Differentiation

Question

Given that y= 1/√(x² +3), show that (x²  +3) dy/dx +xy =0 

 

Thank You!

Answer 1

y=1/√(x²+3)

 

y=(x²+3)^-1/2  (using negative exponents to represent the reciprocal, and fractional exponents for the root)

 

Now use the Chain Rule to differentiate y with respect to x:

 

dy/dx=-1/2(x²+3)^-3/2(2x)

 

        =-x/(x²+3)^3/2

 

Now plugging into the differential equation we have:

 

(x²+3)¹ dy/dx +xy = (x²+3)^2/2[-x/(x²+3)^3/2] +x[1/(x²+3)^1/2]

 

                        = -x/(x²+3)^1/2+x/(x²+3)^1/2

 

                        =0