I am not able to solve this initial value problem plz can u help me on that.?

Question

(y+x^2y) dy/dx = 2x with y(0) = -2

Answer 1

rewrite

y(1+x^2) dy = 2x dx
y dy = 2x/(1 + x^2) dx
(1/2)y^2 = log (x^2 + 1) + C
y^2 = 2log(x^2 + 1) + C1
y = sqrt(2log(x^2 + 1)) + C2
since y(0) = -2
-2 = sqrt(2log(1)) + C2
C2 = -2
so therefore

y = sqrt(2log(x^2 + 1)) - 2

Answer 2

The previous solution is correct to the point where he has

y^2 = 2 log(x^2 + 1) + C1.

But the next step is wrong, because sqrt(a + b) NE sqrt(a) + sqrt(b). Using the above displayed equation with the initial condition, we find C1 = 4, so the answer is
y^2 = 2 log(x^2 + 1) + 4.
We can easily check that by using implicit differentiation:
(2y) dy/dx = 2*2x/(x^2 + 1). A little algebra puts this into the given form.