Dear sir. Your final answer of the Y sub h is C sub 1e^t + C sub 2e^2t. Is it alright that the final answer of the Y sub h, its roots were shuffled? Like mine is C sub 1e^2t + C sub 2e^t. Is this the same?
normally i just look at videos hopelessly and give up. then i found you, thank you for contributing to my 100% on the midterm. you're a lifesaver, i just couldn't learn this one by myself. ily no homo
16:36 yes i feel so accomplished hahah and omaigosh i've been finding a video that explains this well and you're the only one that I UNDERSTOOD SO WELL thank you
The corresponding diffEQ is: L*Q"(t) + Q(t)/C = 0 Characteristic equation: L*r^2 + r/C = 0 Solution for r: r = +/-i/sqrt(L*C) The solution for Q(t), is a linear combination of sine and cosine, with a frequency equal to the imaginary part of r: Q(t) = A*cos(sqrt(1/(L*C))*t) + B*cos(sqrt(1/(L*C))*t) At t=0, Q(0) = Q0, and I(0) = I0, which is the same thing as Q'(0). Find Q'(t), which is also I(t): Q'(t) = sqrt(1/(L*C))*(-A*sin(sqrt(1/(L*C))*t) + B*sin(sqrt(1/(L*C))*t)) At t=0, sine is zero, so the cosine terms govern. This means: A = Q0 B*sqrt(1/(L*C)) = I0, solve for B: B = I0*sqrt(L*C) Thus our result is: Q(t) = Q0*cos(sqrt(1/(L*C))*t) + I0*sqrt(L*C)*sin(sqrt(1/(L*C))*t) I(t) = sqrt(1/(L*C))*(-Q0*sin(sqrt(1/(L*C))*t) + I0*sqrt(L*C)*sin(sqrt(1/(L*C))*t))
Why don't you have to set the particular integral to A*te^t*sin(t)..., the auxiliary equation already has e^t as one of its components. Don't you have to multiply the P.I. by one power of the variable when one of the auxiliary equation's roots is in the RHS?
@1:30 How did you know it was going to be: *yp = Ae^tsin(t) + Be^tcos(t)* ? A bit more explanation is needed there. What happens if the equation is y'' - 3y' + 2y = 3t ? What happens if the equation is y'' - 3y' + 2y = tan(t) ? How does that piece change? Thanks.
you should check his differential equations playlist. there is variation of parameters method for tant and other types of undetermined coefficients methods for 3t, e^t kind of things.
If the function on the right hand side is linearly independent of the homogeneous solution, you directly match it with an arbitrary coefficient, and accompany it with other coefficients on its counterparts (more on that later). If it is a constant multiple of one of your solutions to the homogeneous solution, then you multiply by t until it isn't. For exponentials, the particular solution will also be exponential. For sine or cosine, or a linear combination of both of the same frequency, the particular solution will also be a linear combination of both of the same frequency. Likewise for exponentials, enveloping the sine and cosine. For constants, the particular solution will be a placeholder constant, not necessarily equal to your given constant For polynomials of t, the particular solution will be a polynomial of the same degree, and placeholder coefficients along the way. For anything else that isn't a linear combination of the above, you'd have to use another method, like variation of parameters.
Only if you are "lucky" enough that B=0. For this method, you have to account for the possibility that both sine and cosine will play a role in the result, until you rule out the possibility.
yp only solves a special case of initial conditions, that the coefficients in front of the yh parts diminish to zero. The yh part is the transient response from initial conditions, while the yp part is the steady state response that governs what it will be, once the initial conditions are effectively out of the picture. As an example, given: y" + 4*y' + 3*y = 10*cos(t) The homogeneous solution is: yh = A*e^(-t) + B*e^(-3*t) The particular solution is: yp = 2*sin(t) + cos(t) In the special case that our conditions start on the curve of yp, then the coefficients A and B both become zero, and we don't need the homogeneous part. This would happen if, in this example, y(0) = 1, and y'(0) = 2. What the homogeneous part adds, is the ability for the system to start at any set of initial conditions, and exponentially decay to eventually match the particular solution.
what would the general formula be for yp if the right side is e^x(6cosx +17sinx). I'm struggling with trig functions. I know how to do the derivatives, I'm just not sure what to start with.
