5 pages of notes from my teacher I didn't understand anything . I found this video watched it for 8 minutes and now I know everything . Salute sir , respect to you .
dude honestly, my professor was freaking ambiguous with the way he explained it. literally took 2 days, and still didn't learn crap. found this video, and after 8 minutes,its easy
i was not able to unterstand it by my teacher's notes and many youtube videos, but after watching your video, i got to know you to solve it. thank you so much sir... it was very helpful
Instead of using an integrating factor after you obtained a linear equation, you had a separable equation. It would seem to be simpler to just re-write the equation as: dv/dx - sv = -x -> dv/dx = xv-x = x(v-1) -> dv = x(v-1) dx or dv/(v-1) = x dx. And then Integrate both sides to get: ln|v-1| = x^2/2 + C. Raising both sides to e, you get v-1 = e^(x^2/2)e^C, or v = 1+e^(x^2/2)e^C. Since C is an arbitrary constant, e^C it is merely a variant C .
Instead of taking the integrating factor, you could have easily separated xv - x = dv/dx implies x(v-1) = dv/dx implies Int( x dx ) = Int ((1/v-1) dv) implies x^2/2 = log|v+1| + log c I mean its the easy way ;)
It was also unclear for me at first, but I found out why it is like that in the 2nd and 3rd part of the Bernoulli DE series. The left side of the equation is equal to the derivative of the integration factor multiplied by v. In this case, the left side of the equation on the 3rd line is (d/dx)*[e^(-x^2/2)*v] = [e^(-x^2/2)*v]' = [e^(-x^2/2)]'*v + e^(-x^2/2)*v' = (-x^2/2)'*e^(-x^2/2)*v + e^(-x^2/2)*v' = -x*e^(-x^2/2)*v + e^(-x^2/2)*(dv/dx), which equals to the left side on line 2.
Because the integrating factor is e^integral of p in his equation p=-x. Therefore, he has e raised to whatever the integral of p is. integral of -x =-(1/2)x^2 so integration factor is e^((1/2)x^2)
Hi @2:30, for the subst v=y^(1-n), and (n=2, and because of ref. to EQU y'+xy=xy^2). Can show subst as y=v^-1. My question is in the proper handling of DERIV as y'=-1/v^2 dv, or dy=-dv/v^2, WHY did you show as y'=(-1/v^2)(dv/dx). Not sure why (/dx) was included here.
@ 3:17 you need to multiply, if you divide -V(-2) by -V(-2)= V(-4) if I'm right we should multiply by -V2, so now it should look like -V(2) / (-V(-2))=1
Nice work here! but FYI you lost the singular solution y=0 when you multiplied through by v^2 (which is equivalent to dividing through by y^2). If you look at the original DE, you can see y=0 obeys it (you can also see it fits into the family of graphs at the end), but you won't get it from your present answer unless you allow C to be infinity. Not to take away from the tutorial value of the video, though -- well done.
i didnt do good on all the tests, my best grade was a 62. I got through calc 1 and 2 pretty easily so it was surprising to fail de. I think it was partially due to me not taking calc 3"multivariable calc" before hand. gl.
Why? u = e ^-integral x dx = e ^ passes au-x ^ 2/2 as a whole is that please if anyone can answer me en espanol ¿Por qué? u= e^-integral x dx pasa a u= e^-x^2/2 como resulta esa integral por favor si algien me puede responder
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5 pages of notes from my teacher I didn't understand anything . I found this video watched it for 8 minutes and now I know everything . Salute sir , respect to you .
dude honestly, my professor was freaking ambiguous with the way he explained it. literally took 2 days, and still didn't learn crap. found this video, and after 8 minutes,its easy
thats how i roll
not sure what I would do without your videos
my prof keep bubbling this Bernoulli thing in 3 days of meet, and i dont understand anything until I FOUND this great explantion!! Thanks Sir!
i was not able to unterstand it by my teacher's notes and many youtube videos, but after watching your video, i got to know you to solve it. thank you so much sir... it was very helpful
you made this so easy . i wish i had a teacher like you.
Instead of using an integrating factor after you obtained a linear equation, you had a separable equation. It would seem to be simpler to just re-write the equation as: dv/dx - sv = -x -> dv/dx = xv-x = x(v-1) -> dv = x(v-1) dx or dv/(v-1) = x dx. And then Integrate both sides to get: ln|v-1| = x^2/2 + C. Raising both sides to e, you get v-1 = e^(x^2/2)e^C, or v = 1+e^(x^2/2)e^C. Since C is an arbitrary constant, e^C it is merely a variant C .
Make more Differential Equation videos please! Your videos are so awesome and helpful! :)
Make more videos for Diff Eq. Your explanations are great!
Instead of taking the integrating factor, you could have easily separated xv - x = dv/dx implies x(v-1) = dv/dx implies Int( x dx ) = Int ((1/v-1) dv) implies x^2/2 = log|v+1| + log c
I mean its the easy way ;)
3:50 It's separable; no need for integrating factor..
