Question
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\log\text{x}$
$\frac{\text{dy}}{\text{dx}}=\log\text{x}$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}}=\log\text{x}$
$\Rightarrow\text{dy}=(\log\text{x})\text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int(\log\text{x})\text{dx}$
$\Rightarrow\text{dy}=\int1\times\log\text{x}\text{ dx}$
$\Rightarrow\text{dy}=\log\text{x}\int\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\log\text{x }-\int\frac{\text{x}}{\text{x}}\text{dx}$
$\Rightarrow\text{y}=\text{x}\log\text{x}-\int1\text{dx}$
$\Rightarrow\text{y}=\text{x}\log\text{x}-\text{x}$
$\Rightarrow\text{y}=\text{x}(\log\text{x}-1)+\text{C}$
So, $\Rightarrow\text{y}=\text{x}(\log\text{x}-1)+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\text{x}(\log\text{x}-1)+\text{C}$ where $\text{x}\in\text{R}-\{0\}$ is the solution o the given differential equation.
$\frac{\text{dy}}{\text{dx}}=\log\text{x}$
$\Rightarrow\text{dy}=(\log\text{x})\text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int(\log\text{x})\text{dx}$
$\Rightarrow\text{dy}=\int1\times\log\text{x}\text{ dx}$
$\Rightarrow\text{dy}=\log\text{x}\int\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\log\text{x }-\int\frac{\text{x}}{\text{x}}\text{dx}$
$\Rightarrow\text{y}=\text{x}\log\text{x}-\int1\text{dx}$
$\Rightarrow\text{y}=\text{x}\log\text{x}-\text{x}$
$\Rightarrow\text{y}=\text{x}(\log\text{x}-1)+\text{C}$
So, $\Rightarrow\text{y}=\text{x}(\log\text{x}-1)+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\text{x}(\log\text{x}-1)+\text{C}$ where $\text{x}\in\text{R}-\{0\}$ is the solution o the given differential equation.
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