It looks like the equation reads
2yy' = exp(x - y ²)
(where exp(blah) = e ^(blah))
This DE is separable:
2y dy/dx = exp(x) exp(-y ²)
==> 2y exp(y ²) dy = exp(x) dx
Integrating both sides gives
exp(y ²) = exp(x) + C
The initial condition tells you that y = -2 when x = 4, so that
exp((-2)²) = exp(4) + C
exp(4) = exp(4) + C
==> C = 0
Then the particular solution to this DE is
exp(y ²) = exp(x)
Solving for y as a function of x gives
y ² = x
y = ±√x
But bearing in mind that y = -2 < 0 when x = 4, only the negative square root solution satisfies the DE. So
y(x) = -√x
