Differential Equations

dx/dy=-x/y-x²

(1/x²)dx/dy+1/(xy)=-1

this might help

dy/dx= 1+ x/√1+x² = y/√1+x²

taking derivative,

d²y/dx²= (dy/dx)/√1+x² - yx/(1+x²)^3/2

d²y/dx²= y/(1+x²) - yx/(1+x²)^3/2

also,
yx/(1+x²)^3/2 = dy/dx. x/(1+x²)

so

d²y/dx²= y/(1+x²) - dy/dx x/(1+x²)

thus, d²y/dx²(1+x²) = y-xdy/dx

thus d²y/dx²(1+x²) + xdy/dx = y

Old question.. check solution below..

y = {x+√1+x²}ⁿ

dy/dx= n{x+ √1+x²}^n-1 {1+ x/ √1+x² }

dy/dx = ny/√1+x²

taking derivative,

d²y/dx²= n (dy/dx)/√1+x² - n yx/(1+x²)^3/2

d²y/dx²= n²y/(1+x²) - nyx/(1+x²)^3/2

also,
nyx/(1+x²)^3/2 = dy/dx. x/(1+x²)

so

d²y/dx²= n²y/(1+x²) - dy/dx x/(1+x²)

thus, d²y/dx²(1+x²) = n²y-xdy/dx

thus d²y/dx²(1+x²) + xdy/dx = n²y

Check if i have made any calculation mistake......

ok.. gr8 :)