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(D) Given $y = e^{ax}$.Taking the natural logarithm on both sides,we get $\log y = \log(e^{ax})$.Since $\log(e^{ax}) = ax$,we have $\log y = ax$ ...$(

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$x \frac{dy}{dx} = y \log y$

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(D) Given $y = e^{ax}$.Taking the natural logarithm on both sides,we get $\log y = \log(e^{ax})$.Since $\log(e^{ax}) = ax$,we have $\log y = ax$ ...$(1)$.Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = a$.Now,substitute the value of $a$ from the derivative into equation $(1)$:$\log y = \left( \frac{1}{y} \frac{dy}{dx} \right) x$.Multiplying both sides by $y$,we obtain $y \log y = x \frac{dy}{dx}$,which is $x \frac{dy}{dx} = y \log y$.

The differential equation whose solution is $y=e^{ax}$ is

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