3. Solve the initial value problem. a. 2yy^ prime =e^ x-y^ 2 , given y(4) = - 2 .

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