Question
Solve $2(\text{y}+3)-\text{xy}\frac{\text{dy}}{\text{dx}}=0,$ given that y(1) = -2.

Answer

We have, $2(\text{y}+3)-\text{xy}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow2(\text{y}+3)=\text{xy}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{y}+3}\text{dy}$
$\Rightarrow2\int\frac{\text{dy}}{\text{x}}=\int\Big(1-\frac{3}{\text{y}+3}\Big)\text{dy}$
$\Rightarrow2\log\text{x}=\text{y}-3\log(\text{y}+3)+\text{C}\ ......({\text{i}})$
Given that when x = 1 and y = -2
$\Rightarrow2\log1=-2-3\log(-2+3)+\text{C}$
$\Rightarrow\text{C}=2$
Thus, from Eq. (i)
$2\log\text{x}=\text{y}-3\log(\text{y}+3)+2$
$\Rightarrow2\log\text{x}+3\log(\text{y}+3)=\text{y}+2$
$\Rightarrow\log\text{x}^2(\text{y}+3)^3=\text{y}+2$
$\Rightarrow\text{x}^2(\text{y}+3)^3=\text{e}^{\text{y}+2}$

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