Solve the following differential equation: dy dx = (x+y) - Vidyadip

Question

Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}} = \sec(\text{x}+\text{y})$

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Answer

We have,
$\frac{\text{dy}}{\text{dx}} = \sec(\text{x}+\text{y})$
$\frac{\text{dy}}{\text{dx}} = \frac{1}{\cos(\text{x}+\text{y})}$
Let $\text{ x}+\text{y} = \text{v}$
$\Rightarrow 1+\frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}-1$
$\therefore\frac{\text{dv}}{\text{dx}}-1 = \frac{1}{\cos\text{v}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}} = \frac{\cos\text{v}+1}{\cos\text{v}}$
$\Rightarrow \frac{\cos \text{v}}{\cos\text{v}+1}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int \frac{\cos\text{v}}{\cos\text{v}+1}\text{dv} = \int\text{dx}$
$\Rightarrow \int\frac{\cos\text{v}(1-\cos\text{v})}{1-\cos^2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \int\frac{\cos\text{v}(1-\cos\text{v})}{\sin^2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \int\frac{\cos\text{v}-\cos^2\text{v}}{\sin^2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow\int(\cot\text{v}\text{ cosec}\text{ v}-\cot^2\text{v})\text{dv} = \int\text{dx}$
$\Rightarrow \int(\cot\text{v}\text{ cosec }\text{v}-\text{cosec}^2\text{v}+1)\text{dv} = \int\text{dx}$
$\Rightarrow -\text{cosec }\text{v}+\cot\text{v}+\text{v} = \text{x}+\text{C}$
$\Rightarrow -\text{cosec}(\text{x}+\text{y})+\cot(\text{x}+\text{y})+\text{x}+\text{y} = \text{x}+\text{C}$
$\Rightarrow -\text{cosec}(\text{x}+\text{y})+\cot(\text{x}+\text{y})+\text{y} = \text{C}$
$\Rightarrow \frac{-1+\cos(\text{x}+\text{y})}{\sin(\text{x}+\text{y})}+\text{y}=\text{C}$
$\Rightarrow -\tan\Big(\frac{\text{x}+\text{y}}{2}\Big)+\text{y} = \text{C}$
$\Rightarrow \text{y} = \tan\Big(\frac{\text{x}+\text{y}}{2}\Big)+\text{C}$

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