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

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

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$7$

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(D) Given the differential equation $\frac{dy}{dx} = y + 3$.Separating the variables,we get $\frac{dy}{y + 3} = dx$.Integrating both sides,we obtain $\int \frac{dy}{y + 3} = \int dx$,which gives $\ln|y + 3| = x + C$.Since $y + 3 > 0$,we have $y + 3 = e^{x + C} = e^C \cdot e^x$.Let $e^C = A$,so $y + 3 = A e^x$.Using the initial condition $y(0) = 2$,we substitute $x = 0$ and $y = 2$:$2 + 3 = A e^0 \Rightarrow A = 5$.Thus,the particular solution is $y + 3 = 5 e^x$,or $y = 5 e^x - 3$.To find $y(\ln 2)$,substitute $x = \ln 2$:$y(\ln 2) = 5 e^{\ln 2} - 3$.Since $e^{\ln 2} = 2$,we have $y(\ln 2) = 5(2) - 3 = 10 - 3 = 7$.

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

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