If frac{dy}{dx} = y + 3 and y(0) = 2,then fin | vedclass

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(B) 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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(B) 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 $\log|y + 3| = x + C$. Given the initial condition $y(0) = 2$,substituting $x = 0$ and $y = 2$ into the equation: $\log|2 + 3| = 0 + C \implies C = \log 5$. Thus,the equation becomes $\log(y + 3) = x + \log 5$. Taking the exponential of both sides,$y + 3 = e^{x + \log 5} = 5e^x$. So,$y = 5e^x - 3$. Now,calculating $y(\log 2)$: $y(\log 2) = 5e^{\log 2} - 3$. Since $e^{\log 2} = 2$,we get $y(\log 2) = 5(2) - 3 = 10 - 3 = 7$.

If $\frac{dy}{dx} = y + 3$ and $y(0) = 2$,then find the value of $y(\log 2)$.

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