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.

Main content

Current time:0:00Total duration:7:21

I will now introduce you to the idea of a homogeneous differential equation homogeneous homogeneous is the same word that we use for milk when we say that the milk has been that all the fat clumps have been spread out but the application here at least I don't see the connection homogeneous differential equation and even within differential equations we'll learn later there's a different type of homogeneous differential equation and those are called homogeneous linear differential equations but they they mean something actually quite different but anyway for this purposes we're going to I'm going to show you how modulus differential equations and the what we're dealing with are going to be first-order equations and what is a homogeneous differential equation mean well let's say I had just a a regular first-order differential equation that could be written like this so dy/dx is equal to some function some function of x and y and let's say we you know we try to do this and it's not separable and it's not exact what we what we what we learn is that when we if it can be homogeneous if this is a homogeneous differential equation that we can make a variable substitution and that variable substitution allows this quick equation to turn into a separable one but before I need to show you that I need to tell you what what does it mean to be homogeneous well if I can algebraically manipulate this right side of this equation so that I can actually rewrite it instead of a function of x and y if I could rewrite this differential equation so that dy/dx is equal to some function I don't know let's call that I don't know we'll call that G or we'll call it capital F if I can rewrite it algebraically so it's a function of Y divided by X then I can make a variable substitution that makes it that makes it separable so right now it seems all confusing let me show you an example so let's say that let's say that and I'll just show you the example show you so modulus and then we'll just do the substitution so let's say that my differential equation is d the the derivative of Y with respect to X is equal to X plus y over X and you can if you'd like you can try to separate make this a separable but it's not that trivial to solve or at least I'm looking at an inspection it doesn't seem that trivial to solve and as we see right here the you know we have the derivative its equal to some function of x and y and my question is to you can I just algebraically rewrite this so it becomes a function of Y over X well sure if we just divide both of these top terms by X right this this is the same thing as x over X plus y over X right this equation is the same thing as dy over DX is equal to this which is the same thing as rewriting this whole equation I'm going to switch colors arbitrarily as this dy over DX is equal to X divided by X is equal to one if we assume X doesn't equal zero plus y over X so you're probably wondering what did I mean by a function of Y over X well you can see it here when I just algebraically manipulated this equation I got 1 plus y over X so if I said that Y over X is equal to some third variable I this is just a function of that third variable and actually I'm going to do that right now so let's say let's make a substitution for y over X so let's say that V and I'll do V in a different color let's say that V is equal to Y over X or another way if you just multiply both sides by X you could write that Y is equal to X V and we're going to substitute V for Y over X but we're also going to have to substitute dy over DX so let's figure out what that is in terms of the derivatives of V so the derivative of Y with respect to X is equal to what's the derivative of this with respect to X well if we assume that V is also a function of X then we're just going to use a product rule so the derivative of X is 1 times V plus X times the derivative of V with respect to X and now we can substitute this and this back into this equation and we get so dy over DX that is equal to this so we get V e+ X DV DX 3 review with respect to X is equal to that's just the left hand side it's equal to 1 plus y over X but we make we're making this substitution that V is equal to Y over X equal to 1 plus V and now this should be pretty straightforward so let's see we can subtract V from both sides of this equation subtract V from both sides and then what do we have left with we have X DV DX is equal to 1 let's divide both sides by X and we get the derivative of V with respect to X is equal to 1 over X it should maybe start becoming a little bit clearer what the solution here is but let's just keep going forward so if we multiply both sides by DX we get DV is equal to 1 over x times DX now we can take the antiderivative of both sides integrate both sides and we're left with V is equal to the natural log of the absolute value of x plus C and we are kind of done but it'd be nice to get this solution in terms of just Y and X and not have this third variable V here because our original problem was just in terms of Y and X so let's do that what was V we made the substitution that V is equal to Y over X so let's reverse substitute it now or unsubstituted so we get Y over X is equal to the natural log of X plus C some constant multiplied both sides times X and you get Y is equal to x times the natural log of X plus C and we're done we solved that that seemingly inseparable differential equation by recognizing that it was homogenous and making that variable substitution V is equal to Y over X that turned it into a separable a separable equation in terms of V and then we solved it and then we unsubstituted it back and we got the solution to the differential equation you can verify it for yourself that Y is equal to the X natural log of the absolute value of x plus the Oh actually I made a mistake y over X is equal to the natural log of X plus C if I multiply both sides of this equation times X what's the solution it's not just X natural log of X it's it's I have to multiply this times X to right I'm ultimate of property that was a that was an amateur mistake so the correct solution is y is equal to X natural log of the absolute value of x plus x times C and if you wanted to figure out C I would have to give you some initial conditions and then you could solve for C and that would be the particular solution then for this differential equation in the next video I'll just do a couple more of these problems I'll see you then