A Stunning Differential Equation You Can Enjoy Solving

[@amadeus-1011]

this is why maths is like art- i was wondering if you would start a playlist on problem solving techniques or maybe more systematic organisations of math topics/knowledge/theorems?

[@domedebali632]

Total agree with ya mate

[@drpkmath12345]

Yes haha I will make sure to organize and make those playlists my friend👍👍👍

[@LITHICKROSHANMS-gw2lx]

Good solution sir

[@drpkmath12345]

Thanks a lot my friend👍👍❤️

[@LITHICKROSHANMS-gw2lx]

@@drpkmath12345
Most welcome sir

[@Min-cv7nt]

This is definitely the differential equation of the day

[@drpkmath12345]

Haha thanks my friend👍👍👍

[@mathnerd5647]

Nice solution

[@drpkmath12345]

Thanks my friend👍👍👍

[@MrGLA-zs8xt]

Another great video professor

[@drpkmath12345]

Thanks my friend haha👍👍👍

[@TBA_5854]

1/2y + y/2

[@drpkmath12345]

Nice! And you can go further👍👍👍

[@ashishraje5712]

Pls explain u prime square at 2:36 minutes

[@r.maelstrom4810]

We have y^2(y')^2 = (yy')^2 (*) = y^2 - 1.
If we ser u = y^2 - 1 then u' = 2yy'. That is yy'=u'/2. Substitute this in (*) and you have it.

[@ashishraje5712]

​@@r.maelstrom4810I did not get (u' / 2)^2 = u, cud u clarify

[@r.maelstrom4810]

@@ashishraje5712 He sets u = y^2 - 1 which is the RHS of the equation y^2(y')^2 = y^2 - 1. But note that y^2(y')^2 (the LHS) can be rewritten as (yy')^2. What is inside the exponent is yy'. Compare this with the derivative of u: u' = 2yy'. Ergo (yy')^2 = (u'/2)^2.

[@drpkmath12345]

Nice to share it r.maelstrom4810👍👍👍

[@ashishraje5712]

Oh

[@iqtrainer]

Nice and tricky one professor🎉

[@drpkmath12345]

Thanks my friend haha👍👍👍

[@domedebali632]

Very nice solution professor

[@drpkmath12345]

Thanks a lot my friend👍👍👍