Question
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{x}-\frac{1}{\text{x}},\text{x}\ne0$
$\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{x}-\frac{1}{\text{x}},\text{x}\ne0$
✓
Answer
$\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{x}-\frac{1}{\text{x}}$
$\Rightarrow\text{dy}=\Big(\text{x}^2+\text{x}-\frac{1}{\text{x}}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\text{x}^2+\text{x}-\frac{1}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$
Clearly, $\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
$\Rightarrow\text{dy}=\Big(\text{x}^2+\text{x}-\frac{1}{\text{x}}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\text{x}^2+\text{x}-\frac{1}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$
Clearly, $\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{3}-\log|\text{x}|+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
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