If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Differential equations>Unit 1

Lesson 1: Intro to differential equations

# Worked example: linear solution to differential equation

If a particular solution to a differential equation is linear, y=mx+b, we can set up a system of equations to find m and b. See how it works in this video.

## Want to join the conversation?

• How do you know the solution is a linear function?
• Unfortunately, for most differential equations, is a mixture of practice and experience that gives you an idea of what kind of equation might be the solution. There is not a set method in order to find what family of function would make a good solution for a particular differential equation.
• Hello, thanks for all the help:

In this procedure: m=(3m-2)x+3b-5 to be true for all X values, the coefficient before the x value has to be 0.

Can we corroborate this plug in some numbers? I still don't see it.
• With numbers, you could think of this equation as ax + 5 = 5, where you control the variable a, but the variable x is outside your control, and can be any number whatsoever. The only way you can make this equation true is by making the variable a (the one you control), equal to 0, that way the variable that is outside your control stops messing the equation.
• This is the last video before the exercises start. I understand everything up until this video, but I can't answer the questions. I suggest you add more videos to prepare us for the exercises.
• at @ sal asks us to verify it , can anyone explain how to do so ?
• Yes. You can take the solution:
y = 2/3 x + 17/9
and plug it back into the original equation.

From the solution then
dy/dx = 2/3
plugging back into the original:
2/3 = -2x + 3(2/3 x + 17/9) - 5
2/3 = 17/3 - 5
2/3 = 17/3 - 15/3
2/3 = 2/3
So you verified the solution
• At the beginning of the video you said that if some gave a clue which indicated that the solution to this Diff Eq is a linear (y=mx+b). My question is, what if you didnt have any clue and you had to solve this diff qe problem? how would you do it?
• Watch a couple of the later videos on general cases.
• The starting point in solving this problem was knowing that the solution would be linear, of the form y=mx+b ( in the video). How do we know that the (or one) solution would be linear, i.e. why is a linear solution assumed?
• A linear solution is assumed because that is how Sal set out the question, perhaps to make it easier to solve.
• Hey just a quick question
Wouldn't
y = -x^2 + 3yx - 5x
Also be a solution to that differential equation. As taking the derivative of that function would be equal to dy/dx = -2x + 3y - 5?
• Like Sal mentioned, there are multiple solutions to differential equations, so that should be right. I got the same answer as you after taking the integral of the given differential equation on both sides.
• I don't get why m has to equal 3b-5 and therefore the coefficient of x has to be zero. Why can't m>3b-5 for example, resulting in a need for an actual value of x to balance the equation?
• We want to solve for m, the only assumption we can make that we know is true is that if the coefficient on x is zero, the x term disappears. Therefore m = 3b - 1.
• why does that whole equation needs to be equal to "m"