The differential equation corresponding to th | vedclass

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Question

(C) Given the equation $y = e^{mx}$.Taking the natural logarithm on both sides,we get $\log y = mx$.From this,we can express the arbitrary constant as

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$\frac{dy}{dx} = \left( \frac{y}{x} \right) \log y$

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(C) Given the equation $y = e^{mx}$.Taking the natural logarithm on both sides,we get $\log y = mx$.From this,we can express the arbitrary constant as $m = \frac{\log y}{x}$.Now,differentiate $y = e^{mx}$ with respect to $x$:$\frac{dy}{dx} = m e^{mx}$.Since $e^{mx} = y$,we substitute this into the derivative:$\frac{dy}{dx} = m \cdot y$.Substitute the value of $m = \frac{\log y}{x}$ into the equation:$\frac{dy}{dx} = \left( \frac{\log y}{x} \right) \cdot y = \left( \frac{y}{x} \right) \log y$.Thus,the correct option is $C$.

The differential equation corresponding to the primitive $y = e^{mx}$ is obtained by eliminating the arbitrary constant $m$. What is the resulting differential equation?

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