for a bernoulli’s diff equation, you can divide both sides by y^r. And take y^(1-r) as a U. This works because U’ ends up being y^-r. If you follow this far, you can see this is a linear first order diff. equation.
A very strange approach. Bernoulli equation: y' +A(x)*y =B(x)*y^n, n≠0,1 ,substituting z=y^(1-n) reduces to linear. If it is difficult to remember,then divide the original equation by y^n (for n>0, you should not lose the trivial solution y(x)≡0): y'/y^n +A(x)/y^(n-1) =B(x), z=1/y^(n-1)=y^(1-n), z' = (1-n)*y^(-n)*y'= (1-n)*y' /y^n. We obtain the linear equation z'/(1-n) +A(x)*z=B(x), z'+(1-n)*A(x)*z= (1-n)*B(x).
@@noelani976 The substitution z = y^(1-n) for reducing a Bernoulli equation to a linear equation is taught in differential equations courses on pure math degrees too. In fact this is the first time I have seen the approach presented in this video, and I have to say it is much more onerous than the substitution (note that it is a very simple substitution; we're just considering some power y^m instead of y).
Bernoulli differential equation You like integrating factor There is so called separable integrating factor for Bernoulli equation If the Bernoulli equation is in the form y'+A(x)y=B(x)y^{r} mu(x,y)=exp((1-r)Int(A(x),x))y^{-r}
(35:59) I disagree here. If the original equation was y’/x^2 = (y^2+xy-x^2)/(x^2), you’d be correct, but both sides of the actual original equation are continuous and equal at x = 0 when y = (3x+x^3)/(3-x^2).
There's a mistake at 12:08 - to get _y_ you need to multiply _z_ by e^ _x_ , so you should have cancelled the e^(- _x_ ) in the denominator instead of distributing it through. Also you are using the notation y_0 for two different things -- the solution of the homogeneous equation corresponding to a linear equation, and the y value of an initial condition -- and sometimes at the same time, so I would suggest y_h instead of y_0 for the homogeneous solution.
About the last task. If it is possible to solve the differential equation in a general way without using any initial condition, then this is what should be done. Because if you change the initial condition, you will have to go back to solving the equation. In this case, the general solution (can be tested on any popular computer algebra system) is y(x)= {(x+C*x^3)/(1-C*x^2), where C is any constant ⋃ y(x)=-x}. Formally, the solution y(x)=-x can be obtained for C→∞. Now you can set different initial values for both x>0 and x
I agree that there’s a problem with what he described as the interval of validity, but what you described isn’t it. Note that for 2 to be in the interval of validity, the solution would have to be valid for every x between 1 and 2. However, at x=sqrt(3), which is definitely between 1 and 2, y is discontinuous, so the solution isn’t valid there.
I dunno if anyone else is doing the warmup problems but for the second one (x^2*y’+2xy=y^3) I got the following particular solution: y=(5/(4x^3(1+4x^5)))^(1/4) I guess I should plug it in to check it, but I’m exhausted from solving it… and I’m pretty sure I made a mistake somewhere
For some reason this feels like it's going a dozen times farther and faster than my college diff-e-q course. Which is cool, because that was a long time ago in a university far, far away. I do have a question, though. Why do you keep saying "antiderivative" instead of "integral" or "integrate"? Is there a subtle technical difference?
antiderivative means the same thing as indefinite integral, so I guess it saves a syllable (although of course one could just say "integral" with "indefinite integral" being understood, because indefinite integrals are much more common than definite integrals in the study of differential equations)
The general solution of a linear differential equation is the sum of the corresponding homogeneous equation and the so-called particular solution. E.g. for a linear first-order ODE y' + A(x)y = B(x), letting y_h be the homogeneous solution and y_p be the particular solution we have (y_h)' + A(x)y_h = 0 and (y_p)' + A(x)y_p = B(x), and therefore y = y_h + y_p satisfies the equation since (y_h + y_p)' + A(x)(y_h + y_p) = ((y_h)' + A(x)y_h) + ((y_p)' + A(x)y_p) = 0 + B(x) = B(x). However linearity is essential, so adding a homogeneous solution and particular solution for e.g. a Bernoulli equation y' + A(x)y = B(x)y^r (r ≠ 0, 1) won't work, as you can check.
@@Oskar-zt9dc We have y=A/x (if we want simplify the logarithms) and the terms in the first step disappear too. The constant A disappears in the final result.
@@Oskar-zt9dc okay yeah, because the way we use y_0 is just to replace the left side of the homog. eq’n with zero when it reappears in the inhomog. eq’n. I just got confused when he was talking about the constant being absorbed once y_0 multiplies with z
@@mathmajor mostly just want to help on the back end, admin stuff. Yeah it was a fun time with Steve. Where would i DM you? Is your email public? If not I’ll DM you for it.
Thank you so much for making this series! I'm taking differential equations right now and these videos help so much!
for a bernoulli’s diff equation, you can divide both sides by y^r. And take y^(1-r) as a U. This works because U’ ends up being y^-r. If you follow this far, you can see this is a linear first order diff. equation.
A very strange approach.
Bernoulli equation: y' +A(x)*y =B(x)*y^n, n≠0,1 ,substituting z=y^(1-n) reduces to linear. If it is difficult to remember,then divide the original equation by y^n (for n>0, you should not lose the trivial solution y(x)≡0):
y'/y^n +A(x)/y^(n-1) =B(x), z=1/y^(n-1)=y^(1-n), z' = (1-n)*y^(-n)*y'= (1-n)*y' /y^n.
