Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}} = (\text{x}+\text{y})^2$
$\frac{\text{dy}}{\text{dx}} = (\text{x}+\text{y})^2$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}} = (\text{x}+\text{y})^2$
Let $\text{ x} + \text{y} = \text{v}$
$\Rightarrow 1 + \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}} - 1$
$\therefore \frac{\text{dv}}{\text{dx}} - 1 = \text{v}^2$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \text{v}^2 + 1$
$\Rightarrow \frac{1}{\text{v}^2+1}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{v}^2+1}\text{dv} = \int\text{dx}$
$\Rightarrow \tan^{-1}\text{v} = \text{x} + \text{C}$
$\Rightarrow \text{v} = \tan(\text{x}+\text{C})$
$\Rightarrow \text{x}+\text{y} = \tan (\text{x}+\text{C})$
$\frac{\text{dy}}{\text{dx}} = (\text{x}+\text{y})^2$
Let $\text{ x} + \text{y} = \text{v}$
$\Rightarrow 1 + \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}} - 1$
$\therefore \frac{\text{dv}}{\text{dx}} - 1 = \text{v}^2$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \text{v}^2 + 1$
$\Rightarrow \frac{1}{\text{v}^2+1}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{v}^2+1}\text{dv} = \int\text{dx}$
$\Rightarrow \tan^{-1}\text{v} = \text{x} + \text{C}$
$\Rightarrow \text{v} = \tan(\text{x}+\text{C})$
$\Rightarrow \text{x}+\text{y} = \tan (\text{x}+\text{C})$
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