Seriously wish you taught at my university. All of our higher class, calc3 and up, have really crappy instructors that care more about theory and proofs than actually showing us how to solve the problem. So a big thank you for helping those of us that need someone like you to break it down quick and simple in order to understand. If you have a patreon page or something let me know, you deserve some form of compensation for this. If it wasn't for you I would've failed a few different tests.
This is my first time watching one of your videos. I appreciate how you take your time with the problem and that you write very clearly (surprisingly hard to find). Using the two colors made it easier to follow. I learn a lot of my math from UA-cam and this was very helpful. Thank you.
4:55 you could in fact solve for v. Let's say we have it in some final form like: sec(v) + tan(v) = r (r stands for whatever it is on the right-hand side) Then we do this: 1+sin(v) = cos (v) r 1 + (e^iv - e^-iv)/2i = (r/2) (e^iv + e^-iv) Now let e^iv be p. Then we have: 1 + (p + 1/p)/2i = (r/2) (p + 1/p) //multiply by 2 i p 2 i p + p^2 + 1 = r i p^2 + r i This is some polynomial in variable p, solve for p, make out a logarithm out of it and "see" the arctan function in it. Other approach is: rewrite sin(v) - r cos(v) as: sqrt(1+r^2) sin(x + arctan(r)) and you will finally obtain the same arctan formula. Of course, there would be some decisions like which root to take and add + 2*pi*integer somewhere when taking inverse functions, but, youknow, some people say: a differential equation is not complete unless you provide a sufficient set of initial and/or boundary conditions ;) so after you clarify initial condition, there should be no arbitrarity.
Without WolframAlpha, I know how to solve for v. Start by taking the exponential function of both sides, and relabeling the c: sec(v)+tan(v)=C₁*e^(-1/x). Change sec and tan to sin and cos and combine fractions: (1+sin(v))/cos(v)=C₁e^(1/x). Shift the sin and cos to cos and sin: (1+cos(v+π/2))/sin(v+π/2)=C₁e^(−1/x). Reciprocate: sin(v+π/2)/(1+cos(v+π/2))=C₂e^(1/x). Use the tangent half-angle identity: tan(v/2+π/4)=C₂e^(1/x). Take the arctan of both sides. v/2+π/4=arctan(C₂e^(1/x)). Solve for v: v=2arctan(C₂e^(1/x))−π/2. Substitute back in v=y/x²: y/x²=2arctan(C₂e^(1/x))−π/2. Solve for y: y=2x²arctan(C₂e^(1/x))−πx²/2. BAM! Solved for y.
Thank you! Helping me so much in preparing for the final in my DE class. Ever since I was in Calc 2 I have been watching these, and they are so helpful.
Is there a special name for this sort of functions where you make a substitution to solve it? For example, homogenous 1st ODE you will substitute f(y/x)=f(v), v=y/x.
instead of using ln|sec(v)+tan(v)| for integral of sec(v) if we use artanh(sin(v)) (which is equivalent to ln|sec(v)+tan(v)), it makes it easy to solve for y i got y=x²arcsin(tanh(-1/x+c)) as final answer
“There’s no way to isolate the v”. Well even though int{sec z dz} = ln|sec z + tan z|, I think the more proper form to rewrite it with one input & no absolute value; +/-, is that the int{sec z dz} = arctanh(sin z). So with some “function sliding”, y = x^2 arcsin(tanh(c - 1/x)). Cool fun fact the derivative of arcsin(tanh z) = sech z
Could the answer be sec(y/x^2)+ tan(y/x^2) = Ce^(-1/x) ? Note : after removing the absolute value, i put plus or minus on the other side and a “plus or minus” constant is another constant.
instead of using ln|sec(v)+tan(v)| for integral of sec(v) if we use artanh(sin(v)) (which is equivalent to ln|sec(v)+tan(v)|), it makes it easy to solve for y i got y=x²arcsin(tanh(-1/x+c)) as final answer
instead of using ln|sec(v)+tan(v)| for integral of sec(v) if we use artanh(sin(v)) (which is equivalent to ln|sec(v)+tan(v)|), it makes it easy to solve for y i got y=x²arcsin(tanh(-1/x+c)) as final answer
![Exact Equations [ODE]](http://i.ytimg.com/vi/pgJci5CI9n4/mqdefault.jpg)
![Exact Equations [ODE]](/img/tr.png)
Seriously wish you taught at my university. All of our higher class, calc3 and up, have really crappy instructors that care more about theory and proofs than actually showing us how to solve the problem. So a big thank you for helping those of us that need someone like you to break it down quick and simple in order to understand. If you have a patreon page or something let me know, you deserve some form of compensation for this. If it wasn't for you I would've failed a few different tests.
You’re one of the best math lecturers in the world please keep it up .
the marker switch is smooth af, Michael Jackson is proud.
This is my first time watching one of your videos. I appreciate how you take your time with the problem and that you write very clearly (surprisingly hard to find). Using the two colors made it easier to follow. I learn a lot of my math from UA-cam and this was very helpful. Thank you.
