If frac{dy}{dx} = y + 3,y + 3 0 and y(0) = | vedclass

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(C) Given the differential equation $\frac{dy}{dx} = y + 3$. Separating the variables,we get $\frac{dy}{y + 3} = dx$. Integrating both sides,we have $

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$7$

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(C) Given the differential equation $\frac{dy}{dx} = y + 3$. Separating the variables,we get $\frac{dy}{y + 3} = dx$. Integrating both sides,we have $\int \frac{dy}{y + 3} = \int dx + C$. This gives $\log(y + 3) = x + C$ ... $(i)$. Given $y(0) = 2$,substitute $x = 0$ and $y = 2$ into $(i)$: $\log(2 + 3) = 0 + C \Rightarrow C = \log 5$. Substituting $C$ back into $(i)$,we get $\log(y + 3) = x + \log 5$. Taking the exponential of both sides,$y + 3 = e^{x + \log 5} = 5e^x$. Thus,$y = 5e^x - 3$. Now,calculate $y(\log 2)$: $y(\log 2) = 5e^{\log 2} - 3 = 5(2) - 3 = 10 - 3 = 7$.

If $\frac{dy}{dx} = y + 3$,$y + 3 > 0$ and $y(0) = 2$,then $y(\log 2)$ is equal to

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