Question
Some of the curves corresponding to different values of C in the general solution of the differential equation are shown in the graph. Find the particular solution that passes through the point (0, 2).
y(x^2+y) = C
2xy + (x^2+2y)y' = 0
How would I start this problem to solve for the solution? Thanks in advance!
y(x^2+y) = C
2xy + (x^2+2y)y' = 0
How would I start this problem to solve for the solution? Thanks in advance!
Answers
Steve
huh? what's to start?
The 1st equation is the solution to the second. All you have to do is find C such that
y(x^2+y) = C
goes through (0,2).
2(0+2) = C
The 1st equation is the solution to the second. All you have to do is find C such that
y(x^2+y) = C
goes through (0,2).
2(0+2) = C