If frac{dy}{dx} = y + 5 and y(0) = 4, then y( | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = y + 5$.Separate the variables: $\frac{dy}{y + 5} = dx$.Integrate both sides: $\int \frac{dy}{y +

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$13$

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(D) Given the differential equation $\frac{dy}{dx} = y + 5$.Separate the variables: $\frac{dy}{y + 5} = dx$.Integrate both sides: $\int \frac{dy}{y + 5} = \int dx \implies \log|y + 5| = x + C$.Using the initial condition $y(0) = 4$: $\log|4 + 5| = 0 + C \implies C = \log 9$.Thus, $\log|y + 5| = x + \log 9$.Rearranging gives $\log|y + 5| - \log 9 = x \implies \log|\frac{y + 5}{9}| = x$.Exponentiating both sides: $\frac{y + 5}{9} = e^x \implies y = 9e^x - 5$.Now, calculate $y(\log 2)$: $y(\log 2) = 9e^{\log 2} - 5$.Since $e^{\log 2} = 2$, we get $y(\log 2) = 9(2) - 5 = 18 - 5 = 13$.

If $\frac{dy}{dx} = y + 5$ and $y(0) = 4$, then $y(\log 2)$ is equal to:

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