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
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 \[{{y}^{\frac{1}{2}}}=z\Rightarrow \frac{1}{2}{{y}^{-\frac{1}{2}}}\frac{dy}{dx}=\frac{dz}{dx}\] |
| \[2\frac{dz}{dx}+\frac{x}{1-{{x}^{2}}}z=x\Rightarrow \frac{dz}{dx}+\left( \frac{1}{2}\frac{x}{1-{{x}^{2}}} \right)z=\frac{x}{2}\] |
| I. F. \[{{e}^{\int{\frac{1}{2}\left[ \frac{x}{1-{{x}^{2}}} \right]dx}}}={{e}^{-\frac{1}{4}\log (1-{{x}^{2}})}}={{(1-{{x}^{2}})}^{\frac{1}{4}}}\] |
| The solution |
| is \[z{{(1-{{x}^{2}})}^{-\frac{1}{4}}}=\int{\frac{x}{2}{{(1-x)}^{-\frac{1}{4}}}dx+c}\] |
| \[\Rightarrow z{{(1-{{x}^{2}})}^{-\frac{1}{4}}}=\frac{1}{2}\int{{{(1-{{x}^{2}})}^{-\frac{1}{2}}}xdx+c}\] |
| \[\Rightarrow \sqrt{y}{{(1-{{x}^{2}})}^{\frac{1}{4}}}=\frac{1}{2}\left( -\frac{1}{2} \right)\frac{{{(1-{{x}^{2}})}^{3/4}}}{3/4}+c\] |
| \[\Rightarrow \,\,\,\sqrt{y}{{(1-{{x}^{2}})}^{\frac{1}{4}}}=\,\,-\frac{1}{3}{{\left( 1-{{x}^{2}} \right)}^{3/2}}+c\] |
| \[\Rightarrow \sqrt{y}+\frac{1}{3}\left( 1-{{x}^{2}} \right)=c{{(1-{{x}^{2}})}^{\frac{1}{4}}}\] |
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