4:55 you could in fact solve for v. Let's say we have it in some final form like: sec(v) + tan(v) = r (r stands for whatever it is on the right-hand side)
Then we do this: 1+sin(v) = cos (v) r
1 + (e^iv - e^-iv)/2i = (r/2) (e^iv + e^-iv)
Now let e^iv be p. Then we have:
1 + (p + 1/p)/2i = (r/2) (p + 1/p) //multiply by 2 i p
2 i p + p^2 + 1 = r i p^2 + r i
This is some polynomial in variable p, solve for p, make out a logarithm out of it and "see" the arctan function in it. Other approach is:
rewrite sin(v) - r cos(v) as: sqrt(1+r^2) sin(x + arctan(r))
and you will finally obtain the same arctan formula. Of course, there would be some decisions like which root to take and add + 2*pi*integer somewhere when taking inverse functions, but, youknow, some people say: a differential equation is not complete unless you provide a sufficient set of initial and/or boundary conditions ;) so after you clarify initial condition, there should be no arbitrarity.
Without WolframAlpha, I know how to solve for v. Start by taking the exponential function of both sides, and relabeling the c: sec(v)+tan(v)=C₁*e^(-1/x). Change sec and tan to sin and cos and combine fractions: (1+sin(v))/cos(v)=C₁e^(1/x). Shift the sin and cos to cos and sin: (1+cos(v+π/2))/sin(v+π/2)=C₁e^(−1/x). Reciprocate: sin(v+π/2)/(1+cos(v+π/2))=C₂e^(1/x). Use the tangent half-angle identity: tan(v/2+π/4)=C₂e^(1/x). Take the arctan of both sides. v/2+π/4=arctan(C₂e^(1/x)). Solve for v: v=2arctan(C₂e^(1/x))−π/2. Substitute back in v=y/x²: y/x²=2arctan(C₂e^(1/x))−π/2. Solve for y: y=2x²arctan(C₂e^(1/x))−πx²/2. BAM! Solved for y.
🤔
Thank you! Helping me so much in preparing for the final in my DE class. Ever since I was in Calc 2 I have been watching these, and they are so helpful.
Here is the closed-form solution: 2(x^2)arctan(tanh((Cx-1)/(2x))). Not that anyone cares...
DougCube how did u get that
@@AkshayMuraliNerd098 if you express sec(x) and tan(x) in terms of sin or cos, you can isolate the y
Is there a special name for this sort of functions where you make a substitution to solve it?
For example, homogenous 1st ODE you will substitute f(y/x)=f(v), v=y/x.
Thank you for showing how we get the substitution for dy/dx my profs like to skip intermediate steps also loved the flawless marker flipping hahaha
its 5 years later but thank you for the very clear explanation
Beautiful
Well explained
you should do more of these and Bernoulli's equation
Jesus Garcia they r coming this weekend
instead of using ln|sec(v)+tan(v)| for integral of sec(v) if we use artanh(sin(v)) (which is equivalent to ln|sec(v)+tan(v)), it makes it easy to solve for y
i got y=x²arcsin(tanh(-1/x+c)) as final answer
“There’s no way to isolate the v”. Well even though int{sec z dz} = ln|sec z + tan z|, I think the more proper form to rewrite it with one input & no absolute value; +/-, is that the int{sec z dz} = arctanh(sin z). So with some “function sliding”, y = x^2 arcsin(tanh(c - 1/x)). Cool fun fact the derivative of arcsin(tanh z) = sech z
Sec(u)+tan(u)=tan(1/2u+pi/4). Ln|tan(1/2u+pi/4)|=ln(tan(+-1/2u+pi/4))=
-1/x+C tan(+-1/2u+pi/4)=Aexp(-1/x) +-1/2u+pi/4=arctan(Aexp(-1/x))+mpi. u=2arctan(Aexp(-1/x))+(2n-1)pi/2 y=x^2[2arctan(Aexp(-1/x))+(2n-1)pi/2]
if you use arctanh(sin(x)) as a primitive of 1/cos(x), everything become easier : y = x^2 * arcsin(tanh(C-1/x))
He is a living legend 🎉
It's 7 years later... And still thanks...
Amazinggggg
How the hell do you switch markers so fast!?
it is because they're in the same hand same time i was wondering that too lmao
Very nice mr asian
man thanks for saving me for my exam later
Thank you! You're really helping us. god bless you
five years later and ur still helping! 😂
Which type of ODE is this?
BRO THIS GUY IS THE BEST
the best teacher!
Samuel Minea thanks!
This is really helpful ... Thanks for uploading!
gotta love an asian math teacher!
You can solve for v using a Weierstrass substitution.
Could the answer be sec(y/x^2)+ tan(y/x^2) = Ce^(-1/x) ?
Note : after removing the absolute value, i put plus or minus on the other side and a “plus or minus” constant is another constant.
Every thing is good but voice is too low
you are the best!!!
#RespectFromSouthAfrica
Getting some Doctor Who, the Ood Vibes here lol
thank u this was very helpful
legend
thank u sir. i love u sir
Как выразить y(x) в явном виде?
~~~
How to express y(x) explicitly?
instead of using ln|sec(v)+tan(v)| for integral of sec(v) if we use artanh(sin(v)) (which is equivalent to ln|sec(v)+tan(v)|), it makes it easy to solve for y
i got y=x²arcsin(tanh(-1/x+c)) as final answer
future aub 202 students...i feel u
well done !
Thank you!!!
This problem. Was so fking awsmmmmm
🤯
can u do this sum== tany dy/dx+tanx=cosy*cos^2x.
SO WHAT IS Y FUNCTION?????
instead of using ln|sec(v)+tan(v)| for integral of sec(v) if we use artanh(sin(v)) (which is equivalent to ln|sec(v)+tan(v)|), it makes it easy to solve for y
i got y=x²arcsin(tanh(-1/x+c)) as final answer
Ez
good shit
正在写大学的作业哈哈哈 这个真的太给力啦
Riiight?
Why you do too much details when solving problem?. Like simple algebra