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## Differential equations

### Course: Differential equations>Unit 1

Lesson 8: Homogeneous equations

# First order homogenous equations

Introduction to first order homogenous equations. Created by Sal Khan.

## Want to join the conversation?

• Y=(x)Ln|x| +Cx

if C is an arbitrary constant, wouldn't Cx=C? • Does the domain of the answer have to be restricted so that x=/=0 because v=y/x? • Suppose a term in the differential equation is 'xy' , then its degree is 2. Now what is the degree of the term tan(xy) or log(xy) ? • The term "degree" can only be used to qualify polynomials. Expressions that include trigonometric functions are no longer polynomials, and therefore don't have a "degree".

I think you are confusing the term "degree" of a polynomial with a differential equation "linearity". A "linear" differential equation (that has no relation to a "linear" polynomial) is an equation that can be written as:
``dⁿ         dⁿ⁻¹         dⁿ⁻²                 dy――y + A₁(x)――――y + A₂(x)――――y + ⋯ + A[n-1](x)―― + A[n](x)ydx          dx           dx                  dx``

In your example, since `dy/dx = tan(xy)` cannot be rewritten in that form, then it would be a non-linear differential equation (and thus also non-homogenous, as only linear differential equation can be homogenous).
• dy/dx=y^2 this equetion is linear or not • Sal is preventing me from hating maths. It's all clear now, but video quality is pretty low, r u guys thinking of redoing it? • I was not clear of what exactly is a homogenous equation or homogenous function. What is it in simple terms ? • I wouldn't get caught up in the terminology cause it's a bit weird. Really there are 2 types of homogenous functions or 2 definitions. One, that is mostly used, is when the equation is in the form:
ay" + by' + cy = 0
(where a b c and d are functions of some variable, usually t, or constants)
the fact that it equals 0 makes it homogenous. If the equation was
ay" + by' + cy = d
then you'd end up with a result that was the same as the homogenous result PLUS a particular solution.
Now those are mostly what we talk about when we call something homogenous. They are nice because you can usually solve them with a characteristic equation. There's a COMPLETELY DIFFERENT meaning of homogenous also used in DE (just to confuse us I guess), where y' = F(y/x). One example of this type of homogenous function is y' = y/x + 1, which doesn't really equal 0, but this form is called homogenous and you use a change of variables (let v = y/x) to solve it. Again, I wouldn't get caught up in the terminology and what it means to be homogenous, since there are really 2 different definitions. But mostly if something is homogenous, it's a good thing because we have methods to solve it. And if it's not, then we solve it as if it were homogenous ( set it equal to zero instead of whatever it's equal to) and then do some additional calculations for a particular solution based on what it equals.
• At Sal says that the derivative of v with respect to x is equivalent to 1/x. But isn't dv/dx = d(y/x)/dx which is (-y/x^2), not 1/x? I think Sal was thinking of dv/dy.
dv/dy would be d(y/x)/dy equaling 1/x as the y would be derived to 1. • He was not thinking of dv/dy. He differentiated both sides of y=vx with respect to x and solve for dv/dx.

You can also compute dv/dx directly, however dv/dx = d(y/x)dx is not (-y/x^2) like you suggested since y is a function of x. So dv/dx = d(y/x)/dx = [(x(dy/dx)-y]/x^2, by substituting dy/dx with the expression from the original problem, the right hand side simplifies to 1/x.
• So how does this relate to the method of exact equations? Is it that when rearranged where you have Mdx+Ndy=0 and then when you check for exactness; (partial of M)/(dy) = (partial of N)/(dx), this chosen function of (x+y)/(x) fails that part so you need to do substitution? I learned the exact method so far and don't really recognize this method yet.   