Question 1184060
a)  Use the substitution {{{u = x^2 + y^2}}}.   ===> u' = 2x + 2yy'


b)  From (a), u'/2 = x + yy'.  ===> u'/2 = {{{sqrt(u)}}}, or u' = {{{2sqrt(u)}}}.


c) From (b),  {{{du/dx = 2sqrt(u)}}} ===> {{{du/sqrt(u) = 2dx}}} ===> {{{2sqrt(u) = 2x + c}}}


===> {{{ 2sqrt(x^2 + y^2)= 2x + c}}}.  To get the value of c, plug in x = 1 and {{{y=sqrt(3)}}} into the equation.


===>  {{{2sqrt(1^2+sqrt(3)^2) = 2 + c}}} ===> {{{c = 2}}}, and 

{{{sqrt(x^2+y^2) = x + 1}}}  is the solution to the BVP.