Section 2.2 Homogeneous DEs (related to separable DEs) Key Terms: Homogeneous Functions. Degree of a term. First Order DEs of Homogeneous type

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1 Section. Homogeneous DEs (related to separable DEs) Key Terms: Homogeneous Functions Degree of a term First Order DEs of Homogeneous type

2 From experiences in your study of mathematics you have encountered situations wherethe best way to solve a new problem is to reduce (or change) it, in some way, into the form of a problem that you already know how to solve. This is what we will do with homogeneous differential equations. Basic definition of homogeneous: of the same or a similar kind or nature; composed of parts or elements that are all of the same kind. Example: When a cow is milked, and as the milk settles, a layer of cream forms at the top of the milk. This used to be the way people would judge the quality of milk. A thicker layer of cream meant better quality milk, and especially when milk was still normally sold in bottles, you could easily see into the bottle to judge the cream layer. Pasteurization had become standardized for milk in most countries, since heating the milk destroys any bacteria in it, making the milk safer to drink. Homogenized milk was the next step. Since milk is an oil and water combination, it doesn t stay mixed. Homogenized milk is run through tiny tubes, sometimes during the pasteurization process to keep fat and liquid molecules together. Fat molecules are reduced in size and tend to disperse more evenly throughout the milk so that creaming on the top of milk doesn t occur. Source:

3 For math we start by discussing homogeneous functions in two variables x and y. An easy way to think about homogeneous functions is to say they are functions where the sum of the powers of every term are the same. (This works most of the time.) Examples: F(x,y) = 3x 5xy + y The sum of the exponents of x and y in each terms is. So this a homogenous function. G(x, y) = 9x x + 5y The term -x has the sum of its exponents 1, while the other terms have the sum of the exponents. This is not a homogeneous function. Often the sum of the exponents of the variables in a term is called the degree of the term. 3 H(x, y) = 5x y x + 7 Has two terms of degree 3, but one term of degree 0. H is not homogeneous.

4 Definition: A first order DE y = f(x, y) is called a homogeneous DE provided we can express f(x, y) as a function of yx alone. At times it is easier to transform the DE y = f(x, y) into the form M(x,y) + N(x, y)dy = 0 and check to see if M and N are of the same degree. This gets tricky when fractional powers appear in the DE. In that case it is best to use the formal definition. Example: Determine if the following DE are homogeneous. dy x + 3y = xy Several ways to check things: both numerator and denominator of degree here if we divide both numerator and denominator by y we get write in the form (x + 3y ) -xy dy = 0 and M and N are both of degree ; it is homogeneous dy = + If we divide both sides by x we get dy y x y y = + = + 1 x x x x y x y so this DE is homogeneous ( x y) ( x)dy + = This is not homogeneous since the degree of M is 1, but the N has terms of degree 0 and 1.

5 How to solve homogeneous DEs. If you recognize the fact that an equation is homogeneous you can often perform a substitution which will allow you to use separation of variables to solve the equation. dy dv Set y = vx, then v x. = + The transformed DE will be separable. dy x + 3y Example: Solve the following homogeneous DE = xy dy dv dv x + 3v x Set y = vx, then = v + x and the DE becomes v + x = x(vx) Simplify and rearrange. dv x + 3v x 1+ 3v 1+ 3v v 1+ v x = v = v = = x(vx) v v v vdv Separate the variables. 1+ v = x Integrate both sides. 1 ln + v = ln x + C 1+ v = Cx Let v = yx. 1+ = Cx 3 x + y = Cx B&D Sec.. #3

6 Example: Solve the following homogeneous DE dy = y x + xy Check to see if it homogeneous; on the right side divide numerator and denominator by x. We get dy 1 y x = + The right side is a function of yx alone so it is homogeneous. yx dy dv Set y = vx, then v x. = + The transformed DE will be separable. We get dv v v x 1 v Separate the variables 1+ v dv = v v x + = + Simplifying we get 3 1 simplify ( ) dv v v v v v v v x = v = = 1+ v 1+ v 1+ v v v dv = x 1 Integrate to get v ln v = ln x + C Set v = yx Simplifying we get 1 ( ) ln y ln x = ln x + C 1 y ln = ln x + C x Which finally simplifies to 1 ln y = C