Given an RHS of e^x * (6*cos(x) + 17*sin(x)) The general form of the yp particular solution would be: e^x * (A*cos(x) + B*sin(x)) As long as your given linear combination of sine and cosine both have the same frequency, they both correspond to the particular solution being a linear combination of the two. The e^x applies to both of them. Also important is that we don't overlap the homogeneous solutions, otherwise we'd need to multiply by t until it is linearly independent. As an example: y" + y = e^x * (6*cos(x) + 17*sin(x)) While this may look like there is overlap, there really isn't. We'd have to have y" - 2*y' + 2*y = e^x * (6*cos(x) + 17*sin(x)), to have overlap, since the homogeneous solutions of this one, also include the exponential envelope on the trig functions. The homogeneous solution: yh = A*cos(x) + B*sin(x) The particular solution: yp = e^(x) * (C*cos(x) + D*sin(x)) Differentiate twice: yp' = e^x * ((D - C)*sin(x) + (C + D)*cos(x)) yp" = 2*e^x * (D*cos(x) - C*sin(x)) Apply to original diffEQ, and equate coefs to find C & D: 2*e^x * (D*cos(x) - C*sin(x)) + e^(x) * (C*cos(x) + D*sin(x)) = e^x * (6*cos(x) + 17*sin(x)) Cancel common e^x factor: 2* (D*cos(x) - C*sin(x)) + (C*cos(x) + D*sin(x)) = 6*cos(x) + 17*sin(x) Thus, equating coefs we get: 2*D + C = 6 -2*C + D = 17 Solve for C & D: C = -28/5, D = 29/5. Thus the solution is: y = A*cos(x) + B*sin(x) + e^x*(-28/5*cos(x) + 29/5*sin(x))
It's a similar concept. Essentially, to simplify the equation, you break it up into two sub-equations that both also need to be satisfied, and work out how to satisfy both parts individually. And then form a solution that is a linear combination of the two.
Good question. If your term on the RHS of the original equation is 2^x, recognize that this is really the same family of functions as e^x, since it is also exponential. Therefore, the particular solution would involve linear combinations of 2^x, or alternatively, e^(ln(2)*x).
Assuming your underscores mean +, we're given: y"(t) + y'(t) + 2*y = 5*cosh(2*t) You can use undetermined coefficients, since cosh is just a linear combination of two exponentials. I'll assume a particular ansatz, of a linear combination of cosh and sinh. First find the homogeneous solution: yh" + yh' + 2*yh = 0 (r^2 + r + 2)*e^(r*t) = 0 (r^2 + r + 2) = 0 r = (-1 +/- sqrt(1 - 8))/2 r = -1/2 +/- i*sqrt(7)/2 Thus: yh = e^(-t/2) * (A*cos(t*sqrt(7)/2)) + B*sin(t*sqrt(7)/2))) For the particular solution: yp = C*cosh(2*t) + D*sin(2*t) Apply to original diffEQ, and match coefs: yp' = 2*C*sinh(2*t) + 2*D*cosh(2*t) yp" = 4*C*cosh(2*t) + 4*D*sinh(2*t) Thus: 4*C*cosh(2*t) + 4*D*sinh(2*t) + 2*C*sinh(2*t) + 2*D*cosh(2*t) + 2*C*cosh(2*t) + 2*D*sinh(2*t) = 5*cosh(2*t) 4*C + 2*D + 2*C = 5 4*D + 2*C + 2*D = 0 solution for C&D: C = 15/16, D = -5/16 Thus, the solution is: y = e^(-t/2) * (A*cos(t*sqrt(7)/2)) + B*sin(t*sqrt(7)/2))) + 15/16*cosh(2*t) - 5/16*sinh(2*t)
I love how enthusiastic and natural this guy is:)
ngl, I was too worried about the thermal detonator in his hand to focus on the problem
You just helped me so much on my ODEs test. I wish my professor taught the way you did! Great job.
You're welcome! I hope you did well on the exam!
Dear sir. Your final answer of the Y sub h is C sub 1e^t + C sub 2e^2t. Is it alright that the final answer of the Y sub h, its roots were shuffled? Like mine is C sub 1e^2t + C sub 2e^t. Is this the same?
@@Sleepy46169 yes its the same
thank you very much you the best good techniques especially on 2nd derivative it could have been very very long
normally i just look at videos hopelessly and give up. then i found you, thank you for contributing to my 100% on the midterm. you're a lifesaver, i just couldn't learn this one by myself. ily no homo
You're literally the reason I'm gonna pass my exam
he is the best teacher DEquation teacher out there
16:36 yes i feel so accomplished hahah and omaigosh i've been finding a video that explains this well and you're the only one that I UNDERSTOOD SO WELL thank you
I do appreciate you for your service, you have rilly talented in teaching, may God 🙏 bless you
You just salvaged my fast approaching mathematics examination!
Exactly what have been searching for ...thank you so much 🙏🙏, your video have been helping
Thank you sir… you are my mathematics hero
2B or negative 2B, that is the question ...
lmao i enjoyed that
@@dylongoodyear7718 Can I enjoy you?
