $$y(0) = 9$$

Solution

### 571 solution video

video by Krista King Math

$$\displaystyle{ \frac{dy}{dx} = \frac{x^2}{1-y^2} }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = \frac{x^2}{1-y^2} }$$

Solution

### 572 solution video

$$\displaystyle{ \frac{dy}{dx} = \frac{y \cos(x)}{1+2y^2} }$$; $$y(0) = 1$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = \frac{y \cos(x)}{1+2y^2} }$$; $$y(0) = 1$$

Solution

### 574 solution video

$$\displaystyle{ \frac{du}{dt} = \frac{2t+\sec^2(t)}{2u} }$$; $$u(0) = -5$$

Problem Statement

$$\displaystyle{ \frac{du}{dt} = \frac{2t+\sec^2(t)}{2u} }$$; $$u(0) = -5$$

Solution

### 609 solution video

video by Krista King Math

$$\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y} }$$; $$y(0)=1$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y} }$$; $$y(0)=1$$

Solution

### 1842 solution video

video by Dr Chris Tisdell

$$\displaystyle{ \frac{dy}{dx}=e^{x-y} }$$; $$y(0)=\ln(2)$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx}=e^{x-y} }$$; $$y(0)=\ln(2)$$

Solution

### 1843 solution video

video by Dr Chris Tisdell

$$\displaystyle{ \frac{dy}{dt} = \frac{t}{t^2y+y} }$$, $$y(0) = 3$$, $$y > 0$$

Problem Statement

$$\displaystyle{ \frac{dy}{dt} = \frac{t}{t^2y+y} }$$, $$y(0) = 3$$, $$y > 0$$

$$\sqrt{ \ln(t^2+1) + 9 }$$

Problem Statement

$$\displaystyle{ \frac{dy}{dt} = \frac{t}{t^2y+y} }$$, $$y(0) = 3$$, $$y > 0$$

Solution

### 2082 solution video

video by MIP4U

$$\sqrt{ \ln(t^2+1) + 9 }$$

Intermediate Problems

$$dv/ds = (s+1)/(sv+s)$$

Problem Statement

$$dv/ds = (s+1)/(sv+s)$$

$$v = -1 \pm \sqrt{ 2s + 2\ln(s) + C }$$

Problem Statement

$$dv/ds = (s+1)/(sv+s)$$

Solution

From the $$dv/ds$$ term, we can tell that s is the independent variable and v is the dependent variable, so that the solution will be $$v(s)$$. Let's try separation of variables.

 $$\displaystyle{ dv/ds = \frac{(s+1)}{(sv+s)} }$$ $$\displaystyle{ dv/ds = \frac{(s+1)}{s(v+1)} }$$ $$\displaystyle{ (v+1)\frac{dv}{ds} = \frac{(s+1)}{s} }$$ $$\displaystyle{ (v+1) dv = \frac{(s+1)}{s} ds }$$ $$\displaystyle{ \int{v+1 ~dv} = \int{ \frac{s+1}{s} ds } }$$ $$\displaystyle{ \int{v+1 ~dv} = \int{ 1 + \frac{1}{s} ds } }$$ $$\displaystyle{ \frac{v^2}{2} + v = s + \ln(s) + c_1 }$$ $$v^2 + 2v = 2s + 2\ln(s) + 2c_1$$ Adding 1 to both sides after integrating allows us to complete the square. $$v^2 + 2v + 1 = 2s + 2\ln(s) + 1 + 2c_1$$ $$(v+1)^2 = 2s + 2\ln(s) + 1 + 2c_1$$ $$v = \pm \sqrt{ 2s + 2\ln(s) + 1 + 2c_1 } - 1$$ We can combine all constants under the square root into one by letting $$C=1+2c_1$$. This is not required but cleans up the equation. $$v = \pm \sqrt{ 2s + 2\ln(s) + C } - 1$$

Most of this should be easy to follow. There just a few clarifications in order.
1. When taking the square root, don't assume your answer will be positive. Use $$\pm$$ unless you can show that the result will always be positive (or negative).
2. You may not need to solve for $$v$$. Check to see what your instructor requires.

$$v = -1 \pm \sqrt{ 2s + 2\ln(s) + C }$$

$$\displaystyle{ \frac{dy}{dx} = \frac{3x^2+4x+2}{2(y-1)} }$$; $$y(0) = -1$$

Problem Statement

$$\displaystyle{ \frac{dy}{dx} = \frac{3x^2+4x+2}{2(y-1)} }$$; $$y(0) = -1$$

Solution