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!
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
To find the particular solution that passes through the point (0, 2), we need to substitute the coordinates (x, y) = (0, 2) into the given equation y(x^2 + y) = C.
Substituting these values, we get:
2(0)(0^2 + 2) = C
0 = C
So, the value of C in the particular solution is 0.
Now, let's differentiate the given equation implicitly with respect to x to find the derivative y':
2xy + (x^2 + 2y)y' = 0
Now, we can substitute the value of C = 0 into this equation and solve for y':
2xy + (x^2 + 2y)y' = 0
2xy + (x^2 + 2y)y' = 0
Substituting C = 0, we get:
2xy + (x^2 + 2y)y' = 0
Now, let's substitute the y-coordinate of the given point (0, 2), which is y = 2, and solve for y':
2(0)(2) + (0^2 + 2(2))y' = 0
0 + (0 + 4)y' = 0
4y' = 0
y' = 0/4
y' = 0
So, the derivative y' of the particular solution that passes through the point (0, 2) is 0.
Therefore, the particular solution is given by:
y = 2
Thus, the particular solution that passes through the point (0, 2) is y = 2.