24 First-Order Differential Equations

First Order Differential Equations Ultimate Calculus Tutorial! The topics include separable differential equations, first-order linear differential equations, exact differential equations, almost exact differential equations, homogeneous differential equations, Bernoulli differential equations, Clairaut differential equations, Riccati differential equations, special substitution, and special integrating factors.

timestamps,
0:00 24 first order differential equations
Q1, 1:57
Q2, 12:08
Q3, 28:47
Q4, 36:16
Q5, 43:27
Q6, 55:50
Q7, 1:01:39
Q8, 1:07:32, mistake at 1:10:59, please jump to 1:17:28
Q9, 1:28:10
Q10, 1:32:35
Q11, 1:40:47
Q12, 1:48:37
Q13, 2:01:00, Clairaut differential equation,
Q14, 2:21:19
Q15, 2:30:10
Q16, 2:38:56, logistic differential equation,
Q17, 2:49:20, Gompertz differential equation,
*Struggling.... (because of a typo in my question) from 3:02:20 to 3:21:15
**How to create your own almost exact differential equation? 3:21:15
Actually solved Q18, 3:29:20 YAYYYY (my THIRD try!!)
Q19, 3:39:53
Q20, 3:45:25
Q21, 3:49:25
Q22, 4:00:15 Riccati differential equation (I messed up. Please use y2=y1*v instead of y1+v)
Q23, 4:38:54
Q24, 4:46:54 This is actually *also Bernoull* LOL! We can write it as dy/dx+1/x*y=x*y^-1
I will redo this one in all three ways!!

Want more diff eq? check out...

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[blackpenredpen]

Sorry for all the mistakes! Most of them were just minor but fatal 😡 ...

but this was so much FUN!!!!

Anyway, stay safe and healthy! Best wishes to everyone!

Take care
24/(1-x)

[drpeyam]

Haha, the hero we need in this time of need 😍

[riccardoimolesi]

Gf: I'm sure he's cheating on me
Me and the boys:

[younesabid]

And it all started a year ago with the 100 integrals madness...

[BhanuSingh-mcfy]

This guy is seriously amazing ....
Totally dedicated to do the best and needful work
His work is truly appreciative....

[stk]

Hi,
Please atleast drink water.
Staying hydrated is crucial in such times.

[kwertie]

I LOVE how varied your questions are and it shows that you took time to pick out the questions too. This just inspires me to do math. You're an amazing person bprp.

[MuPrimeMath]

Wow, that's a lot of different types of equations!

Btw, if anyone is interested, my differential equation videos are mostly about higher-order differential equations.

[jingchen]

To whoever is reading this, I hope you will excel at math!!!

[pvhdvut]

