3.1.2.4 Example 4

\begin{align*} y^{\prime } & =\frac {1}{2\sqrt {x}}\\ y\left ( 0\right ) & =1 \end{align*}

In standard form \(y^{\prime }-p\left ( x\right ) y=q\left ( x\right ) \). Hence \(p=\frac {-1}{2\sqrt {x}},q=0\). Domain of \(p\) is \(x>0\) (to avoid complex numbers) and the domain for \(q\) is all \(x\). Combining these gives \(x>0\). Since IC includes \(x=0\) then the theory does not apply. Solving the ode gives

\[ y=\sqrt {x}+c \]

At \(\left ( x_{0},y_{0}\right ) \) the above gives

\[ 1=c \]

Hence solution is

\[ y=\sqrt {x}+1\qquad x>0 \]

So here solution exists and is unique. Even though theory did not apply.