Differential Equations Question 224
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
Options:
A) $ \sqrt{y}+\frac{1}{3}(1-x^{2})=c{{(1-x^{2})}^{\frac{1}{4}}} $
B) $ y{{(1-x^{2})}^{\frac{1}{4}}}=c(1-x^{2}) $
C) $ \sqrt{y}{{(1-x^{2})}^{\frac{1}{4}}}=\frac{1}{3}(1-x^{2})+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] 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 $ Put $ {y^{\frac{1}{2}}}=z\Rightarrow \frac{1}{2}{y^{-\frac{1}{2}}}\frac{dy}{dx}=\frac{dz}{dx} $ $ 2\frac{dz}{dx}+\frac{x}{1-x^{2}}z=x\Rightarrow \frac{dz}{dx}+( \frac{1}{2}\frac{x}{1-x^{2}} )z=\frac{x}{2} $ I. F. $ {e^{\int{\frac{1}{2}[ \frac{x}{1-x^{2}} ]dx}}}={e^{-\frac{1}{4}\log (1-x^{2})}}={{(1-x^{2})}^{\frac{1}{4}}} $ The solution is $ z{{(1-x^{2})}^{-\frac{1}{4}}}=\int{\frac{x}{2}{{(1-x)}^{-\frac{1}{4}}}dx+c} $
$ \Rightarrow z{{(1-x^{2})}^{-\frac{1}{4}}}=\frac{1}{2}\int{{{(1-x^{2})}^{-\frac{1}{2}}}xdx+c} $
$ \Rightarrow \sqrt{y}{{(1-x^{2})}^{\frac{1}{4}}}=\frac{1}{2}( -\frac{1}{2} )\frac{{{(1-x^{2})}^{3/4}}}{3/4}+c $
$ \Rightarrow \sqrt{y}{{(1-x^{2})}^{\frac{1}{4}}}=-\frac{1}{3}{{( 1-x^{2} )}^{3/2}}+c $
$ \Rightarrow \sqrt{y}+\frac{1}{3}( 1-x^{2} )=c{{(1-x^{2})}^{\frac{1}{4}}} $
BETA