We obtain the linear equation z'/(1-n) +A(x)*z=B(x), z'+(1-n)*A(x)*z= (1-n)*B(x).
That is very nice, I think I kinda like your approach more the Michaels
@@Oskar-zt9dc If this was MY approach, I would walk with my chest out.))
But this is a standard, well-known approach, and I was taught that way.
I guess you are or were an engineering student, Michael is different; he does a lot of Pure Mathematics.
@@noelani976 The substitution z = y^(1-n) for reducing a Bernoulli equation to a linear equation is taught in differential equations courses on pure math degrees too. In fact this is the first time I have seen the approach presented in this video, and I have to say it is much more onerous than the substitution (note that it is a very simple substitution; we're just considering some power y^m instead of y).
@@schweinmachtbree1013 Okay, noted!
Thank you so much sir for teaching us such important topics at free cost. And obviously you go through these topics thoroughly.
Bernoulli differential equation
You like integrating factor
There is so called separable integrating factor for Bernoulli equation
If the Bernoulli equation is in the form
y'+A(x)y=B(x)y^{r}
mu(x,y)=exp((1-r)Int(A(x),x))y^{-r}
I have never heard the term "separable integrating factor" before - does it go by any other name?
@@schweinmachtbree1013 this is integrating factor but in the same form as RHS of separable equation
mu(x,y)=f(x)g(y)
Thank you!
Thank you!!!
(35:59) I disagree here. If the original equation was y’/x^2 = (y^2+xy-x^2)/(x^2), you’d be correct, but both sides of the actual original equation are continuous and equal at x = 0 when y = (3x+x^3)/(3-x^2).
I looked up the name on the shirt... man, that is AWESOME music. Great taste!!
There's a mistake at 12:08 - to get _y_ you need to multiply _z_ by e^ _x_ , so you should have cancelled the e^(- _x_ ) in the denominator instead of distributing it through.
Also you are using the notation y_0 for two different things -- the solution of the homogeneous equation corresponding to a linear equation, and the y value of an initial condition -- and sometimes at the same time, so I would suggest y_h instead of y_0 for the homogeneous solution.
About the last task. If it is possible to solve the differential equation in a general way without using any initial condition, then this is what should be done.
Because if you change the initial condition, you will have to go back to solving the equation.
In this case, the general solution (can be tested on any popular computer algebra system) is y(x)= {(x+C*x^3)/(1-C*x^2), where C is any constant ⋃ y(x)=-x}. Formally, the solution y(x)=-x can be obtained for C→∞.
Now you can set different initial values for both x>0 and x
I agree that there’s a problem with what he described as the interval of validity, but what you described isn’t it. Note that for 2 to be in the interval of validity, the solution would have to be valid for every x between 1 and 2. However, at x=sqrt(3), which is definitely between 1 and 2, y is discontinuous, so the solution isn’t valid there.
I dunno if anyone else is doing the warmup problems but for the second one (x^2*y’+2xy=y^3) I got the following particular solution:
y=(5/(4x^3(1+4x^5)))^(1/4)
I guess I should plug it in to check it, but I’m exhausted from solving it… and I’m pretty sure I made a mistake somewhere
yea something's wrong there. i got sqrt(5x/(2+8x⁵)) and verified it on wolframalpha
For some reason this feels like it's going a dozen times farther and faster than my college diff-e-q course. Which is cool, because that was a long time ago in a university far, far away. I do have a question, though. Why do you keep saying "antiderivative" instead of "integral" or "integrate"? Is there a subtle technical difference?
antiderivative means the same thing as indefinite integral, so I guess it saves a syllable (although of course one could just say "integral" with "indefinite integral" being understood, because indefinite integrals are much more common than definite integrals in the study of differential equations)
Isn't the general solution the sum of the solution for the homogeneous equation and the solution found?
The general solution of a linear differential equation is the sum of the corresponding homogeneous equation and the so-called particular solution. E.g. for a linear first-order ODE y' + A(x)y = B(x), letting y_h be the homogeneous solution and y_p be the particular solution we have (y_h)' + A(x)y_h = 0 and (y_p)' + A(x)y_p = B(x), and therefore y = y_h + y_p satisfies the equation since (y_h + y_p)' + A(x)(y_h + y_p) = ((y_h)' + A(x)y_h) + ((y_p)' + A(x)y_p) = 0 + B(x) = B(x). However linearity is essential, so adding a homogeneous solution and particular solution for e.g. a Bernoulli equation y' + A(x)y = B(x)y^r (r ≠ 0, 1) won't work, as you can check.
I don’t understand the justification for why the +C can be dropped at 14:48. How exactly can it be absorbed?
It's because u just need one function that satisfies the DE, so you can cancel stuff. It's like you just can substitute whatever you want.
@@Oskar-zt9dc We have y=A/x (if we want simplify the logarithms) and the terms in the first step disappear too. The constant A disappears in the final result.
@@Oskar-zt9dc okay yeah, because the way we use y_0 is just to replace the left side of the homog. eq’n with zero when it reappears in the inhomog. eq’n. I just got confused when he was talking about the constant being absorbed once y_0 multiplies with z
Do you like phoebe bridgers? If so you’re the coolest guy I know
I would like help with either channel hmu
What do you have in mind? Send me an email/dm. Btw: I saw you did a video with Steve Shives -- I am a big fan of his channel.
@@mathmajor mostly just want to help on the back end, admin stuff. Yeah it was a fun time with Steve. Where would i DM you? Is your email public? If not I’ll DM you for it.
@@NotoriousSRG you can find my email in my website