17Calculus Differential Equations - Separation Of Variables
classification: this technique applies to first-order, separable differential equations |
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An equation that allows us to separate the variables is called a separable differential equation. |
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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 |
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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 |
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After you work through the practice problems we recommend learning about first-order, linear differential equations next.
Practice
Unless otherwise instructed, solve these separable differential equations and find the particular solution, if possible. Give your answers in exact terms.
Basic |
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\(\displaystyle{ \frac{dy}{dx} = xy }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = xy }\)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = y^2 (1+x^2) }\); \( y(0) = 1 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = y^2 (1+x^2) }\); \( y(0) = 1 \)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = \cos(x) }\); \( y(0) = -1 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = \cos(x) }\); \( y(0) = -1 \)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = x/y^2 }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = x/y^2 }\)
Solution |
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563 video
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\(\displaystyle{ \frac{dy}{dx} \cdot \frac{y^3+y}{x^2+3x} = 1 }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} \cdot \frac{y^3+y}{x^2+3x} = 1 }\)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = \frac{\cos(x)}{y-1} }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = \frac{\cos(x)}{y-1} }\)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = e^{4x-y} }\); \( y(0) = 5 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = e^{4x-y} }\); \( y(0) = 5 \)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y-1} }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y-1} }\)
Solution |
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\(\displaystyle{ \frac{y+2}{x^2-x+2} \frac{dy}{dx} = \frac{x}{y} }\); \( y(1) = 2 \)
Problem Statement |
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\(\displaystyle{ \frac{y+2}{x^2-x+2} \frac{dy}{dx} = \frac{x}{y} }\); \( y(1) = 2 \)
Solution |
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\(\displaystyle{ y \frac{dy}{dx} = x^2 + \sech^2(x) }\); \( y(0) = 2 \)
Problem Statement |
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\(\displaystyle{ y \frac{dy}{dx} = x^2 + \sech^2(x) }\); \( y(0) = 2 \)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = \frac{x^2+1}{x^2(3y^2+1)} }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = \frac{x^2+1}{x^2(3y^2+1)} }\)
Solution |
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569 video
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\(\displaystyle{ \frac{dy}{dx} = \frac{4-2x}{3y^2-5} }\); \( y(1) = 3 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = \frac{4-2x}{3y^2-5} }\); \( y(1) = 3 \)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = 2 \sqrt{y} }\); \( y(0) = 9 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = 2 \sqrt{y} }\); \( y(0) = 9 \)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = \frac{x^2}{1-y^2} }\)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = \frac{x^2}{1-y^2} }\)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = \frac{y \cos(x)}{1+2y^2} }\); \( y(0) = 1 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = \frac{y \cos(x)}{1+2y^2} }\); \( y(0) = 1 \)
Solution |
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\(\displaystyle{ \frac{du}{dt} = \frac{2t+\sec^2(t)}{2u} }\); \( u(0) = -5 \)
Problem Statement |
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\(\displaystyle{ \frac{du}{dt} = \frac{2t+\sec^2(t)}{2u} }\); \( u(0) = -5 \)
Solution |
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\(\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y} }\); \( y(0)=1 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y} }\); \( y(0)=1 \)
Solution |
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\(\displaystyle{ \frac{dy}{dx}=e^{x-y} }\); \( y(0)=\ln(2) \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx}=e^{x-y} }\); \( y(0)=\ln(2) \)
Solution |
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\(\displaystyle{ \frac{dy}{dt} = \frac{t}{t^2y+y} }\), \( y(0) = 3 \), \( y > 0 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dt} = \frac{t}{t^2y+y} }\), \( y(0) = 3 \), \( y > 0 \)
Final Answer |
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\( \sqrt{ \ln(t^2+1) + 9 } \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dt} = \frac{t}{t^2y+y} }\), \( y(0) = 3 \), \( y > 0 \)
Solution |
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Final Answer |
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\( \sqrt{ \ln(t^2+1) + 9 } \) |
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Intermediate |
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\( dv/ds = (s+1)/(sv+s) \)
Problem Statement |
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\( dv/ds = (s+1)/(sv+s) \)
Final Answer |
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\( v = -1 \pm \sqrt{ 2s + 2\ln(s) + C } \)
Problem Statement |
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\( dv/ds = (s+1)/(sv+s) \)
Solution |
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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.
Final Answer |
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\( v = -1 \pm \sqrt{ 2s + 2\ln(s) + C } \) |
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\(\displaystyle{ \frac{dy}{dx} = \frac{3x^2+4x+2}{2(y-1)} }\); \( y(0) = -1 \)
Problem Statement |
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\(\displaystyle{ \frac{dy}{dx} = \frac{3x^2+4x+2}{2(y-1)} }\); \( y(0) = -1 \)
Solution |
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You CAN Ace Differential Equations
Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, solve these separable differential equations and find the particular solution, if possible. Give your answers in exact terms.
