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Separable equations (old)

An old introduction video to separable differential equations. Created by Sal Khan.

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• At , when Sal throws the constant of integration in on the right with the 'x' terms, could he have just as well put it on the left and come up with -3 as an answer so C would be +/- 3 instead of just 3?
• If someone puts the C on the RHS, like Sal did, then the answer will be C=3, and this will be correct for the way that person wrote it, i.e., with C on the RHS. Writing C=-3 would be incorrect in this case.

If you put the C on the LHS, then the answer will indeed be C=-3, and this will be correct for the way you wrote it, i.e., with the C on the LHS. Writing C=3 would be incorrect in this case.

So technically, you can't write C=+/-3, because 3 would only be correct if you introduced C on the RHS, and -3 would only be correct if you had introduced C on the LHS.
• What does it actually mean to 'multiply both sides by dx'? It makes sense algebraically, but in the first place, dx is not a number.
• Does whether an equation is linear impact the method of solution to that equation, or are linear and nonlinear simply ways to classify types of differential equations?
• It matters. You would use a different method to solve a nonlinear than you would to solve a linear one. If you watch more of the videos you will see why! Hope this helps!
• Why can't I play this video?
• when finding the constant we are given that y(0)=-1. Why does Sal then plug in -1 for all the y's and 0 for the x's? shouldn't it be the other way around?
(1 vote)
• y(0) = -1 is written in f(x) notation. x is independent variable and y is dependent variable. In this case 0 is x and y is -1 and the variables are substituted with their respective values.
• dy/dx = sqrt(y)/sqrt(x) => 1/sqrt(y) dy = 1/sqrt(x) dx => 2 * y^(0.5) = 2 *x^(0.5) + c=> y = x + 2 * x ^0.5*C + C
• I need a little help with a physical DGL describing a capacitor:

R*Q'(t) + Q(t)/C = 0, where Q(t)=C*U(t), C is capacity, U is voltage and R resistance

I can write this as RC*U'(t) = - U(t), which then leads to U'(t)/ U(t) = -1/RC. Then I integrate on both sides, which gives, according to my books, log(absolute value(U(t))) = -t/RC + constant.

My question is, how do you get the logarithm in there? Is integral[ f'(t)/f(t) ] = log [ f(t) ] and therefore the same as integral[ 1/f(t) ]?
• Yeah I figured it out by myself, haven't logged into Khan Academy since, but thank you anyway :)
(1 vote)
• what is the convention that groups the 2 to the (y-1)? In other words, why is it incorrect to simply multiply both sides by (y-1) instead of 2(y-1)?
• Both ways are correct, you can leave the 2 in the denominator of the right hand side and integrate it with the rest of the x dependant parts. The result will be the same regardless of where that 2 ends up at the moment of separation.

I guess Sal decided to take the 2 with the (y-1) to avoid having a denominator in the right hand side.
• what is implicit form of a differential equation?
• While solving the differential equation you used the relation (x/y)^1/2 = (x^1/2)/(y^1/2) which is not true as when x is negative and y is negative the first expression is defined but the second is not?