I started learning Differential Equations just a week ago. And "2u" joke is just fine for me! This funny way is what makes learning more engaging and entertaining
Really nice! I was thinking to myself, "how's SyberMath going to solve a differential equation that isn't homogeneous?" and OF COURSE the answer is, "find a way to make it homogeneous". Just a reminder, an equation is homogeneous if you can replace every "x" with "cx" and every "y" with "cy" and all the c's cancel each other out.
The delight of the neophytes is so touching. They saw the application of one of the standard methods from the first-order ODE theory. And if so, dy/dx= - (e^x+y+3)/(x-y+5) .
The general method to solve this is the change of variables x=u+h, y=v+k imposing h+k+3=0, h-k+5=0, hence h=-4, k=1. You then get an homothetic equation dv/du = f(v/u) which is solved by the next change of variable t=v/u. This works only when the determinant of the linear forms is non-zero. In the degenerate case where it is zero, e.g. the 2 linear forms are proportional, the change of variable u= one of the linear forms without the constant brings you to a separable first order ode.
What SyberMath isn't teaching is the thousand and one techniques to solve differential equations. What he's teaching is how to wrestle an equation into submission so that it can then be subject to differential equation techniques. So that's the thing to get from these videos IMHO: be creative in your approach.
@@kingbeauregard after a year of watching this channel, I 100% agree with this! All of sybermaths videos demonstrate amazing creativity, and I can say I’ve learned a lot.
FYI, probably should be inferred that x, y are *real* Long story short, if complex solutions are acceptable, I found a complex (perhaps singular?) solution: *y = 1 ± i (x+4)* (didn't check for homogenous solution and +C) The substitution I used is most likely incorrect: z = x - y + 5 and go from there, and after proper subs, one could assume that the solution is a polynomial, can be shown deg [z] = 1 z = ax + b OR assume that y = ax+b, which is even faster to solve, and Bob's your uncle :D
@@kodirovsshik I don’t know if I could plug it back since it’s quite complicated, but I cross-multiplied and integrated both sides. From there, I separated the y terms from the x terms which is how I got my answer.
@@scottleung9587 > but I cross-multiplied and integrated both sides How would you do that? One would need to take integrals of x dy and y dx, which is not xy+C, because x and y are not independend
Hello everyone. Someone can help I define perfect difference factor number (PDFn) like 392 which has the follow properety: 392 factors is: (n, 392/n) namely: 2×196 =4×98 =7×56 =8×49= 14×28 the differences of factors (n,392/n) is 194, 94, 48 , 41 , 14 the sum of difference factors(plus first factor 1) is 1+194+94+48+41+14=392 Then 392 is PDFn. its the first number with this properety , the next is over 10^10. Is there next one number??
No, that wouldn't work because the 3 and 5 on the top and the bottom will ruin the substitution for you. Basically an x will be factored out of y (which you want to substitute) and x itself (on the top and the bottom) but not out of both of the numbers 3 and 5. If you couldn't understand what I'm saying you can write it on a paper and hopefully you'll get my point. Cheers!
@@yassinezanned9837 been struggling to apply y=vx and I’ve been stuck since yesterday. 3 and 5 are making things ever harder. It doesn’t even lend itself to partial fraction for easy manipulation
I started learning Differential Equations just a week ago.
And "2u" joke is just fine for me! This funny way is what makes learning more engaging and entertaining
Really nice! I was thinking to myself, "how's SyberMath going to solve a differential equation that isn't homogeneous?" and OF COURSE the answer is, "find a way to make it homogeneous". Just a reminder, an equation is homogeneous if you can replace every "x" with "cx" and every "y" with "cy" and all the c's cancel each other out.
The delight of the neophytes is so touching. They saw the application of one of the standard methods from the first-order ODE theory. And if so, dy/dx= - (e^x+y+3)/(x-y+5) .
Furthermore, you can put the logarithms on the same side to get arctan[(y-1)/(x+4)]=ln[(x+4)^2 + (y-1)^2]^1/2 + C.
But no isolation of 'y'?
The general method to solve this is the change of variables x=u+h, y=v+k
imposing h+k+3=0, h-k+5=0, hence h=-4, k=1.
You then get an homothetic equation dv/du = f(v/u) which is solved by the next change of variable t=v/u.
This works only when the determinant of the linear forms is non-zero. In the degenerate case where it is zero, e.g. the 2 linear forms are proportional, the change of variable u= one of the linear forms without the constant brings you to a separable first order ode.
