We have #y'=-xy+sqrty# or #dy/dx=y^(1/2)-xy#.
Let #v=y^(1/2)# and #(dv)/dx=1/2y^(-1/2)dy/dx#
Now
#y^(-1/2)dy/dx=1-xy^(1/2)=1-vx#
so
#(dv)/dx=1/2(1-vx)=1/2-1/2vx#
#rArr(dv)/dx+1/2vx=1/2#
This is now a linear first-order ordinary differential equation of the form
#(dv)/(dx)+vP(x)=Q(x)#
We solve this using an integrating factor:
#mu=e^(int1/2xdx)=e^(1/4x^2)#
So the general solution is
#ve^(1/4x^2)=int1/2e^(1/4x^2)dx=1/2(sqrtpi "erfi"(1/2x)+"c")#
Solving for #v#
#v=1/2e^(-1/4x^2)(sqrtpi"erfi"(1/2x)+"c")#
But #v=sqrty# so
#sqrty=1/2e^(-1/4x^2)(sqrtpi"erfi"(1/2x)+"c")#
and
#y=1/4e^(-1/2x^2)(sqrtpi"erfi"(1/2x)+"c")^2#
Now, we have the condition #y(0)=0#
So
#y(0)=1/4e^0(sqrtpi"erfi"(0/2)+"c")^2=0#
#rArr1/4c^2=0rArrc=0#
So the final solution is
#y=y(0)=1/4pie^(-1/2x^2)"erfi"^2(1/2x)#
