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You CAN Ace Differential Equations

Topics You Need To Understand For This Page

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17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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how to take good notes

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

Final Answer

\( y = e^{-2x} \)

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 \( y(0)=1 \),
\( 1 = Ae^0 ~~~ \to ~~~ A = 1 \)

Final Answer

\( y = e^{-2x} \)

close solution

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

Problem Statement

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

Final Answer

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

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 \( y(0)=2 \),
\( 2 = 1-B ~~~ \to ~~~ B = -1 \)

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

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

Final Answer

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

close solution

\( y' = y - y^2 \)

Problem Statement

Find the general solution to \( y' = y - y^2 \)

Final Answer

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

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 \(y\).
\(\begin{array}{rcl} \displaystyle{\frac{y}{1-y}} & = & Ae^x \\ y & = & Ae^x (1-y) \\ & = & Ae^x - yAe^x \\ y + yAe^x & = & Ae^x \\ y(1+Ae^x) & = & Ae^x \\ y & = & \displaystyle{\frac{Ae^x}{1+Ae^x}} \end{array}\)

Final Answer

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

close solution

\(\displaystyle{ \frac{dy}{dx} = xy }\)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = xy }\)

Solution

559 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Solution

560 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Solution

562 solution video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

563 solution video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

564 solution video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

565 solution video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

566 solution video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

568 solution video

video by Krista King Math

close solution

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

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

close solution

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

Problem Statement

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

Solution

561 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Solution

569 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

570 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

571 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

572 solution video

video by Khan Academy

close solution

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

Problem Statement

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

Solution

574 solution video

video by Khan Academy

close solution

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

Problem Statement

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

Solution

609 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

1842 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Solution

1843 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

2082 solution video

video by MIP4U

Final Answer

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

close solution

Intermediate Problems

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

Problem Statement

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

Final Answer

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

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.

Final Answer

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

close solution

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

Problem Statement

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

Solution

573 solution video

video by Khan Academy

close solution
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