Solve y' + 4xy^2 = 0. What is the value of y(2) if y({0.61} )={3.53}?

Question

im lost as ever

Answer 1

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