Separable differential equation 2.2#17

Learn how to solve a separable differential equation. This is usually the first kind of differential equations that we learn in an ordinary differential equations class.

To learn how to solve different types of differential equations:

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#DifferentialEquations

[OonHan]

You must constantly remember your constant rules

[dahazz]

Hey man, doing a great job here! I think I can solve almost any separable and firs order linear equation. But I have no idea  how to distinguish between them. Like this one for example, it looks linear and separable at the same time. Since the equation can be also written as y'+yx^3=x^3. Which is linear since P(t)=x^3 and q(t)=x^3. Can we solve it by using an Integrating factor? Please let me know, thank you.

[SmileyMPV]

4:18
Why should the +- be constant? Maybe whether it is + or - is dependent on x meaning +-c3 is not constant.
I do see that y would not be differentiable if the sign flips, but in general this does not work.

[Azouzazu]

4:40 Plus minus a constant I S N O T always equal to a constant in functional equations. Consider the following counterexample.
Let y be a real differentiable function of x, y(1)=1 and holds that:
|y|=c|x³|, for all real x.
The only solutions to this problem are the functions:
y=x³ for all real x and
y=|x³| for all real x
(This can be proven rigorously by finding the zeroes of y, which is only 0, and by finding the sign of y in both the intervals (-infty, 0) and (0, +infty). The sign of y is constant in each interval since y is continuous).
BUT if we try to solve the problem using that ±constant=constant we get the following:
|y|=c|x³| for all real x
=>y=±c|x³| for all real x
Let ±c=C thus: y=C|x³|
By substituting x=1 and y=1 we get:
1=C|1|=>C=1
Therefore y=|x³| for all real x
CONTRADICTION! because we know that y=x³ for all real x is a function that satisfies all the conditions too.
Thus ± a constant I S N O T a constant.
In fact blackpenredpen was lucky here because there was only one function satisfying the differential equation. If the equation 1-y=0 had at least one solution for y then blackpenredpen would have incomplete conclusions. But blackpenredpen arbitrarily (and wrongly) implied that 1-y=0 has no solutions by dividing by (1-y) in both sides.

[frede]

But what if y is -1.5, for example? The expression 2e^(-x^4/4)+1 can't be negative, because none of the terms can be negative. But this would work if +/-C_3 could be two constants: C_4=C_3 or C_5=-C_3. You said that +/- C_3 was only one constant C_4. But if you let it be C_4 OR C_5, then the constant in front of e^(-x^4/4) could be 2 when y is positive, and -2 if y is
negative. Then there would be no problem with negative y-values.

[TheGrimReaperMc]

Isnt this like saying that c4 = c3 AND c4 = -c3 ? how can a constant be something and minus that something at the same time? (except 0 of course)

[MrZauberwuerfel]

do we not need to define c3 as >0? because e^c2 has to be positive

[illyakuzmych]

why add a constant to the right side of the integration but not the left?

[shobhitmishra]

This post was clearly made by the Leibniz Notation Gang

[oziwan]

We have + or - c3 but c3 was an e^c2 so c3 wasnt ANY constant, it was a positive constant: so why can u change it by c4

[frede]

When you let C_4 be C_3, then you also "destroyed" the possibility that y
could be smaller than -1

[patrickhodson]

You just used a _blue_ pen!
*dislikes video and unsubs from channel*