The method of substitution is used when a function and its derivative can be found in the function we are to integrate. Here in this case, the derivative of 2y is 2 w.r.t y and that of e^y^2 is 2ye^y^2. So, it is wrong to use the method of substitution here because the rule that should be hold when applying it is not found in the function.
Well it's true, but you can as well use u integration, and that's what I did, which is very fast and simple. Let me explain how Integral of 2ye^(y^2) Let u = y^2 du/dy = 2y, dy = du/2y. Therefore Integral of 2ye^(y^2) becomes Integral of 2ye^(u).du/2y Integral of e^u dy Which is equal to: e^u + c, you sub u = y^2 back. e^y^2. Hope you get it. Thanks
@eucliddankwah1557 I think I've just explained to you how I used u integration to do it, when you have a question like this ay times and exponential function, my method usually works. But not all cases. So refine your solution process on integration by part to see if you will get what I had. I think the problem is your solution process.
Hi can you please give an example of if the integration factor is not subject to x , you have mentioned the second case but you did not show it please do because I could not understand anything from my tutor you are marvellous
In the 2nd eg, when integrating Ndy; the first term was a product of functions of y thus 2ye^y^2 so I was expecting the integration of the first term of N to be done by parts...? My humble concern..
I am trying to solve this y dx + (x2y3 + x)dy = 0 I know this equation can be solved by inspection.(the integrating factor is x^-2 y^-2) But I try to use your method, I cannot find a function that's form by x or y only.(case1 and case2 are both cannot be fullfilled) Do I make anything wrong？Or just there's an exemption of this method？ Thanks
@@SkanCityAcademy_SirJohn sorry,maybe I type the question wrong and made some misunderstanding. the question should be ydx + ((x^2)(y^3)+ x)dy=0 and the textbook just told me this can be solved by inspection and give the answer of the integrating factor = (x^-2)(y^-2)
Many thanks you are a star , please do mention why the integration factor in case of x you divide by M , and in case of y you have divided by -M only this point is not clear to me I need to know the reason thanks
Extremely well explained sir, genuinely cleared my doubts about non exact DEs. Subscribed 🤝
You are most welcome.
Thanks so much for watching. Keep watching for more
The method of substitution is used when a function and its derivative can be found in the function we are to integrate. Here in this case, the derivative of 2y is 2 w.r.t y and that of e^y^2 is 2ye^y^2. So, it is wrong to use the method of substitution here because the rule that should be hold when applying it is not found in the function.
Great input, however I disagree with you
Why didn't you use integration by parts when integrating 2ye^y²
We are integrating two products and one of the products is not a constant
Well it's true, but you can as well use u integration, and that's what I did, which is very fast and simple.
Let me explain how
Integral of 2ye^(y^2)
Let u = y^2
du/dy = 2y, dy = du/2y.
Therefore
Integral of 2ye^(y^2) becomes
Integral of 2ye^(u).du/2y
Integral of e^u dy
Which is equal to: e^u + c, you sub u = y^2 back.
e^y^2.
Hope you get it. Thanks
@@SkanCityAcademy_SirJohn Yh but you ve taken the 2y as a constant
I used integration by parts and didn't get the same answer as yours
@eucliddankwah1557 I think I've just explained to you how I used u integration to do it, when you have a question like this ay times and exponential function, my method usually works. But not all cases. So refine your solution process on integration by part to see if you will get what I had. I think the problem is your solution process.
@@SkanCityAcademy_SirJohn okay
Thanks
@@eucliddankwah1557 you are welcome
Thanks 🙏🏾
That's real stuff right there ur too gud nd u left me with no choice but to follow nd share this stuff 🤞
Awwww thanks so much
Hi can you please give an example of if the integration factor is not subject to x , you have mentioned the second case but you did not show it please do because I could not understand anything from my tutor you are marvellous
I'm so so grateful, but i also think the one on the right is supposed to be integration by parts. Thanks
You are most welcome. I will be glad if you specify the time in the video for easy reference.
9:37
really helpful
Very good video. Thank you.
You are most welcome
In the 2nd eg, when integrating Ndy; the first term was a product of functions of y thus 2ye^y^2 so I was expecting the integration of the first term of N to be done by parts...? My humble concern..
I see
Please similar concerns here
This was super helpful, thank you very much !!
Thanks so much
that was very helpful and great explanation🤩
Thanks so much
could you please talk about case 3 also? i really need it please
Will try that...
if there is a link you could post for me to download your book and read from, would be very much appreciated. Thanks
Advanced engineering math, Allan Jeffery
@@SkanCityAcademy_SirJohn got it, thanks!😍
I am trying to solve this
y dx + (x2y3 + x)dy = 0
I know this equation can be solved by inspection.(the integrating factor is
x^-2 y^-2)
But I try to use your method, I cannot find a function that's form by x or y only.(case1 and case2 are both cannot be fullfilled)
Do I make anything wrong？Or just there's an exemption of this method？
Thanks
Yes, I think there is something wrong with your solution. Try it again and let's see
Integrating factor is u(y) = 1/y
@@SkanCityAcademy_SirJohn sorry,maybe I type the question wrong and made some misunderstanding.
the question should be
ydx + ((x^2)(y^3)+ x)dy=0
and the textbook just told me this can be solved by inspection and give the answer of the integrating factor = (x^-2)(y^-2)
@williamstechtips9726 oh okay
Many thanks you are a star , please do mention why the integration factor in case of x you divide by M , and in case of y you have divided by -M only this point is not clear to me I need to know the reason thanks
I understand you, but that has been a formula proven and defined for us, so we just use that, not to border ourselves much
the last integration , confused me 😵💫