Question
The expression which is the general solution of the differential equation $$\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}}y = x\sqrt y $$ is :
A.
$$\sqrt y + \frac{1}{3}\left( {1 - {x^2}} \right) = c{\left( {1 - {x^2}} \right)^{\frac{1}{4}}}$$
B.
$$y{\left( {1 - {x^2}} \right)^{\frac{1}{4}}} = c\left( {1 - {x^2}} \right)$$
C.
$$\sqrt y {\left( {1 - {x^2}} \right)^{\frac{1}{4}}} = \frac{1}{3}\left( {1 - {x^2}} \right) + c$$
D.
none of these
Answer :
$$\sqrt y + \frac{1}{3}\left( {1 - {x^2}} \right) = c{\left( {1 - {x^2}} \right)^{\frac{1}{4}}}$$
Solution :
Divide the equation by $$\sqrt y ,$$ we get $${y^{ - \frac{1}{2}}}\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}}{y^{\frac{1}{2}}} = x$$
$$\eqalign{
& {\text{Put }}{y^{\frac{1}{2}}} = z \cr
& \Rightarrow \frac{1}{2}{y^{ - \frac{1}{2}}}\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}} \cr
& \Rightarrow 2\frac{{dz}}{{dx}} + \frac{x}{{1 - {x^2}}}z = x \cr
& \Rightarrow \frac{{dz}}{{dx}} + \left( {\frac{1}{2}\frac{x}{{1 - {x^2}}}} \right)z = \frac{x}{2} \cr
& {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{1}{2}} \left[ {\frac{x}{{1 - {x^2}}}} \right]dx}} = {e^{ - \frac{1}{4}\log \left( {1 - {x^2}} \right)}} = {\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} \cr
& {\text{The solution is}} \cr
& z{\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} = \int {\frac{x}{2}{{\left( {1 - x} \right)}^{ - \frac{1}{4}}}dx + c} \cr
& \Rightarrow z{\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} = \frac{1}{2}\int {{{\left( {1 - {x^2}} \right)}^{ - \frac{1}{4}}}x\,dx + c} \cr
& \Rightarrow \sqrt y {\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} = \frac{1}{2}\left( { - \frac{1}{2}} \right)\frac{{{{\left( {1 - {x^2}} \right)}^{\frac{3}{4}}}}}{{\frac{3}{4}}} + c \cr
& \Rightarrow \sqrt y {\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} = - \frac{1}{3}{\left( {1 - {x^2}} \right)^{\frac{3}{4}}} + c \cr
& \Rightarrow \sqrt y + \frac{1}{3}\left( {1 - {x^2}} \right) = c{\left( {1 - {x^2}} \right)^{\frac{1}{4}}} \cr} $$