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### Course: Differential equations>Unit 1

Lesson 8: Homogeneous equations

# First order homogeneous equations 2

Another example of using substitution to solve a first order homogeneous differential equations. Created by Sal Khan.

## Want to join the conversation?

• At how come we can just drop the absolute value function from the cx term?
• I'm sure Sal just forgot to put an absolute value here. We can not drop it. The reason is that C is an ARBITRARY constant. Arbitrary means that it can be ANYTHING and it should NOT depend on whether x is positive or negative. Let me explain. If we drop the absolute value then the choice of C, positive or negative, puts restrictions on a domain of the function. If C>0 then x must be >0 too, if C<0 then x must be <0. It's a huge problem. Doing so we voluntary deprive yourself from a half of solutions! But if we keep the absolute value then sign of x does not depend on choice of C. Here is an example: suppose a pair (a,b) is a solution to our diff. equation. If we save the absolute value, then pair (-a,b) is also solution, but if we drop the absolute value it's not.
• I have been struggling with understanding exactly what we are doing in these videos on diff equations. It si like I see how to do it, but I'm bot sure what I am doing. But is it correct to say that we are presented with some expression containing the derivative of a function, and the solution is an expression with no derivative, only the functions themselves? So once I have reached an expression without any dx/dy, y' etc, I am done?
• Hey Tommy, you have it right.
Heres an analogy that may be helpful. In algebra we have algebraic equations, e.g. 2x+5=-1 , the solution of which is a number. In this case x=-3.
In differential calculus we have differential equations, e.g., y'(x)=y(x), the solution of which is a function.
The ultimate goal is to write that function in terms of simple operations (+,-,*,/,sin,cos,log,...).
Sometimes there is no solution. Sometimes there is a solution but it can't be represented in terms of simple operations. In an introductory class you will only get a glimpse of the universe of differential equations.
• Why don't we solve for Y to complete the problem? I understand we can easily do that and I also understand the final result of this video, but why do we leave it in the form someFunction(x,y) = 0 rather than solving Y explicitly?
• It's not as easy to "solve for y" as you may think. The final answer in this video (x^2 + y^2 − cx^3 = 0) is expressed as an implicit function of y. Clearly, we have y^2 = cx^3 − x^2, which is still valid, but not quite "solved for y." You might be tempted to say that y = sqrt(cx^3 − x^2), but that's only the positive square root, so you're losing half the values!

To demonstrate, I plotted both the explicit form and the not-quite-correct explicit form (https://i.imgur.com/y50vYNL.png). You can see that the curves are the same, but that the explicit form only has the positive values.

Now, in this example, you could say (hand-wavily) that y = ± sqrt(cx^3 − x^2). This isn't really a function, because there are two y-values for each x-value in the domain. On the other hand, you could say that y_1 = sqrt(cx^3 − x^2) and y_2 = −sqrt(cx^3 − x^2); together, these two functions represent the entire solution set.

In general, the implicit function theorem (https://en.wikipedia.org/wiki/Implicit_function_theorem) states (loosely speaking) that any implicit function (i.e., f(x,y) = 0) can be expressed as the "combination" of some set of functions {y_1(x), y_2(x), …}. In this case, we only needed two. In other cases, we might need more. For example, the implicit function sin(y) = 0 is true for y = 0, 2π, −2π, 4π, −4π, …, so we would need infinitely many explicit functions: {y_1 = 0, y_2 = 2π, y_3 = −2π, y_4 = 4π, y_5 = −4π, …}.

Finally, for some applications, you don't need a (set of) explicit function(s). For example, you may just want to test whether a given value is on the curve. In this case, you could easily just evaluate f(x, y) for a given point and test whether it's zero.

(Bonus: imagine trying to come up with the explicit functions for this graph! https://i.imgur.com/G20kEbR.png)
• At couldn't he have taken everything as a power of e. to cancel the natural log. Or would it have become e^(ln(x) + C) which is the same as e^(ln(x))*e^c which is cx.

So it would be the same solution, so, nvm. Just figured that in the fly. Was my logic correct?
• Your logic is correct, I regualary use the approach you suggested in my Differential Equations class and the teachers does it too. I think Sal just took a different approach
• Do we always substitute v = y/x and write the function in terms of y/x?
• If we have a differential equation of this form, then this is the first thing you should try (if no other obvious and simpler method is apparent).
• where is the homogeneous equations practice ??
• i'm having trouble defining the proper method to solve this eq: y'-1=e^(x+2y)
any help would be appreciated