Question 14)
I'll write dy/dx as y'
y' = (x-y)/(x+y) I did it slightly differently.
y' = (x+y - y - y)/x+y, add y and subtract y on numerator
y' = 1 -2y/(x+y), split into two fractions
y' = 1 - 2(y/(x+y)), bring the 2 "outside the fraction"
y' = 1 - 2(y+x-x/(x+y)), add and subtract x on numerator
y' = 1 - 2(1 - x/(x+y)), split the inside into two fractions
y' = 1 - 2 + 2x/(x+y), expansion
y' = -1 + 2x/(x+y)
Let v = x+y
=>y = v-x, make y the subject
=>y' = v' - 1, differentiate both sides
v' - 1 = -1 + 2x/(v), substitution
v' = 2x/v, cancel the -1 on both sides.
dv/dx = 2x/v, using different notation
vdv = 2xdx, bring the v's with the dv's and the x's with the dx's
v^2/2 = x^2 + C_1, integrate both sides {C_1 is a real number}
v^2 = 2x^2 + 2C_1, multiply both sides by 2
(x+y)^2 = 2x^2 + C_2, substitute x and y back in instead of v{C_2 = 2C_1}
x^2 + 2xy + y^2 = 2x^2 + C_2, expansion. I basically have the same answer as you now.
However, I'm going to do something extra here.
y^2 + 2xy - x^2 + C_3 = 0, bring everything to the left hand side. {C_3 = -C2}
On the left hand side "we have a quadratic, so I'll now use the quadratic formula to find the roots of y"
y = (-2x +/- sqrt((2x)^2 - 4(1)(-x^2+C_3)))/2, using the quadratic formula
y = (-2x +/- sqrt(4x^2 + 4x^2 - 4C_3))/2, simplifying
y = (-2x +/- sqrt(8x^2 - 4C_3))/2, further simplifying
y = (-2x +/- sqrt(4(2x^2 - C_3)))/2, factorising inside the sqrt
y = (-2x +/- 2sqrt(2x^2 + C_4))/2, simplification {C_4 = -C_3}
y = (2(-x +/- sqrt(2x^2 + C_4)))/2, factorisation of the numerator
y = - x +/- sqrt(2x^2 + C), cancel out the 2 on numerator and denominator, {C = C_4}
^I guess which branch to take will depend on the initial conditions.
Checking answer via differentiation:
First take y = -x + sqrt(2x^2 + C)
=> y' = -1 + (4x)/(2sqrt(2x^2+C)), differentiate both sides
=> y' = -1 + 2x/sqrt(2x^2+C) [eqn 1]
Recall y = -x + sqrt(2x^2 + C)
=> y+x = sqrt(2x^2)+C [eqn 2]
sub [eqn 2] into [eqn 1]
=> y' = -1 + 2x/(x+y) [***]
=> y' = (-(x+y) + 2x)/(x+y), getting a common denominator
=> y' = (- x - y + 2x)/(x+y), expansion on numerator
=> dy/dx = (x - y)/(x+y), simplification. Matches the differential equation at the start.
Therefore y = -x + sqrt(2x^2 + C)
Checking other solution:
Now take y = - x - sqrt(2x^2+C)
=> y' = -1 - (4x)/(2sqrt(2x^2+C)), differentiate both sides
=> y' = -1 - 2x/(sqrt(2x^2 +C) [eqn 3]
recall y = -x - sqrt(2x^2 + C)
=> sqrt(2x^2 + C) = -x - y [eqn 4]
Sub [eqn 4] into [eqn 3]
y' = -1 - 2x/(-x-y)
y' = -1 - 2x/-(x+y), factorising the negative on the denominator
y' = -1 + 2x/(x+y), cancellation
but this equation is the same as [***]
which implies dy/dx = (x - y)/(x+y)
Check complete.

[PlayGuy]

Its march 19 2023 exactly 3 years after😄 and i got a differential equation exam tomorrow👍. Thank you

[aryanks]

I love the amount of effort you put into explaining everything. I'm sure you could do this in 3 times less time otherwise. After I have finished studying diff eq, at at least high school level, I'll attempt to follow through this all at once

[mohammedm]

I’m taking ODE this semester and this serves for a great review.

Thanks math dad

[georgesadler]

MR. blackpenredpen, thank you for an incredible journey through Introductory Differential Equations. Finding and correcting errors is another great tool in advanced mathematics. The overall video/lecture improves my understanding of Differential Equations.

[pklearning-mathandstatisti]

Wow, great concentration! I would not imagine solving differential equation for 5 hours without eating anything or drinking water. This video should inspire many students majoring in math. Great!

[jpdemont]

Outstanding Professor Chow. Don't ever hide or apologize for mistakes. The fact that you make mistakes gives the rest of us mortals -- who make many more mistakes -- perspective and hope that perhaps someday, with lots of practice, that we can become as proficient as you. Also, it is very important that you mention often that math in general, and differential equations in particular, are very difficult.

[waterninja]

For 6 it can be simplified to dy/dx = (x - xy)/(x^2 + 1) and then to (1/(1 - y)) dy = (x/(x^2 + 1)) dx, making the problem a lot simpler without the need for an integrating factor.

[hadymudkipper]

I truly am fascinated at how you attack those problems with such ease ! I hope I become as good as you one day. The hero we need in these tough times ! Sending support from Egypt !

[CliffSedge-nufv]

I'm halfway through, just finished #15. Now I feel like we're warmed up and ready for this crazy back half.
(I'm a math teacher too and using this to find good questions for my students.)

[isaacaguilar]

For 12, 1:52:41 after Bernoulli’s substitution, its easier to just separate variables instead of an integration factor.