A) \[f(x)=y(x+c)\]
B) \[f(x)=cxy\]
C) \[f(x)=c(x+y)\]
D) \[yf(x)=cx\]
Correct Answer: A
Solution :
| [a] We have \[\frac{dy}{dx}=\frac{f'(x)}{f(x)}y-\frac{{{y}^{2}}}{f(x)}\] |
| \[\Rightarrow \frac{dy}{dx}-\frac{f'(x)}{f(x)}y=-\frac{{{y}^{2}}}{f(x)}\] |
| Divide by \[{{y}^{2}}{{y}^{-2}}\frac{dy}{dx}-{{y}^{-1}}\frac{f'(x)}{f(x)}=-\frac{1}{f(x)}\] |
| Put \[{{y}^{-1}}=z\Rightarrow -{{y}^{-2}}\frac{dy}{dx}=\frac{dz}{dx}\] |
| \[-\frac{dz}{dx}-\frac{f'(x)}{f(x)}(z)=-\frac{1}{f(x)}\] |
| \[\Rightarrow \frac{dz}{dx}+\frac{f'(x)}{f(x)}z=\frac{1}{f(x)}\] |
| I.F. \[={{e}^{\int{\frac{f'(x)}{f(x)}dx}}}={{e}^{\log f(x)}}=f(x)\] |
| \[\therefore \] The solution is \[z(f(x))=\int{\frac{1}{f(x)}(f(x))dx+c}\] |
| \[\Rightarrow {{y}^{-1}}(f(x))=x+c\Rightarrow f(x)=y(x+c)\] |
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