Well, the answer will indeed be different, but, most importantly, you should understand that there is always a constant of integration after you perform integration. Meanwhile the constant value is not important here. What matters is c is a constant, and ln(c) is also a constant. For simplicity, you can let c1 be ln(c).
I want to get something right here, so if you find out the equation is homogeneous then we do substitution before preceeding to the method of separations.
You are most welcome and thanks for watching too. Because we integrated 1/x which is In(x) it's just fine to make the constant of integration In (c) instead of c. They are not the same, I mean In(c) is not = c, but they are all constants.
The homogenous equation is not separable, but to make it separable, we need to do substitution y=vx and y' = v + xv' to make the differential equation separable in terms of v and x
I doubt your answer for Example 2, becase you was suppoused to replace back c^2, when using the initial condition, for now you just found the value of c_1, the one you introduced.
I get your point. But the solution is correct, once we have introduced c1 as c1 = c², c² will surely be c1 = 2. Try substituting it at the line after where we let c² = c1. You are going to get the same expression finally in terms of y².
You out here saving lives. Really glad i found this channel thanks a lot fam
You are very welcome, Kaka. Thanks so much too. Where do you watch from?
@@SkanCityAcademy_SirJohn I'm from 🇰🇪
@kakabraza8869 oh really, Kenya. Nice.
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if we use the constant c instead of ln(c) the answer will be different . so which one is correct
Well, the answer will indeed be different, but, most importantly, you should understand that there is always a constant of integration after you perform integration.
Meanwhile the constant value is not important here.
What matters is c is a constant, and ln(c) is also a constant.
For simplicity, you can let c1 be ln(c).
Why is lnc on the left and not the right?
I want to get something right here, so if you find out the equation is homogeneous then we do substitution before preceeding to the method of separations.
Exactly
The degree of the third example was not stated? Or its not needed in this instance ?
No please, no need
Thanks so much for your videos, but please why do you use ln(c) instead c after the integration, is it the same?
You are most welcome and thanks for watching too. Because we integrated 1/x which is In(x) it's just fine to make the constant of integration In (c) instead of c. They are not the same, I mean In(c) is not = c, but they are all constants.
Please what if we have more than two terms containing x and y. How will you go about it?
You first need to check if it's homogeneous, if not, you need to use a different differential method...
@@SkanCityAcademy_SirJohn oh ok thank you
My guy you're very good 😮
Keep it up bro
Thanks so much. Keep watching for more
Thanks a lot❤🎉
You are most welcome
please can v be replace by any other variable in the equation y=vx
Yes please, you can do that
great explanations
Thanks
why is it that you made y=vx?
The homogenous equation is not separable, but to make it separable, we need to do substitution y=vx and y' = v + xv' to make the differential equation separable in terms of v and x
thanks💓💓💓
You are most welcome
I doubt your answer for Example 2, becase you was suppoused to replace back c^2, when using the initial condition, for now you just found the value of c_1, the one you introduced.
I get your point. But the solution is correct, once we have introduced c1 as c1 = c², c² will surely be c1 = 2.
Try substituting it at the line after where we let c² = c1.
You are going to get the same expression finally in terms of y².
Any ways, it's a good observation you made. I wish you could try it and get back to mee