If dydx=y+3>0 and y(0)=2, then y(ln⁡2) is equal to :

dydx=y+3∫dyy+3=∫dx⇒ln⁡(y+3)=x+cln⁡5=0+c∴ln⁡(y+3)=x+ln⁡5If x=ln2 then ln⁡(y+3)=ln⁡2+ln⁡5⇒y+3=10⇒y=7

If dydx=y+3>0 and y(0)=2, then y(ln⁡2) is equal to :

dydx=y+3∫dyy+3=∫dx⇒ln⁡(y+3)=x+cln⁡5=0+c∴ln⁡(y+3)=x+ln⁡5If x=ln2 then ln⁡(y+3)=ln⁡2+ln⁡5⇒y+3=10⇒y=7

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

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Detailed Solution

$\begin{array}{l} \frac{dy}{dx} = y + 3 \\ \int \frac{dy}{y + 3} = \int dx \Rightarrow \ln (y + 3) = x + c \\ \ln 5 = 0 + c \\ ∴ \ln (y + 3) = x + \ln 5 \\ I f x = \ln 2 t h e n \ln (y + 3) = \ln 2 + \ln 5 \Rightarrow y + 3 = 10 \Rightarrow y = 7 \end{array}$