MISTAKE!!! At 6:30 I use (x^2 + xy) when I should have used the expression WITH the integrating factor which is: (x^3 + x^2y). Other than that, the process is correct. To correct this error, simply set x^3 + x^2y + g'(y) = x^3 + x^2y at 6:36
Haha I was just working through the problem and when I went to check that I did it right I was like wait what at this part XD Thanks for the correction!!
why did you use the first experessios of psi "y" that we had before multiplying by our integratig factor instead of the other one?, i mean, before multiplying by these factor the equation was not exact. If you do like i say, the derivate of g(y) is equal to 0, ad in effect g(y)=c
isaacdesantigo isaac ah! you are right. I should have set psi_y = x^3+x^2y . good catch. I'm not immune to mistakes haha thanks for commenting about this
Your solution is actually wrong you missed x to multiply on LHS I can't figure out in which step you have made the mistake. But solving the solution again you won't get the initial differential equation we have.
MISTAKE!!! At 6:30 I use (x^2 + xy) when I should have used the expression WITH the integrating factor which is: (x^3 + x^2y). Other than that, the process is correct. To correct this error, simply set x^3 + x^2y + g'(y) = x^3 + x^2y at 6:36
Haha I was just working through the problem and when I went to check that I did it right I was like wait what at this part XD
Thanks for the correction!!
Yeah so the general solution should really be:
x^3y+(x^2y^2)/2=C
really amazing....cleared my week points.
Since x^3 + x^2y + h’(y) = x^3 +x^2y , h’(y) = 0 and h(y) = C correct?
Yes I think that's right
If we divide the equation by xy we find that the equation is homogeneous. Why do you say its not homogeneous?
It looks like you are correct!
why did you use the first experessios of psi "y" that we had before multiplying by our integratig factor instead of the other one?, i mean, before multiplying by these factor the equation was not exact. If you do like i say, the derivate of g(y) is equal to 0, ad in effect g(y)=c
isaacdesantigo isaac ah! you are right. I should have set psi_y = x^3+x^2y . good catch. I'm not immune to mistakes haha thanks for commenting about this
Your solution is actually wrong you missed x to multiply on LHS I can't figure out in which step you have made the mistake. But solving the solution again you won't get the initial differential equation we have.