Very nice! My intro to differential equations starts next Wednesday, so this video is very timely!
What SyberMath isn't teaching is the thousand and one techniques to solve differential equations. What he's teaching is how to wrestle an equation into submission so that it can then be subject to differential equation techniques. So that's the thing to get from these videos IMHO: be creative in your approach.
@@kingbeauregard after a year of watching this channel, I 100% agree with this! All of sybermaths videos demonstrate amazing creativity, and I can say I’ve learned a lot.
Thank you! 🥰
You can not isolate y,but you can find a private solution: y= ix +(1+4i). Then y'=i, and ((1+i)x+(4+4i))/((1-i)x+(4-4i))=i
I just like the way he solve Diophantine equations
this solving was very educational.
Hey, I like the jokes about 2u or 2b or not 2b.
Thanks a million! This is the best video I've seen so far.
Thank you! ❤
Great presentation, thanks a lot👍
Glad you liked it!
Hey SyberMath, awesome video!
Thank you! ❤
you can combine 2 ln form
I hate the condition tan-1 , prefer arctan specially when there is powers in the equation
Arc gang here
@@kodirovsshik
Where is he?☺
Hi Your exercises are pretty difficult always. However thanks.I appreciate your cleverness.
You are welcome!
FYI, probably should be inferred that x, y are *real*
Long story short, if complex solutions are acceptable, I found a complex (perhaps singular?) solution:
*y = 1 ± i (x+4)* (didn't check for homogenous solution and +C)
The substitution I used is most likely incorrect:
z = x - y + 5 and go from there,
and after proper subs, one could assume that the solution is a polynomial, can be shown deg [z] = 1
z = ax + b
OR
assume that y = ax+b, which is even faster to solve,
and Bob's your uncle :D
Thanks subermath
No problem Necmettin hoca 😄
You change the postulate of the problem and direct it upon your wishes
Really nice
Thanks a lot
01:24
I got y=+-sqrt(6x-x^2+25)+5...didn't need any of that substitution or trig crap.
You have probably made a mistake somewhere as this doesn't seem to solve the equation stated. Have you plugged it back to check the correctness?
@@kodirovsshik I don’t know if I could plug it back since it’s quite complicated, but I cross-multiplied and integrated both sides. From there, I separated the y terms from the x terms which is how I got my answer.
@@scottleung9587 cross products won't work since the dx and dy would distribute in a way that makes them inseparable.
@@scottleung9587 > but I cross-multiplied and integrated both sides
How would you do that? One would need to take integrals of x dy and y dx, which is not xy+C, because x and y are not independend
Thank you sir
Of course!
Excellent
Thank you! ❤
You're definitely next level, excellent teaching and explanations
Wow, thank you!!! 🥰🧡
Or you can take x-y+5=t and differentiate with respect to x and substitute and we will.get the values
Perfect!
Thank you! ❤
Hello everyone.
Someone can help
I define perfect difference factor number (PDFn) like 392 which has the follow properety:
392 factors is: (n, 392/n) namely:
2×196 =4×98 =7×56 =8×49= 14×28
the differences of factors (n,392/n) is
194, 94, 48 , 41 , 14
the sum of difference factors(plus first factor 1) is
1+194+94+48+41+14=392
Then 392 is PDFn.
its the first number with this properety , the next is over 10^10.
Is there next one number??
Nice explanation. But what a messy answer!
I literally understood nothing in this video cuz I don't know Integration yet.
Just a year more to come to Integration chapter....
O lol, i was so weak just one year ago
Would the substitution of y=vx work in converting it to homogenous ODE?
No, that wouldn't work because the 3 and 5 on the top and the bottom will ruin the substitution for you. Basically an x will be factored out of y (which you want to substitute) and x itself (on the top and the bottom) but not out of both of the numbers 3 and 5. If you couldn't understand what I'm saying you can write it on a paper and hopefully you'll get my point. Cheers!
@@yassinezanned9837 been struggling to apply y=vx and I’ve been stuck since yesterday. 3 and 5 are making things ever harder. It doesn’t even lend itself to partial fraction for easy manipulation
@@charlesgodswill6161 Exactly, that wouldn't help you to do partial fraction, thus the substitution wouldn't work in this case. Have a great day 😊
So what are gonna find, what is the answer?
Is non-homogeneous but reducible to homogeneous
👍