Question

The integrating factor of the differential equation.
\((1-y^{2})\frac{dx}{dy}+yx=ay\)\((-1<y<1)\)\(is\)

• A. \(\frac{1}{y^{2}-1}\)
• B. \(\frac{1}{\sqrt{y^{2}-1}}\)
• C. \(\frac{1}{1-y^{2}}\)
• D. \(\frac{1}{\sqrt{1-y^{2}}}\)

Answer 1

The Correct Option is D

The given differential equation is:

\((1-y^{2})\frac{dx}{dy}+yx=ay\)

\(⇒\frac{dy}{dx}+\frac{yx}{1-y^{2}}=\frac{ay}{1-y{^2}}\)

This is a linear differential equation of the form:

\(\frac{dx}{dy}+py=Q\)\((where p=\frac{y}{1-y^{2}} and Q=\frac{ay}{1-y^{2)}}\)

The integrating factor(I.F.)is given by the relation,

\(e^{\int{pdx}}\)

\(∴I.F.\)=\(e^{\int{pdy}}=e^{\int{\frac{y}{1-y^{2}}}dy}=\)\(e^{-\frac{1}{2}log(1-y{^2})}=\)\(e^{log[\frac{1}{\sqrt{1-y^{2}}}]}=\)\(\frac{1}{\sqrt{1-y^{2}}}\)

Hence,the correct answer is D.