Solve the following differential equations: xy dy dx =(x+2)(y+2),… - Vidyadip

Question

Solve the following differential equations:
$\text{xy}\frac{\text{dy}}{\text{dx}}=(\text{x}+2)(\text{y}+2),\text{y}(1)=-1$

✓

Answer

$\text{xy}\frac{\text{dy}}{\text{dx}}=(\text{x}+2)(\text{y}+2),\text{y}(1)=-1$
$\Rightarrow\frac{\text{y}}{\text{y}+2}\text{dy}=\frac{\text{x}+2}{\text{x}}\text{dx}$
$\Rightarrow\frac{\text{y}+2-2}{\text{y}+2}\text{dy}=\frac{\text{x}+2}{\text{x}}\text{dx}$
$\Rightarrow\Big(1-\frac{2}{\text{y}+2}\Big)\text{dy}=\Big(1+\frac{2}{\text{x}}\Big)\text{dx}$
$\Rightarrow\Big(1-\frac{2}{\text{y}+2}\Big)\text{dy}=\Big(1+\frac{2}{\text{x}}\Big)\text{dx}$
Integrating both sides, we get
$\int\Big(1-\frac{2}{\text{y}+2}\Big)\text{dy}=\int\Big(1+\frac{2}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|+\text{C}...(1)$
We know that at $\text{x}=1,\text{y}=-1$
Substituting the valuse of x and y in (1), we get
$-1-2\log|1|=1+2\log|1|+\text{C}$
$\Rightarrow-1=1+\text{C}$
$\Rightarrow\text{C}=-2$
Substituting the value of C in (1), we get
$\text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|-2$
Hence, $\text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|-2$ is the required solution.

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