A) \[\frac{dy}{dx}=\frac{y}{x}\left( 1-\log \frac{y}{x} \right)\]
B) \[\frac{dy}{dx}=\frac{y}{x}\log \left( \frac{y}{x} \right)+1\]
C) \[\frac{dy}{dx}=\frac{y}{x}\left( 1+\log \frac{y}{x} \right)\]
D) \[\frac{dy}{dx}+1=\frac{y}{x}\log \left( \frac{y}{x} \right)\]
E) \[\frac{dy}{dx}=\frac{x}{y}\left( 1+\log \frac{y}{x} \right)\]
Correct Answer: C
Solution :
Given, \[y=x{{e}^{cx}}\] On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}={{e}^{cx}}+x{{e}^{cx}}.c=\frac{y}{x}+y.c\] \[\therefore \]\[y=x{{e}^{cx}}\Rightarrow \log y=\log x+cx\] \[\Rightarrow \] \[\log \frac{y}{x}=cx\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{y}{x}+\frac{y}{x}\log \left( \frac{y}{x} \right)\] \[=\frac{y}{x}\left( 1+\log \left( \frac{y}{x} \right) \right)\]
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