hey this is an incredible video, great explanation and very easy to follow along. my college profs skipped non-separable differentials completely and this helped so much, wonderful vid man
Nice video! I don't want to be picky since this is a tricky question with subtle nuances, but at 1:43 there is a little more to consider. The indefinite integral of 1/x is actually ln |x| +C, so in your solution you have assumed x>0, and the answer you provide is only valid for x>0. It is possible (by using ln|x|) to have a solution valid on both sides of zero. Like I said, this is a tricky question, and this oversight is often made. Thanks for all of your videos!
@@TheLazyEngineer X is multiplied by an exponential. Okay, where did the X go? At least after integrating the exponential we still have the X multiplied by it untouched
@@workoutwithfayomiIm assuming you are talking about integral(x*exp(-x^2/2)). I used a change of variables and integrated with respect to u. I don’t see anything wrong at the moment. Take the derivative of the result, and you recover the integrand.
I dont understand how did you write u=y, v=x etc. When i tried to place them and understand it wont go as you explain.(v.dv/dx)+u=2/v how is this similar to vu'+v'u=(uv)'? How did you find xy?????????? I think you try so hard to make videos short, you miss important points.
![Exact Equations [ODE]](http://i.ytimg.com/vi/pgJci5CI9n4/mqdefault.jpg)
Good stuff
Thank you for great explanation and going through those concrete examples. I don't like when it's too abstract.
hey this is an incredible video, great explanation and very easy to follow along. my college profs skipped non-separable differentials completely and this helped so much, wonderful vid man
This video is neat and easy to understand! Thank you for that. :)
I love your videos
Nice video! I don't want to be picky since this is a tricky question with subtle nuances, but at 1:43 there is a little more to consider. The indefinite integral of 1/x is actually ln |x| +C, so in your solution you have assumed x>0, and the answer you provide is only valid for x>0. It is possible (by using ln|x|) to have a solution valid on both sides of zero. Like I said, this is a tricky question, and this oversight is often made. Thanks for all of your videos!
Wouldn't the integrating factor also be multiplied by a constant since there's an integrating constant
At 6:27, you are meant to do integration by part, but you threw the X attached to the exponential function
How so?
@@TheLazyEngineer X is multiplied by an exponential. Okay, where did the X go? At least after integrating the exponential we still have the X multiplied by it untouched
@@workoutwithfayomiIm assuming you are talking about integral(x*exp(-x^2/2)). I used a change of variables and integrated with respect to u. I don’t see anything wrong at the moment. Take the derivative of the result, and you recover the integrand.
@@TheLazyEngineer No he is actually talking about the x on the left handside of equation which i being multiplied by exp.
@@rotatrylevogaming Not seeing it.
I dont understand how did you write u=y, v=x etc. When i tried to place them and understand it wont go as you explain.(v.dv/dx)+u=2/v how is this similar to vu'+v'u=(uv)'? How did you find xy?????????? I think you try so hard to make videos short, you miss important points.
Thanku ❤️ sir
i never comment on these math videos, but i had to stop and take my time to say that had i not found this video I would have dropped my class
Glad to hear! If you ever have any questions just drop a comment!
shouldn't {xy'+y dx = (y'x^2)/2+yx?
Please you are too fast