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A first-order ordinary differential equation having the form $\frac{dy}{dx}=f(x,y),y\left(0\right)=y_0$ is referred to as an initial value problem.

The solution of an initial value problem is a particular curve that satisfies the given differential equation and passes through the point$(x_0,y_0)$ .

The particular solution is determined from the general solution. By plugging the initial condition into the general condition, the numerical value of the integrating constant is determined. It results a particular curve belonging to the family of curves.

The given equation is

$\frac{1}{2}\frac{dy}{dx}=\sqrt{1+y}\cos x.........\left(1\right)$

After separating the variables, equation (1) can be written as

role="math" localid="1664355046862" $\frac{dy}{2\sqrt{1+y}}=\cos xdx...........\left(2\right)$

Integrate both sides of equation (2). It results,

$$\begin{gathered} 鈭�\frac{dy}{2\sqrt{1+y}}=鈭�\cos xdx \\ \sqrt{1+y}=\sin x+C[C=IntegratingCons\tan t] \\ y={(\sin x+C)}^2-1 \end{gathered}$$

Now, apply the initial condition by plugging x=$蟺$,y=0 in the general solution. It follows,

0=0+$C^2$-1

The equation is $y={(\sin x+1)}^2-1$.

Hence, the solution of the given initial value problem is$y={(sinx+1)}^2-1$ .