A) \[y=\log ({{x}^{2}}+cx)\]
B) \[\log \,\,y=x\left( c{{x}^{2}}+\frac{1}{2} \right)\]
C) \[x=\log \,\,y\left( c{{x}^{2}}+\frac{1}{2} \right)\]
D) None of these
Correct Answer: C
Solution :
| [c] Divide the equation by \[y{{(log\,y)}^{2}}\] |
| \[\frac{1}{y{{(log\,\,y)}^{2}}}\frac{dy}{dx}+\frac{1}{\log y}\cdot \frac{1}{x}=\frac{1}{{{x}^{2}}}\] |
| Put \[\frac{1}{\log \,y}=z\Rightarrow \frac{-1}{y{{(log\,\,y)}^{2}}}\frac{dy}{dx}=\frac{dz}{dx}\] |
| Thus we get, \[-\frac{dz}{dx}+\frac{1}{x}\cdot z=\frac{1}{{{x}^{2}}}\], linear in z |
| \[\Rightarrow \frac{dz}{dx}+\left( -\frac{1}{x} \right)z=-\frac{1}{{{x}^{2}}}\] |
| I.F. \[={{e}^{-\int{\frac{1}{x}dx}}}={{e}^{-\log \,\,x}}=\frac{1}{x}\] |
| \[\therefore \] The solution is,\[z\left( \frac{1}{x} \right)=\int{\frac{-1}{{{x}^{2}}}\left( \frac{1}{x} \right)dx+c}\] |
| \[\Rightarrow \frac{1}{\log y}\left( \frac{1}{x} \right)=\frac{-{{x}^{-2}}}{-2}+c\Rightarrow x=\log y\left( c{{x}^{2}}+\frac{1}{2} \right)\] |
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