Here is my approach: (1) to simplify the denominator, sub u = yln(x), (2) convert the equation to u which can be solved by separating the variables. the result is ln(u) - u = ln(ln(x)) + c1, (3) now back-sub to obtain y = Cx^y
Congratulations on this video, you have amazed me again. By contributing something, in 08:32 you give as a solution y=k*x^y , doing some operations this can be converted into x=(R*y)^1/y with R a constant, which is a cleaner solution than yours because it is in the form x=g(y). If R=1 x=y^1/y and then the SyberMarth function S can be defined as the inverse of x^1/x, then the solution of the case R=1 would be y=S(x) Again, congratulations on your video, I really liked it.
Nice! I got the solution y=-W(clnx)/lnx. First make the sub z=ylnx and you end up with: z'lnx=1/x(z+z^2/(1-z)) . By seperation and integration you get -z e^-z=ln(clnx) so y=-W(clnx)/lnx. The solution is the same😊.
Props to bprp.😊
Cool!
Here is my approach: (1) to simplify the denominator, sub u = yln(x), (2) convert the equation to u which can be solved by separating the variables. the result is ln(u) - u = ln(ln(x)) + c1, (3) now back-sub to obtain y = Cx^y
Congratulations on this video, you have amazed me again.
By contributing something, in 08:32 you give as a solution y=k*x^y , doing some operations this can be converted into x=(R*y)^1/y with R a constant, which is a cleaner solution than yours because it is in the form x=g(y).
If R=1 x=y^1/y and then the SyberMarth function S can be defined as the inverse of x^1/x, then the solution of the case R=1 would be y=S(x)
Again, congratulations on your video, I really liked it.
Nice! I got the solution y=-W(clnx)/lnx. First make the sub z=ylnx and you end up with: z'lnx=1/x(z+z^2/(1-z)) . By seperation and integration you get -z e^-z=ln(clnx) so y=-W(clnx)/lnx. The solution is the same😊.
Interesting one! Sad that the audio is ahead of the video though :(
I don't know why
If y=k*x^y, then unless k=y=0, x=(y/k)^1/y.
y'(x-xylnx)=y^2...y'x-y'xylnx=y^2...divido per xy..y'/y=y'lnx+y/x..(lny)'=(ylnx)'..lny=ylnx+c. .y=kx^y