Separable Differential Equation (introduction & example)
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- Опубликовано: 18 сен 2024
- Learn how to solve a separable differential equation. This is usually the first kind of differential equation that we learn in an ordinary differential equations class. We will also the differential equation dy/dx=x+xy^2 with the initial condition y(0)=-1.
To learn how to solve different types of differential equations:
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#DifferentialEquations
The piano playing in the background is the secret behind building a math genius.
@Λ bummer
C is actually infinitely recurring since tangent is periodic. C = -pi/4 + pi•k, where k is an integer.
Good catch!
I was confused about this sense there is multiple solutions to tan(x)=-1
you describe everything so clearly that even watching at double speed I can follow everything
Thank you so much for the Diff Eq series!! I've learned so much more in this 9 minute video than I have in watching 50 mins of my professor talk.
Man, youre just getting even better topics! I am actually studying Differential Equations this semester(Computer Engineering) and I bought Wiliam' Boyce's book(Its awesome). Imagine my surprise to find out that youre making a series about D.E! I was a huge fan of you now I am a disciple, my friend! Thanks for helping me.
I'm taking diff eq this spring semester coming up! Will you be making more diff eq videos??? Also I got a B in calc 3!!!! You helped me alot while i was taking calc 2 haha
I will try to make as many vids as possible before Spring, but I do have to teach a class in the Winter, so we will see!
THANKYOU SO MUCH SIRRRR!!! YOU HELPED ME ALOT, PLS DONT STOP MAKING VIDEO TUTORIALS LIKE THIS, WE REALLY APPRECIATED IT! MORE POWER WOHO!
I love how you can tell he enjoys what he is doing so much.
Thank you so much. Tomorrow is my midterm exam. You helped me too much.
This was extremely helpful. THANK YOU
Well, now I can't wait for differential equations next fall! Great video as always!
THANKS!
YOU.ARE.THE.BEST.
Brilliant review. Just what I needed.
blackpenredpen ily
dude you are an absolute beast!
Using the identity tan(θ+kπ) = tanθ
You should be able to write a general solution given initial condition y(0) = -1
I passed my Integral Calculus subject with the help of ur videos. Help this will help me as well. thank u :)
thank U so much to be successful🥰
thanks for the help
man i would love if you do some about partial differential equations
Thank you for the calc help :)
I'm struggling with what feels like some ambiguity here: arctan(y)=1/2 x^2 +c. I'm used to arctan only taking values from - to +pi/2, but the RHS can clearly take a much larger range of values. If we consider arctan +n(pi), does that not create ambiguity for equality when taking tan of each side?
Did you do one where there is dy/dx on both sides like this y-x dy/dx = 3-2x^2 dy/dx? Where is it?
Found ‘em
Who Am I? Excellent!
I have Diff EQs this term. Thanks for this :)
you're welcome
Man, I wish this was on last semester made it so much easier.
Can you prove that any function with two separate inputs x and y can be represented as a product of a function of x and another function of y?
Without such a proof I do not feel able to disregard the possibility of a function sufficiently complicated to make such a separation impossible. If a proof is too complicated perhaps you could demonstrate a few examples that don't seem obviously separable, such as (ln|x+y|)^sin(x-y).
Not all differential equations are separable. That is, not all functions f(x, y) can be written as g(x)·h(y).
What about (3pi)/4?
Very cool thanks
Good day can i help please 🥺 in this equation
How to find separable Differential equation
dz-(4x³-4x³z+x³z²)dx+(4y-1)(2-z)²dv=0
Thank you !!!! :-)
you are perfect🙏🙏🙏🙏🙏
amazing
just stumbled across this video. Amazing explanation but had a question hopefully someone can answer. Isnt the tan of -1 = 3pi/4? I mean its -pi/4 too but its also 3pi/4. Does it matter which one we take? Thanks
I have the solution as y(x)=tan(x^2/2+3π/4+nπ) for n=0,1,2,3….
