A) \[f(x)=y(x+c)\]
B) \[f(x)=cxy\]
C) \[f(x)=c(x+y)\]
D) \[yf(x)=cx\]
Correct Answer: A
Solution :
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 \[z\left( f(x) \right)=\int{\frac{1}{f(x)}}\,\left( f(x) \right)dx+c\] \[\Rightarrow \,\,{{y}^{-1}}\left( f(x) \right)=x+c\Rightarrow f(x)=y(x+c)\]
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