If dydx=y+3>0 and y(0)=2, then y(log2) is equal to

Here, dydx=y+3>0 and y(0)=2⇒ ∫dyy+3=∫dx⇒ log|y+3|=x+CSince, y(0) = 2⇒ loge|2+3|=0+C∴ C=loge5 ⇒ loge|y+3|=x+ loge5When x=loge2 ⇒ loge|y+3|=loge2+loge5=loge10 ⇒ y+3=10 ⇒ y=7

If dydx=y+3>0 and y(0)=2, then y(log2) is equal to

Here, dydx=y+3>0 and y(0)=2⇒ ∫dyy+3=∫dx⇒ log|y+3|=x+CSince, y(0) = 2⇒ loge|2+3|=0+C∴ C=loge5 ⇒ loge|y+3|=x+ loge5When x=loge2 ⇒ loge|y+3|=loge2+loge5=loge10 ⇒ y+3=10 ⇒ y=7

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If $\frac{d y}{d x} = y + 3 > 0 a n d y (0) = 2, t h e n y (\log 2)$ is equal to

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Detailed Solution

Here, $\frac{d y}{d x} = y + 3 > 0 a n d y (0) = 2$
$\Rightarrow \int \frac{d y}{y + 3} = \int d x$
$\Rightarrow \log | y + 3 | = x + C$
Since, y(0) = 2

$\Rightarrow \log_{e} | 2 + 3 | = 0 + C$
$∴ C = \log_{e} 5 \Rightarrow \log_{e} | y + 3 | = x + \log_{e} 5$
When $x = \log_{e} 2$
$\Rightarrow \log_{e} | y + 3 | = \log_{e} 2 + \log_{e} 5 = \log_{e} 10 \Rightarrow y + 3 = 10 \Rightarrow y = 7$