Question
Solve the following differential equation
$(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}=1$
$(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}=1$
✓
Answer
We have,
$(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\text{dy}=\frac{1}{\text{x}^2+1}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\frac{1}{\text{x}^2+1}\Big)\text{dx}$
$\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$
so, $\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
$(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\text{dy}=\frac{1}{\text{x}^2+1}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\frac{1}{\text{x}^2+1}\Big)\text{dx}$
$\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$
so, $\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
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