Question 352078
Oh well if you need it now.
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{{{dy/y^2=2xdx}}}
Integrate both sides.
{{{int((1/y^2),dy)=int((2x),dx)}}}
{{{-y^(-1)=x^2}}}
{{{-1/y=x^2+C}}}
Use the initial condition to solve for C.
{{{-1/y=x^2+C}}}
{{{-1/1=0^2+C}}}
{{{C=-1}}}
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{{{-1/y=x^2+1}}}
{{{y=-1/(x^2+1)}}}