Question
Solve the following differential equation
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{xy}$
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{xy}$
✓
Answer
We have
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{xy}$
$\Rightarrow(\text{x}-1)\text{dy}=2\text{xy dx}$
$\Rightarrow\frac{2\text{x}}{(\text{x}-1)}\ \text{dx}=\frac{1}{\text{y}}\text{ dy}$
Integrating both sides, we get
$2\int\frac{\text{x}}{(\text{x}-1)}\ \text{dx}=\int\frac{1}{\text{y}}\ \text{dy}$
$\Rightarrow2\int\frac{\text{x}-1+1}{\text{x}-1}\ \text{dx}=\int\frac{1}{\text{y}}\ \text{dx}$
$\Rightarrow2\int\text{dx}+2\int\frac{1}{\text{x}-1}\text{dx}=\int\frac{1}{\text{y}}\ \text{dy}$
$\Rightarrow2\text{x}+2\log|\text{x}-1|=\log|\text{y}|+\text{C}$
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{xy}$
$\Rightarrow(\text{x}-1)\text{dy}=2\text{xy dx}$
$\Rightarrow\frac{2\text{x}}{(\text{x}-1)}\ \text{dx}=\frac{1}{\text{y}}\text{ dy}$
Integrating both sides, we get
$2\int\frac{\text{x}}{(\text{x}-1)}\ \text{dx}=\int\frac{1}{\text{y}}\ \text{dy}$
$\Rightarrow2\int\frac{\text{x}-1+1}{\text{x}-1}\ \text{dx}=\int\frac{1}{\text{y}}\ \text{dx}$
$\Rightarrow2\int\text{dx}+2\int\frac{1}{\text{x}-1}\text{dx}=\int\frac{1}{\text{y}}\ \text{dy}$
$\Rightarrow2\text{x}+2\log|\text{x}-1|=\log|\text{y}|+\text{C}$
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