It was also unclear for me at first, but I found out why it is like that in the 2nd and 3rd part of the Bernoulli DE series. The left side of the equation is equal to the derivative of the integration factor multiplied by v. In this case, the left side of the equation on the 3rd line is (d/dx)*[e^(-x^2/2)*v] = [e^(-x^2/2)*v]' = [e^(-x^2/2)]'*v + e^(-x^2/2)*v' = (-x^2/2)'*e^(-x^2/2)*v + e^(-x^2/2)*v' = -x*e^(-x^2/2)*v + e^(-x^2/2)*(dv/dx), which equals to the left side on line 2.
Because the integrating factor is e^integral of p
in his equation p=-x. Therefore, he has e raised to whatever the integral of p is.
integral of -x =-(1/2)x^2 so integration factor is e^((1/2)x^2)
he calculated y', which is dy/dx so he also took the derivative of v with respect to x, hence dv/dx.
Thx man, saved my assignment due tomorrow.
I am falling in love with the shape of graph
Thank u so so much I from Algeria 💗💗🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿
Indeed, the Bernoulli Differential Equation is also a Separable Differential Equation :dy/dx=xy(y-1).
was finding the integrating factor the only method to solving this ?
Hi @2:30, for the subst v=y^(1-n), and (n=2, and because of ref. to EQU y'+xy=xy^2). Can show subst as y=v^-1. My question is in the proper handling of DERIV as y'=-1/v^2 dv, or dy=-dv/v^2, WHY did you show as y'=(-1/v^2)(dv/dx). Not sure why (/dx) was included here.
Dear professor could you post videos on ricatti ode
Thank you so much for this great and short video. Helped me a lot.
@ 3:17 you need to multiply, if you divide -V(-2) by -V(-2)= V(-4)
if I'm right we should multiply by -V2, so now it should look like -V(2) / (-V(-2))=1
You're a lifesaver!
I never did bernoulli equations and understood it. I also got a application of this.
i'm solving a bernoulli diff. eqn. now. the P(x)v = 6/(xv) . since the v^(-1), what term of P(x) should be taken to make an integrating factor? :(
For the first time I understood bernoulis equations
We found this helpful, cheers mate ;)
Thank you . now i know to solve the Bernoulli defferential equation
Nice work here! but FYI you lost the singular solution y=0 when you multiplied through by v^2 (which is equivalent to dividing through by y^2). If you look at the original DE, you can see y=0 obeys it (you can also see it fits into the family of graphs at the end), but you won't get it from your present answer unless you allow C to be infinity. Not to take away from the tutorial value of the video, though -- well done.
BEST VIDEO EVER
THANK YOU
if you had 2vx for the 2nd term would the integrating factor be e^(2x)dx?or just e^(x)dx?
Thank you for the great demonstration sir
It became du substitution. u = -x^2/2 and its du = -xdx which is exactly the same in the right side of the equation.
Great explanation, i wish if you are my math professor :)
Thanks a lot !
well im getting this one wrong on my test which is in about two hours. DE is taking my mind to brown town.
How'd it go? Mine is in 13 hours, this is gonna be a fun all-nighter lol.
i didnt do good on all the tests, my best grade was a 62. I got through calc 1 and 2 pretty easily so it was surprising to fail de. I think it was partially due to me not taking calc 3"multivariable calc" before hand. gl.
MarK Zambelli I'm taking both at the same time in the summer semester, so the class moves twice as fast. This shit sucks major balls.
Thanks for the tip. I will now take calc 3 first.
Brilliant work! thanks!
can you please tell me which toll is using to rite like paper and pen writing i need this too please tell me
I'm not understand how on the right side of the equation at, 4:03, you have (v^-2)
It is actually (xv^-1)/(-v^-2)= -xv
you just saved me bro thanks
because the integral of x is (1/2)(x^2)........he is just integrating -x and
its (-1/2)(x^2)
This helped me a lot... Thank you...
idk why but it would not let me use ^ in that last post
Beautifully done!
very good thankyou so much
you're the man
Why z = 1-n instead of just z = n
Thanks a lot man, really helped me out!
Thanks for the help! Awesome explanation :)
i wish i can give a million likes to this video, thx =D
very helpful and nice quote
very helpful explination! thanks
p(x) could be 1 ?
Thanks
Why? u = e ^-integral x dx = e ^ passes au-x ^ 2/2 as a whole is that please if anyone can answer me
en espanol
¿Por qué? u= e^-integral x dx pasa a u= e^-x^2/2 como resulta esa integral por favor si algien me puede responder
v = y ^(n-1) , you shouldn't upload a video when your own concepts are not clear
thanks you so much u really helped me alot
Thank.
You.
Master.
very helpful..Thanks
Thank you .. Very good way!
Thanks man
Thank you!
this only applicable if y^n=y^2.....^ ^
THANK YOU
Why nobody talk about Mohammed altamimi who solve this!!!!
thank you sir.
awesome thank you!
Thank you :)
The graph resembles the female reproductive system featuring uterus, uterine wall and falopian tube.
thank you so much:))
thank you,,,
thanks alot
Anas Bourini
haha me too I spent three days.
Great explanation! Thanks!