Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}} = \tan(\text{x}+\text{y})$
$\frac{\text{dy}}{\text{dx}} = \tan(\text{x}+\text{y})$
✓
Answer
$\frac{\text{dy}}{\text{dx}} = \tan(\text{x}+\text{y})$
Let $\text{x}+\text{y} = \text{v}$
$1+\frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\frac{\text{dv}}{\text{dx}}-1 = \tan\text{v}$
$\frac{\text{dv}}{\text{dx}} = 1+\tan\text{v}$
$1+\frac{1}{1+\tan\text{v}}\text{dv} = \text{dx}$
$\frac{\cos\text{v}}{\cos\text{v}+\sin\text{v}}\text{dv} = \text{dx}$
$\Big(\frac{2\cos\text{v}}{\cos\text{v}+\sin\text{v}}\Big)\text{dv} = 2\text{dx}$
$\big(\frac{\cos\text{v}+\sin\text{v}+\cos\text{v}-\sin\text{v}}{\cos\text{v}+\sin\text{v}}\big)\text{dv} = 2\text{dx}$
$\int\text{dv}+\int\big(\frac{\cos\text{v}-\sin\text{v}}{\cos\text{v}+\sin\text{v}}\big)\text{dv}=2\int\text{dx}$
$\text{v}+\log|\cos\text{v}+\sin\text{v}| = 2\text{x}+\text{C}$
$\text{x}+\text{y}+\log|\cos(\text{x}+\text{y})+\sin(\text{x}+\text{y})| = 2\text{x}+\text{C}$
$\text{y}-\text{x}+\log|\cos(\text{x}+\text{y})+\sin(\text{x}+\text{y})|=\text{C}$
Let $\text{x}+\text{y} = \text{v}$
$1+\frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\frac{\text{dv}}{\text{dx}}-1 = \tan\text{v}$
$\frac{\text{dv}}{\text{dx}} = 1+\tan\text{v}$
$1+\frac{1}{1+\tan\text{v}}\text{dv} = \text{dx}$
$\frac{\cos\text{v}}{\cos\text{v}+\sin\text{v}}\text{dv} = \text{dx}$
$\Big(\frac{2\cos\text{v}}{\cos\text{v}+\sin\text{v}}\Big)\text{dv} = 2\text{dx}$
$\big(\frac{\cos\text{v}+\sin\text{v}+\cos\text{v}-\sin\text{v}}{\cos\text{v}+\sin\text{v}}\big)\text{dv} = 2\text{dx}$
$\int\text{dv}+\int\big(\frac{\cos\text{v}-\sin\text{v}}{\cos\text{v}+\sin\text{v}}\big)\text{dv}=2\int\text{dx}$
$\text{v}+\log|\cos\text{v}+\sin\text{v}| = 2\text{x}+\text{C}$
$\text{x}+\text{y}+\log|\cos(\text{x}+\text{y})+\sin(\text{x}+\text{y})| = 2\text{x}+\text{C}$
$\text{y}-\text{x}+\log|\cos(\text{x}+\text{y})+\sin(\text{x}+\text{y})|=\text{C}$
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