QUESTION: Solve y' + 4xy^2 = 0. What is the value of y(2) if y({0.61} )={3.53}?
This one is separation of variables.
dy/dx + 4xy^2 = 0
dy/dx = -4xy^2
(1/y^2)dy/dx = -4x
Now integrate both sides wrt x, to give
int (1/y^2) dy = int -4x dx, (Noting that int (dy/dx) dx = int dy)
The above integration then results in
-1/y = -2x^2 + c
y = 1/(2x^2 - c)
Initial condition
y(0.61) = 3.53,
Therefore, 3.53 = 1/(2*(0.61)^2 - c)
0.7742 - c = 1/3.53 = 0.28327
c = 0.7742 - 0.28327
c = 0.4609
y(x) = 1/(2x^2 - 0.4609)
When x = 2,
y(2) = 1/(8 - 0.4609)
y(2) = 0.13264