Differential Equations....?

Question

Consider the differential equation :
(1+y^3)y' = x^2
Determine a region of the xy-plane for which the differential equation would have a unique solution whose graph passes through a point (x0, y0) in the region (where 0 is subscript).

Please show all workings.
Thanks.

Answer 1

let y'=dy/dx then

The given equation can be writen in the form

x^2 dx=(1+y^3) dy

integrate both sides

x^3/3=y+y^4 /4

4x^3=12y+3y^4

Answer 2

dy/dx = (x^2)(8 + y) First, we separate the variables via multiplying via dx and dividing via (8 + y): dy/(8 + y) = (x^2)dx combine the two factors, remembering the left will use a organic log and the main suitable could have a relentless: ln(8 + y) = (a million/3)x^3 + C develop e to the capability of the two factors: 8 + y = Ae^((a million/3)x^3) Subtract 8 and we've the final style: y = Ae^((a million/3)x^3) - 8 Now, we've the factor (0, 3), so plug those in to discover a and the particular answer: 3 = Ae^((a million/3)(0)^3) - 8 resolve for A: 11 = Ae^((a million/3)(0)) 11 = Ae^(0) A = 11 So, the particular answer is: y = 11e^((a million/3)x^3) - 8

Answer 3

I loved diff class...but it has been 2 years since I 've done equations like that. I wish you luck on finding the answer...but if you got this far, I would suggest you look at the procedures as explained in your textbook.