@@nsifrahman9031 You can enjoy me!
every time I have a 2B i say "or not to be," it's a terrible compulsion and I wish I could stop.
A coil of inductance L Henry and a capacitor of C Farad are connected in series if I=I0, Q=Q0 when t=0 Find: Q and I when t
The corresponding diffEQ is:
L*Q"(t) + Q(t)/C = 0
Characteristic equation:
L*r^2 + r/C = 0
Solution for r:
r = +/-i/sqrt(L*C)
The solution for Q(t), is a linear combination of sine and cosine, with a frequency equal to the imaginary part of r:
Q(t) = A*cos(sqrt(1/(L*C))*t) + B*cos(sqrt(1/(L*C))*t)
At t=0, Q(0) = Q0, and I(0) = I0, which is the same thing as Q'(0).
Find Q'(t), which is also I(t):
Q'(t) = sqrt(1/(L*C))*(-A*sin(sqrt(1/(L*C))*t) + B*sin(sqrt(1/(L*C))*t))
At t=0, sine is zero, so the cosine terms govern. This means:
A = Q0
B*sqrt(1/(L*C)) = I0, solve for B: B = I0*sqrt(L*C)
Thus our result is:
Q(t) = Q0*cos(sqrt(1/(L*C))*t) + I0*sqrt(L*C)*sin(sqrt(1/(L*C))*t)
I(t) = sqrt(1/(L*C))*(-Q0*sin(sqrt(1/(L*C))*t) + I0*sqrt(L*C)*sin(sqrt(1/(L*C))*t))
The hardest part for this guy was distributing the - into the (A + B) at 7:35😂
all the way from the u world to differential equations, we been together my friend... thank you
it is a good video, thanks. Already subscribed, hoping to watch more!
zhou coooool thanks!!!
You are awesome. Thank you very much
I love how you handled the second derivative
It so amazing please more videos please for the subject Deferential equation please.
I have many playlists on that already. Check out my channel home page.
Wow! This is so helpful. I wish I found this video earlier! Thank you !!!
Dude you are fantastic thank you so much
Joshua Stier
My pleasure to help!!!
14:49 “I didn’t even use a calculator so you guys can do this too” says the human calculator
this guy is a real chad
your techniques are very helpful
You the best 👍👍👍✅🏆🥇
Ive been looking for examples of
X^n*sin(x)
Or X^n*cos(x)
But haven't seen any on youtube. Any one know some links? Thanks.
Very well ....super bro... thank u so much
you are serving the humanity dude!
is there any previous illustration video such #18 or 4?
I used De-operator method and it was much easier and fast too. I had the same answer
Why don't you have to set the particular integral to A*te^t*sin(t)..., the auxiliary equation already has e^t as one of its components. Don't you have to multiply the P.I. by one power of the variable when one of the auxiliary equation's roots is in the RHS?
I have the same question!!!
@@amzayd he made a mistake
@@zingaphiranana8425 Yes he did cause at 12:25 he plugged in -A + B but it should be -A + 3B, so A and B are miscalculated.
A million times better than my professor
Thank u so so much for helping me...now i understand it clearly
Thanks a lot. The video helped me a lot
you are awesome
I UNDERSTOOD EVERYTHING!!! GOD BLESS YOUUU
Thanks
thank you kind sir
Thank you for this video
@1:30 How did you know it was going to be: *yp = Ae^tsin(t) + Be^tcos(t)* ? A bit more explanation is needed there.
What happens if the equation is y'' - 3y' + 2y = 3t ?
What happens if the equation is y'' - 3y' + 2y = tan(t) ?
How does that piece change? Thanks.
you should check his differential equations playlist. there is variation of parameters method for tant and other types of undetermined coefficients methods for 3t, e^t kind of things.
I think it's from a table given.. that what i use actually
If the function on the right hand side is linearly independent of the homogeneous solution, you directly match it with an arbitrary coefficient, and accompany it with other coefficients on its counterparts (more on that later). If it is a constant multiple of one of your solutions to the homogeneous solution, then you multiply by t until it isn't.
For exponentials, the particular solution will also be exponential.
For sine or cosine, or a linear combination of both of the same frequency, the particular solution will also be a linear combination of both of the same frequency. Likewise for exponentials, enveloping the sine and cosine.
For constants, the particular solution will be a placeholder constant, not necessarily equal to your given constant
For polynomials of t, the particular solution will be a polynomial of the same degree, and placeholder coefficients along the way.
For anything else that isn't a linear combination of the above, you'd have to use another method, like variation of parameters.
