help with implicit differentiation?

Question

Find y = f(x) by solving the differential equation dy/dx =2xy with the initial condition f(0) = 2

Answer 1

I have already solved this problem for you....

dy.dx = 2xy

(1/y) dy = 2x dx <--- separation of variables

ln y = x^2 + c <--- integrates both sides

y = e^(x^2)*e^c <--- takes the exponential of both sides

y = k* e^(x^2) <--- let k = e^c be the fixed undetermined coefficient

f(0) = 2 ---> (0,2) is the given initial condition

2 = k*e^(0^2) = k* e^(0) = k*1 = k

y = f(x) = 2 * e^(x^2)

CHECK:

y' = 2*e^(x^2) * 2x <--- derivative involves chain rule

= y*2x = 2xy <--- substitutes the quantity in bold above with y=e^(x^2), the solution function

dy/dx = 2xy <--- replaces y' with dy.dx; which is the original differential equation;