Question
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=1-\text{x + y}-\text{xy}$
$\frac{\text{dy}}{\text{dx}}=1-\text{x + y}-\text{xy}$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}}=1-\text{x + y}-\text{xy}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\text{y}-\text{x}(1+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\text{y})(1-\text{x})$
$\Rightarrow\frac{1}{1+\text{y}}\text{dy}=(1-\text{x})\text{dx}$
Integrating both sides, we get
$\int\frac{1}{1+\text{y}}\text{dy}=\int(1-\text{x})\text{dx}$
$\Rightarrow\log|1+\text{y}|=\text{x}-\frac{\text{x}^2}{2}+\text{C}$
Hence, $\log|1+\text{y}|=\text{x}-\frac{\text{x}^2}{2}+\text{C}$ is the required solution.
$\frac{\text{dy}}{\text{dx}}=1-\text{x + y}-\text{xy}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\text{y}-\text{x}(1+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\text{y})(1-\text{x})$
$\Rightarrow\frac{1}{1+\text{y}}\text{dy}=(1-\text{x})\text{dx}$
Integrating both sides, we get
$\int\frac{1}{1+\text{y}}\text{dy}=\int(1-\text{x})\text{dx}$
$\Rightarrow\log|1+\text{y}|=\text{x}-\frac{\text{x}^2}{2}+\text{C}$
Hence, $\log|1+\text{y}|=\text{x}-\frac{\text{x}^2}{2}+\text{C}$ is the required solution.
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