A) \[\frac{1-\sqrt{x}}{1+\sqrt{x}}\]
B) \[\frac{1+\sqrt{x}}{1-\sqrt{x}}\]
C) \[\frac{1-x}{1+x}\]
D) \[\frac{\sqrt{x}}{1-\sqrt{x}}\]
E) \[\frac{\sqrt{x}}{1+\sqrt{x}}\]
Correct Answer: B
Solution :
Given differential equation is \[\frac{dy}{dx}+\frac{y}{(1-x)\sqrt{x}}=1-\sqrt{x}\] Here, \[P=\frac{1}{(1-x)\sqrt{x}}\] \[\therefore \] \[IF={{e}^{\int{P}\,dx}}={{e}^{\int{\frac{1}{(1-x)\sqrt{x}}}dx}}\] Put\[\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\] \[\therefore \] \[IF={{e}^{\int{\frac{2}{1-{{t}^{2}}}dt}}}\] \[={{e}^{\frac{2}{2}\log \left( \frac{1+t}{1-t} \right)}}=\frac{1+t}{1-t}\] \[=\frac{1+\sqrt{x}}{1-\sqrt{x}}\]
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