00:01
Hello everyone here the function elasticity function your fix is given the question is to find the explicit formula y equal to c of x.
00:08
This is the thing we have to find.
00:11
So e of x we already know your fix is nothing but x by y dy by dx which is given as 20 x minus y 2 y minus 10 x.
00:20
So this implies dy by dx will be 20 x minus y divided by 2 y minus 10 x into y into x.
00:36
So put y equal to v x then differentiating with respect to x will be having v plus x dv by dx.
00:46
So if this v plus x dv by dx substitute dy by dx dy by dx is 20 x minus y y 2 y minus 10 x x.
00:58
So in the place so in the place of y put x v x so 20 x minus v x into v x divided by 2 v x minus 10 x into x.
01:12
So if i take x common i will be having 20 minus v into v x divided by x 2 v minus 10 x.
01:21
So this xx and this xx will get cancelled.
01:24
So i have left with 20 minus v into v divided by 2 v minus 10.
01:32
So i am writing this as 20 v minus v square 2 v minus 10 here x dv by dx this v will come to this side.
01:42
So minus v so 20 v minus v square minus 2 v minus 10 into v divided by 2 v minus 10 that i take an lcm.
01:55
So 20 v minus v square minus 2 v square minus 10 v divided by 2 v minus 10.
02:03
So i will be having 20 v minus v square minus 2 v square plus 10 v divided by 2 v minus 10.
02:12
So simplifying this i will be having minus dv square plus 30 v divided by 2 v minus 10.
02:20
So if i take c common, i will be having 10 v minus v square divided by 2 v minus 10.
02:31
So now we use separation of variables that is we move this 2 v minus 10 to that side that is 2 v minus 10 divided by 10 v minus v square dv here.
02:43
We will be having 3 dx by x.
02:46
So integrating on both sides integrate on both sides.
02:52
So if you integrate, i mean if you differentiate this denominator, we will be getting the numerator.
02:59
So such a function is nothing but minus log of 10 v minus v square.
03:06
Why i am putting minus means if i differentiate i will get minus 2 v plus 10 but here it is minus.
03:12
So minus here this is 3 log x minus log c1.
03:18
So what happened means minus log of 10 v minus v square here this can be written as log x cube minus log c1.
03:28
So minus log of 10 v minus v square if i take minus common, i will be log c1 minus log x cube.
03:36
So this minus and minus will get cancelled log of 10 v minus v square into this can be written as log of c1 by x cube.
03:47
So this log and log will get cancelled.
03:49
So we will be having 10 v minus v square c1 by x cube...