Question
Solve the following differential equation
$(1+\text{x}^2)\text{dy}=\text{xy dx}$

Answer

 We have

$(1+\text{x}^2)\text{dy}=\text{xy dx}$

$\Rightarrow\frac{1}{\text{y}}\text{dy}=\frac{\text{x}}{1+\text{x}^2}\ \text{dx}$

Integrating both sides, we get

$\int\frac{1}{\text{y}}\text{dy}=\int\frac{\text{x}}{1+\text{x}^2}\ \text{dx}$

Substituting 1+ x2 = t, we get

$2\text{x dx}=\text{dt}$

$\therefore\int\frac{1}{\text{y}}\text{dy}=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$

$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|\text{t}|+\log\text{C}$

$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|1+\text{x}^2|+\log\text{C}$

$\Rightarrow\log|\text{y}|=\log\Big[\text{C}\sqrt{1+\text{x}^2}\Big]$

$\Rightarrow\text{y}=\text{C}\sqrt{1+\text{x}^2}$

Hence, $\text{y}=\text{C}\sqrt{1+\text{x}^2}$ is the required solution. 

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