Solve the following differential equation dy dx + 1+y^2 y =0 - Vidyadip

Question

Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}+\frac{1+\text{y}^2}{\text{y}}=0$

✓

Answer

We have
$\frac{\text{dy}}{\text{dx}}+\frac{1+\text{y}^2}{\text{y}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\Big(\frac{(1+\text{y}^2}{\text{y}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=-\frac{\text{y}}{1+\text{y}^2}$
$\Rightarrow\text{dx}=\Big(-\frac{\text{y}}{1+\text{y}^2}\Big)\text{dy}$
Integrating both sides, we get
$\int\text{dx}=\int\Big(-\frac{\text{y}}{1+\text{y}^2}\Big)\text{dy}$
$\Rightarrow\text{x}=\int\Big(-\frac{\text{y}}{1+\text{y}^2}\Big)\text{dy}$
Putting $1 + y^2 = t$ we get
$2y\ dy\ dt$
$\therefore\text{x}=-\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$\Rightarrow\text{x}=-\frac{1}{2}\log|\text{t}|+\text{C}$
$\Rightarrow\text{x}=-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$
$\Rightarrow\text{x}+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$
Hence, $\text{x}+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$ is the required solution.

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