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(B) Given the differential equation $\log \left(\frac{dy}{dx}\right) = 2x - 5y$. By definition of logarithm,we have $\frac{dy}{dx} = e^{2x - 5y} = e^{

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$5e^{2x} - 2e^{5y} = 3$

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(B) Given the differential equation $\log \left(\frac{dy}{dx}\right) = 2x - 5y$. By definition of logarithm,we have $\frac{dy}{dx} = e^{2x - 5y} = e^{2x} \cdot e^{-5y}$. Separating the variables,we get $e^{5y} \, dy = e^{2x} \, dx$. Integrating both sides,$\int e^{5y} \, dy = \int e^{2x} \, dx$. This gives $\frac{e^{5y}}{5} = \frac{e^{2x}}{2} + C$. Multiplying by $10$,we get $2e^{5y} = 5e^{2x} + 10C$,or $2e^{5y} - 5e^{2x} = K$. Using the initial condition $y(0) = 0$,we substitute $x = 0$ and $y = 0$: $2e^{5(0)} - 5e^{2(0)} = K \implies 2(1) - 5(1) = K \implies K = -3$. Thus,$2e^{5y} - 5e^{2x} = -3$,which can be rewritten as $5e^{2x} - 2e^{5y} = 3$.

The solution of $\log \left(\frac{dy}{dx}\right) = 2x - 5y$ with the initial condition $y(0) = 0$ is:

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