Question
Solve the following differential equations:$\text{xy}\frac{\text{dy}}{\text{dx}}=\text{y}+2,\text{y}(2)=0$
✓
Answer
We have,
$\text{xy}\frac{\text{dy}}{\text{dx}}=\text{y}+2,\text{y}(2)=0$
$\Rightarrow\frac{\text{y}}{\text{y}+2}\text{dy}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{y}}{\text{y}+2}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\int\frac{\text{y}+2-2}{\text{y}+2}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\int\text{dy}-2\int\frac{1}{\text{y}+2}\text{dy}=\log\text{x + C}$
$\Rightarrow\text{y}-2\log|\text{y}+2|=\log|\text{x}|+\text{C}\dots(1)$
It is given that at $\text{x}=2,\text{y}=0.$
Substituting the valuse of x and y in (1), we get
$-2\log2-\log2=\text{C}$
$\Rightarrow-\log(2^2\times2)=\text{C}$
$\Rightarrow\text{C}=-\log8$
Substituting the value of C in (1), we get
$\text{y}-2\log|\text{y}+2|=\log|\text{x}|-\log8$
$\Rightarrow\text{y}-2\log|\text{y}+2|=\log\Big|\frac{\text{x}}{8}\Big|$
Hence, $\text{y}-2\log|\text{y}+2|=\log\Big|\frac{\text{x}}{8}\Big|$ is the required solution.
$\text{xy}\frac{\text{dy}}{\text{dx}}=\text{y}+2,\text{y}(2)=0$
$\Rightarrow\frac{\text{y}}{\text{y}+2}\text{dy}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{y}}{\text{y}+2}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\int\frac{\text{y}+2-2}{\text{y}+2}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\int\text{dy}-2\int\frac{1}{\text{y}+2}\text{dy}=\log\text{x + C}$
$\Rightarrow\text{y}-2\log|\text{y}+2|=\log|\text{x}|+\text{C}\dots(1)$
It is given that at $\text{x}=2,\text{y}=0.$
Substituting the valuse of x and y in (1), we get
$-2\log2-\log2=\text{C}$
$\Rightarrow-\log(2^2\times2)=\text{C}$
$\Rightarrow\text{C}=-\log8$
Substituting the value of C in (1), we get
$\text{y}-2\log|\text{y}+2|=\log|\text{x}|-\log8$
$\Rightarrow\text{y}-2\log|\text{y}+2|=\log\Big|\frac{\text{x}}{8}\Big|$
Hence, $\text{y}-2\log|\text{y}+2|=\log\Big|\frac{\text{x}}{8}\Big|$ is the required solution.
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