Answer
$$y = - \frac{1}{{2x}}\cos {x^2} - \frac{1}{{2x}}$$
Work Step by Step
$$\eqalign{
& y' + \left( {\frac{1}{x}} \right)y = \sin {x^2},{\text{ }}\left( {\sqrt \pi ,0} \right) \cr
& \frac{{dy}}{{dx}} + \left( {\frac{1}{x}} \right)y = \sin {x^2} \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = \frac{1}{x}{\text{ and }}Q\left( x \right) = \sin {x^2} \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\frac{1}{x}dx} }} = {e^{\ln x}} = x \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& x\left( {\frac{{dy}}{{dx}} + \left( {\frac{1}{x}} \right)y} \right) = x\sin {x^2} \cr
& x\frac{{dy}}{{dx}} + y = x\sin {x^2} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {xy} \right] = x\sin {x^2} \cr
& {\text{Integrate both sides}} \cr
& xy = \frac{1}{2}\int {\sin {x^2}\left( {2x} \right)} dx \cr
& xy = - \frac{1}{2}\cos {x^2} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {\sqrt \pi ,0} \right) \cr
& 0 = - \frac{1}{2}\cos {\left( {\sqrt \pi } \right)^2} + C \cr
& 0 = - \frac{1}{2}\left( { - 1} \right) + C \cr
& C = - \frac{1}{2} \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = - \frac{1}{{2x}}\cos {x^2} - \frac{1}{{2x}} \cr
& \cr
& {\text{Graph}} \cr} $$
