Please can you make a video about mastering Mathematics from the beginning and suggest books to read and study , and I hoped. If you can give recommendations for IYMC
@@me0449 see I am a high school student at grade 10 , so the most branch of mathematics I concerned with it is precalculus , Trigonometry and Algebra. I intended to participate in IYMC , so I wanted to learn skills like making a mathematical proof , logic and problem solving, so I needed a book that I can start with. Further more When I had a look at the branches of mathematics I found they were too many , so I confused and I didn't know which branch should I start with , some people told me to start with discrete mathematics and others told me that discrete mathematics isn't important for you and I should start with prealgebra, so I am really confused.
If you consider x in terms of y, you get x=p*e^y*e^(1-t) And if you consider y in terms of x, you get y=q*ln(x) + t - 1 (p and q being two more constants)
The second equation, dy/dx = y + 1 may be transformed into: dz/dx = z (where z=y+1). This is a standard first order linear homogeneous o.d.e, which has a classic general solution: z = C exp(t) , or : y = C exp(t) - 1.
Yes, if only because, even if you find an equation relating x and y directly, you still need to constrain what x and y can be by solving the differential equations.
I was thinking we would eliminate t and solve y in terms of x. Divide the 2 equations dx/dy = xy/(y+1) => dx/x = ydy/(y+1) => lnx = c + y - ln(y+1) => x = n*(e^y)/(y+1)
Wouldn't it be a lot simpler to start off by dividing the 2nd equation by the 1st equation? That would get rid of't' and just have dy/dx in terms of the variables 'x' and 'y'.
Given dx/dt = xy [1] dy/dt = y+1 [2] Divide [2] by [1] dy/dx = (y+1)/(xy) This is separable: dy y/(y+1) = dx/x dy [1 - 1/(y+1)] = dx/x Integrate both sides, take a constant into ln(x) y - ln(y+1) = ln(kx) Solve for x x = exp(y) /[k(y+1)] You can get x as a function of y, but not easily vice versa. Anyway the problem is asking for x and y as functions of t. Conceptually we are working on the (x, y) plane. The point (x, y) moves according to [1] and [2].
Chain rule : y+1 = dy/dt = (dy/dx)*(dx/dt) = (dy/dx)*xy --->
Separation : (y/y+1)*dy = (1-1/(y+1))dy = dx/x ---->
Integration : y - ln(|y+1| + c1 = ln(|x|) + c2 ---> x = +- c3*e^y / (y+1)
You're missing the point. The idea is to solve for x and y in terms of t
Please can you make a video about mastering Mathematics from the beginning and suggest books to read and study , and I hoped. If you can give recommendations for IYMC
what field of math are u interested in
@@me0449 see I am a high school student at grade 10 , so the most branch of mathematics I concerned with it is precalculus , Trigonometry and Algebra. I intended to participate in IYMC , so I wanted to learn skills like making a mathematical proof , logic and problem solving, so I needed a book that I can start with. Further more When I had a look at the branches of mathematics I found they were too many , so I confused and I didn't know which branch should I start with , some people told me to start with discrete mathematics and others told me that discrete mathematics isn't important for you and I should start with prealgebra, so I am really confused.
What is IYMC?
@@aashsyed1277 International youth mathematics competition.
Start with Geometry Revisited by Coxeter. I also recommend finding Kiran Kedlaya's PDFs online for more training.
If you consider x in terms of y, you get x=p*e^y*e^(1-t)
And if you consider y in terms of x, you get y=q*ln(x) + t - 1
(p and q being two more constants)
Thanks for these videos, im a 10th grader but learn a lot from these videos!
The second equation, dy/dx = y + 1 may be transformed into: dz/dx = z (where z=y+1). This is a standard first order linear homogeneous o.d.e, which has a classic general solution: z = C exp(t) , or :
y = C exp(t) - 1.
x=c1e^(ce^t-t), y=ce^t-1
y=-W(C/x) - 1
Please make a video on integration and related short trickx
this concept is new to me
It's a beautiful day when I get the answer and I use your same methodology.
Perfect!
Substitute z=y+1
=> dz/dt = z
=> z = k*e^t
=> y = k*e^t - 1
Great syber !!!!!!!
Thanks, aash! Long time, no see 😄
Would it be wrong if I got rid of the parametric independent variable t??
Yes, if only because, even if you find an equation relating x and y directly, you still need to constrain what x and y can be by solving the differential equations.
I was thinking we would eliminate t and solve y in terms of x.
Divide the 2 equations
dx/dy = xy/(y+1)
=> dx/x = ydy/(y+1)
=> lnx = c + y - ln(y+1)
=> x = n*(e^y)/(y+1)
What you did was also nice 😅
Thanks! 😁
Wouldn't it be a lot simpler to start off by dividing the 2nd equation by the 1st equation? That would get rid of't' and just have dy/dx in terms of the variables 'x' and 'y'.
Agreed
Not necessarily in this case.
In my opinion, solving the equation: dy/dx = (y+1) / (xy) is not simpler than solving the given system as it is.
You need to find x and y in terms of t, not y in terms of x
@@SyberMath Ohhhh ok that makes more sense, thanks 👍
Got it@@SyberMath . Thanks for clarifying the problem statement. Please continue to inspire us with such challenges.
Bazı şeyleri anlamlandırmamda çok yardımcı oldu, emeğinize sağlık :)
Sevindim. Tesekkurler!
thanks🙃
No problem 😊
the bounds of integration(notation)
was easy but nice 👍
Glad you liked it!
The solution can be more simpler I guess.
If we separate dt and equate the the dt of both sides. I ended up with [y=((k-x)/x))]
Inyour case, dy/dt = -k(k - x)/x², which doesn't satisfy the given value of dy/dt = y + 1.
@@GirishManjunathMusic I think you've diffrentiate in respective for y, you should apply implicit differentiation for both y and x.
@@omograbi I differentiated y wrt t like you're suppsoed to do for a separated equation like this.
Given
dx/dt = xy [1]
dy/dt = y+1 [2]
Divide [2] by [1]
dy/dx = (y+1)/(xy)
This is separable:
dy y/(y+1) = dx/x
dy [1 - 1/(y+1)] = dx/x
Integrate both sides, take a constant into ln(x)
y - ln(y+1) = ln(kx)
Solve for x
x = exp(y) /[k(y+1)]
You can get x as a function of y, but not easily vice versa. Anyway the problem is asking for x and y as functions of t. Conceptually we are working on the (x, y) plane. The point (x, y) moves according to [1] and [2].
@@pwmiles56 yes that's correct, thank you, I misscalculated the integration.
dx=xy*dt
define d is HOUR.
dx = HOUR basic.
dx = 1 o'clock.
xy*dt = 1
second in 1 o'clock
dt is second by inlaw Sine0
dt=3600
xy*3600 = 1
xy=1/3600=0.1 sec
inlaw paspal dy/dt is minute
dx/dt is second
x=0.1/minute
y = 1 was time
0.1=1+1=2
0.1=2 Answer.
Why u took "e" on both side
e^ln(x) = x
I can't understand this video cuz I haven't started Differentiation and Integration yet 😅
Now is the time!
@@SyberMath It will be started in like November or something
But I have done some basics of both
@@tbg-brawlstars 12 me ho n abhi aap
Jee k liye prepare kr rhe ho may be..
@@chandrashekharmehta6121 Abhi Boards chal rahe hai....11th mei nhi aaya but still thoda bahut kar liya aage ka
And Yes JEE prep
@@tbg-brawlstars Ohh nice yarr 🌟😀👍
👍