Actually you should write the solution as y=tan(1/2x^2-п/4+пn), sinve the period of tan is п. However all those functions are the same, so i guess it doesnt matter and you can take any value for n
Good day can i help please 🥺 in this equation
dz-(4x³-4x³z+x³z²)dx+(4y-1)(2-z)²dv=0
how do u get tan inverse from 1 over 1 + y square?
you differentiate y = arctan u.
u = tan y
du/dx = d/dx (tan y) = sec² y dy/dx
dy/dx = 1/sec² du/dx
= (1/1+tan² y) du/dx from the identity (sec² = 1+tan² y)
=(1/1+u²) du/dx
the antiderivative of 1/(1+x²) is arctan(x)
Why C is not 3pi/4? Or 7pi/4
Fantastic
I know it doesn't matter, but c can be more than just pi/4.
Ognjen Kovačević da, pi/4+k*pi, gde je k ceo broj
Luka Popovic Tačno tako.
It's actually not that bad 😂😂
I'm glad to u
Why isn't the general solution y = tan( (x^2)/2 + C_1 ) + C_2 ?
great video by the way :D thank you for the content!
In the initial problem you had that
dy/dx =x+xy^2
Now when we differentiate the final result
y=tan((1/2)x^2+C)
dy/dx=(1+tan^2((1/2)x^2+C)x
Now when we compare the result with our derivative, they have to be totally equal
x+xy^2= (1+tan^2((1/2)x^2+C)x
Divide both sides by x
1+y^2=1+tan^2((1/2)x^2+C)
y^2=tan^2((1/2)x^2+C)
Now if you substitute our result into this, it will work, but if our solution had another C outside of tan you'd have a function plus a constant is equal to itself. That is, that constant has to be zero.
I hope that this helps.
It does! Thank you!!
tan^-1 it's arctan?
looked at the video, but I can't find a solution for my problem , don't find how to separate them.
y' = exp(-y/x) + y/x ==> How do i solve this,, probably simple solution but can't find it ......
tried by playing with ln(y') = (-y/x + ln y/x)
but then how to integratie to find Y ??
greetz
gunter hoflack ln x/y = ln y - ln x
What about this example
(x^2-y^2)y'=x-y
(x^2-y^2)dy/dx = x-y
(x-y)(x+y)dy/dx = x-y
(x+y)dy/dx = 1
x*dy/dx + y*dy/dx = 1
int(x*dy) + int(y*dy) = int(1*dx)
xy + (y^2)/2 = x + c
y = -x + sqrt(x^2 + 2(x+c))
Im not sure tho
I think that is correct.
where did tan^-1 comes from
The integral of 1/(x^2 + 1) dx is arctan(x)
You can prove this as follows:
Let y = arctan(x)
Take the tangent of both sides:
tan(y) = x
Use implicit differentiation to find dy/dx:
sec(y)^2 * dy/dx = 1
Solve for dy/dx:
dy/dx = cos(y)^2
Recall that y = arctan(x):
dy/dx = cos(arctan(x))^2
Construct a right triangle with angle arctan(x), base of 1, and height of x. This means the hypotenuse is:
sqrt(x^2 + 1^2
We'd like to find the base over hypotenuse, to find cos(arctan(x)), which therefore is:
cos(arctan(x)) = 1/sqrt(x^2 + 1)
We'd really like to find cos(arctan(x))^2, which therefore is:
cos(arctan(x))^2 = 1/(x^2 + 1)
Thus:
d/dx arctan(x) = 1/(x^2 + 1)
Can anyone explain me, He wrote 'y(0)=1' in the thumbnail and he's putting the initial condition at 'y(0)= -1 ' like wtf?
I can't stop smiling
🔥
great
0:10
King
I can prove you can't solve any dif equation with this form y'+ln|sinx|y = g(x) simple the special integration factor is non-algebraic