Question
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
$\Rightarrow\text{dy}=(\tan^{-1}\text{x})\text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int(\tan^{-1}\text{x})\text{dx}$
$\Rightarrow\text{y}=\int1\times\tan^{-1}\text{x}\text{ dx}$
$\Rightarrow\text{y}=\tan^{-1}\text{x}\int\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x }-\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\int\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$
So, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is defined for all $\text{x}\in\text{R}$
Hence, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is the solution o the given differential equation.
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
$\Rightarrow\text{dy}=(\tan^{-1}\text{x})\text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int(\tan^{-1}\text{x})\text{dx}$
$\Rightarrow\text{y}=\int1\times\tan^{-1}\text{x}\text{ dx}$
$\Rightarrow\text{y}=\tan^{-1}\text{x}\int\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x }-\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\int\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$
So, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is defined for all $\text{x}\in\text{R}$
Hence, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is the solution o the given differential equation.
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