B=-1/2 not 1/2. Otherwise -A-B=0 isn't satisfied. Final answer should read: y=C1e^t+C2-(1/2)(e^t)sin(t)-(1/2)(e^t)cos(t)
-(-1/2)-(1/2) = 0
oh my god..you make it easy for me..and you r soo funny
You are soooooo goooood
Tq so much bro very gud explanation
I can't find any video on the internet doing a combination of polynomial, exponential, and sin and cosine :((
great work. no need to attend to attend to my lecture
Excellent video. Thanks!
In particular sol.
Can I consider yp= Ae^t sin(t) not yp= Ae^t sin(t) + Be^t cos(t) ?
Because A&B are abitary const.
Only if you are "lucky" enough that B=0. For this method, you have to account for the possibility that both sine and cosine will play a role in the result, until you rule out the possibility.
My God!!!! you're just amazing ,thx
u are the best
Can u solve some hardest problems in nth order eqn
Not all heroes wear capes... Thanks
great😀
How could differential Ae^t(sint) still be Ae^t(cost)+Ae^t(sint)?????dont u have to differentiate e^t into te^t ???
Can you please do one with the differential operator
Pretty please with chocolate frosting on the top
Operator Method:
Auxiliary equation: r^2-3r+2=0
(r-1)(r-2)=0
r=1,2
Complementary solution y_0=C_1 e^x+C_2 e^2x
(D-1)(D-2)y=e^x sinx
Particular solution y_p=1/(D-1)(D-2) (e^x sinx)
=e^x 1/(D+1-1)(D+1-2) (sinx)
=e^x 1/D(D-1) (sinx)
=e^x 1/((D^2-D) ) (sinx)
=e^x 1/((-1-D) ) (sinx)
=〖-e〗^x ((D-1))/((D^2-1) ) (sinx)
=〖-e〗^x ((D-1))/((-1-1) ) (sinx)
=〖-e〗^x ((D-1))/((-2) ) (sinx)
=e^x/2 (cos x-sinx)
General solution y=y_0+y_p=C_1 e^x+C_2 e^2x+e^x/2 (cos x-sinx)
Another useful video. But I think it's quite dangerous that you're holding a thermal detonator haha
Very nice
thank you brother.you are supper
好像就我一个中国人哈哈，祝您2020新年快乐！
Thanks bro.
Thank you well explained.
wow, EXTREMLY helpful!!!!!
Perfect thanks
That was awesome
bro you are the best thank you so much man!
hey man, how do you guess there is a cost??
thanku
Thank you so much, this helped me a great amount !
Unbelievable!
I feel so accomplished indeed my man
thank you so much, my friend!!
thank u so much
I just don´t get why we need the Yh solution. Isn´t Yp enough? What does the Homogenous solution add?
yp only solves a special case of initial conditions, that the coefficients in front of the yh parts diminish to zero. The yh part is the transient response from initial conditions, while the yp part is the steady state response that governs what it will be, once the initial conditions are effectively out of the picture.
As an example, given:
y" + 4*y' + 3*y = 10*cos(t)
The homogeneous solution is:
yh = A*e^(-t) + B*e^(-3*t)
The particular solution is:
yp = 2*sin(t) + cos(t)
In the special case that our conditions start on the curve of yp, then the coefficients A and B both become zero, and we don't need the homogeneous part. This would happen if, in this example, y(0) = 1, and y'(0) = 2.
What the homogeneous part adds, is the ability for the system to start at any set of initial conditions, and exponentially decay to eventually match the particular solution.
Yp= t e^t ( A Cos(t)+B sin(t)) IS true ?! Because e^t IS in Yh and in f(t) = e^t sin(t) ?! Please answer me
Thank You.
ارجو حل الكثير من المعادلات التفاضليه
Thank you so much
hello blackpenredpen... is the integral sqrt[(x^4-1)/(c-x)] dx solvable???
No
I sincerely hate these problems
what would the general formula be for yp if the right side is e^x(6cosx +17sinx). I'm struggling with trig functions. I know how to do the derivatives, I'm just not sure what to start with.
Given an RHS of e^x * (6*cos(x) + 17*sin(x))
The general form of the yp particular solution would be:
e^x * (A*cos(x) + B*sin(x))
As long as your given linear combination of sine and cosine both have the same frequency, they both correspond to the particular solution being a linear combination of the two. The e^x applies to both of them. Also important is that we don't overlap the homogeneous solutions, otherwise we'd need to multiply by t until it is linearly independent.
As an example:
y" + y = e^x * (6*cos(x) + 17*sin(x))
While this may look like there is overlap, there really isn't. We'd have to have y" - 2*y' + 2*y = e^x * (6*cos(x) + 17*sin(x)), to have overlap, since the homogeneous solutions of this one, also include the exponential envelope on the trig functions.
