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(D) Given differential equation is $\log\left(\frac{dy}{dx}\right) = x$. By definition of logarithm,we have $\frac{dy}{dx} = e^x$. Separating the vari

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$y = e^x$

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(D) Given differential equation is $\log\left(\frac{dy}{dx}\right) = x$. By definition of logarithm,we have $\frac{dy}{dx} = e^x$. Separating the variables,we get $dy = e^x dx$. Integrating both sides,we get $\int dy = \int e^x dx$,which gives $y = e^x + C$. Given that when $x = 0, y = 1$,substituting these values into the equation: $1 = e^0 + C$. Since $e^0 = 1$,we have $1 = 1 + C$,which implies $C = 0$. Therefore,the particular solution is $y = e^x$.

The particular solution of the differential equation $\log\left(\frac{dy}{dx}\right) = x$,when $x = 0, y = 1$ is .....

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