The general solution of the differential equa | vedclass

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(D) Given the differential equation: $\frac{dx}{dt} = \frac{x \log x}{t}$ Separating the variables,we get: $\int \frac{dx}{x \log x} = \int \frac{dt}{

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$e^{ct} = x$

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(D) Given the differential equation: $\frac{dx}{dt} = \frac{x \log x}{t}$ Separating the variables,we get: $\int \frac{dx}{x \log x} = \int \frac{dt}{t}$ Let $u = \log x$,then $du = \frac{1}{x} dx$. Substituting this into the integral: $\int \frac{du}{u} = \int \frac{dt}{t}$ Integrating both sides: $\log |u| = \log |t| + \log |c|$ $\log |\log x| = \log |tc|$ Taking the exponential of both sides: $\log x = tc$ Therefore,$x = e^{tc}$.

The general solution of the differential equation $\frac{dx}{dt} = \frac{x \log x}{t}$ is

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