The homogeneous solution:
yh = A*cos(x) + B*sin(x)
The particular solution:
yp = e^(x) * (C*cos(x) + D*sin(x))
Differentiate twice:
yp' = e^x * ((D - C)*sin(x) + (C + D)*cos(x))
yp" = 2*e^x * (D*cos(x) - C*sin(x))
Apply to original diffEQ, and equate coefs to find C & D:
2*e^x * (D*cos(x) - C*sin(x)) + e^(x) * (C*cos(x) + D*sin(x)) = e^x * (6*cos(x) + 17*sin(x))
Cancel common e^x factor:
2* (D*cos(x) - C*sin(x)) + (C*cos(x) + D*sin(x)) = 6*cos(x) + 17*sin(x)
Thus, equating coefs we get:
2*D + C = 6
-2*C + D = 17
Solve for C & D:
C = -28/5, D = 29/5.
Thus the solution is:
y = A*cos(x) + B*sin(x) + e^x*(-28/5*cos(x) + 29/5*sin(x))
You mentioned it was the superposition principle. Is it the same method as used for circuit analysis?
It's a similar concept. Essentially, to simplify the equation, you break it up into two sub-equations that both also need to be satisfied, and work out how to satisfy both parts individually. And then form a solution that is a linear combination of the two.
Amazing
the literal goat
banging video
thank you very helpful
nice sir thank u so much
maVuti maths 3 where are you? send a shout out!
thank you sir!
Thanks
channel name: blackpenredpen
Blue pen: Am I a joke to you??
thanks for uploading this!
WHAT IS THE GUESS FOR A CONSTANT TO THE POWER OF X?? eg. 2 + 2^x
Good question. If your term on the RHS of the original equation is 2^x, recognize that this is really the same family of functions as e^x, since it is also exponential. Therefore, the particular solution would involve linear combinations of 2^x, or alternatively, e^(ln(2)*x).
Do you have any higher order examples?
Like what order? I have a video on the 6th order with complex numbers
blackpenredpen I feel like I’m taking to a celebrity. You are one of my favorite UA-camrs! I was thinking something like y’’’+8y’’+19y’+12y=270e^(2t)
Also adding some initial conditions and getting the full y(t)
Bai maayo kayka this helps a lot T.T
thanks bro
I'm taking calc2 again this summer and diffeq this fall. This video is scary!
Dan Morgan hahahha summer..I hate summer
What should be the method of solving this equation
D2y/dt2 _ dy/dt _2y= 5 cosh(2t)....by undetermined method....kindly help plzzz
Assuming your underscores mean +, we're given:
y"(t) + y'(t) + 2*y = 5*cosh(2*t)
You can use undetermined coefficients, since cosh is just a linear combination of two exponentials. I'll assume a particular ansatz, of a linear combination of cosh and sinh.
First find the homogeneous solution:
yh" + yh' + 2*yh = 0
(r^2 + r + 2)*e^(r*t) = 0
(r^2 + r + 2) = 0
r = (-1 +/- sqrt(1 - 8))/2
r = -1/2 +/- i*sqrt(7)/2
Thus:
yh = e^(-t/2) * (A*cos(t*sqrt(7)/2)) + B*sin(t*sqrt(7)/2)))
For the particular solution:
yp = C*cosh(2*t) + D*sin(2*t)
Apply to original diffEQ, and match coefs:
yp' = 2*C*sinh(2*t) + 2*D*cosh(2*t)
yp" = 4*C*cosh(2*t) + 4*D*sinh(2*t)
Thus:
4*C*cosh(2*t) + 4*D*sinh(2*t) + 2*C*sinh(2*t) + 2*D*cosh(2*t) + 2*C*cosh(2*t) + 2*D*sinh(2*t) = 5*cosh(2*t)
4*C + 2*D + 2*C = 5
4*D + 2*C + 2*D = 0
solution for C&D:
C = 15/16, D = -5/16
Thus, the solution is:
y = e^(-t/2) * (A*cos(t*sqrt(7)/2)) + B*sin(t*sqrt(7)/2))) + 15/16*cosh(2*t) - 5/16*sinh(2*t)
if there is no sinx on the right side,but have "2+3x-e^2x" on the right side,what will happen?
linear and exponent function at the same time then evaluate them separately then sum up your solution y=y1+y2+y3 as your particular solution
I have a question. Shouldnt the Yp be te^t(sin t) ??
Yah. I'm confused too. Looks like it's a resonance case.
Super super super
Thanks dude!