Question 1025928
Use the chain rule,
{{{y^2(tan(x)sec(x)dx)+sec(x)(2ydy)=0}}}
Factor out sec(x).
{{{y^2tan(x)dx=-2ydy}}}
{{{dy/dx=-(ytan(x))/2}}}
Now differentiate again using the chain rule,
{{{(d2y)/(dx2)=-(1/2)(y*sec^2(x)+tan(x)(dy/dx))}}}
Substitute,
{{{(d2y)/(dx2)=-(1/2)(y*sec^2(x)+tan(x)(-(ytan(x))/2))}}}
{{{(d2y)/(dx2)=-(y/4)(2sec^2(x)-tan^2(x)))}}}