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 $1+ x^2 = t,$ 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.
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\frac{\text{x}}{1+\text{x}^2}\ \text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\frac{\text{x}}{1+\text{x}^2}\ \text{dx}$
Substituting $1+ x^2 = t,$ 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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