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Question

Mathematics Question on Differential equations

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

A

1y2−1\frac{1}{y^{2}-1}

B

1y2−1\frac{1}{\sqrt{y^{2}-1}}

C

11−y2\frac{1}{1-y^{2}}

D

11−y2\frac{1}{\sqrt{1-y^{2}}}

Answer

11−y2\frac{1}{\sqrt{1-y^{2}}}

Explanation

Solution

The given differential equation is:

(1−y2)dxdy+yx=ay(1-y^{2})\frac{dx}{dy}+yx=ay

⇒dydx+yx1−y2=ay1−y2⇒\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∫pdxe^{\int{pdx}}

∴I.F.∴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.