Solve the following differential equation: dy dx (x-y)=1

Question

Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}\cos(\text{x}-\text{y})=1$

✓

Answer

$\frac{\text{dy}}{\text{dx}}\times\cos(\text{x}-\text{y})=1$
Let $\text{x}-\text{y}=\text{v}$
$1-\frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dv}}{\text{dx}}$
So,
$\Big(1-\frac{\text{dv}}{\text{dx}}\Big)\cos\text{v}=1$
$1-\frac{\text{dv}}{\text{dx}}=\sec\text{v}$
$1-\sec\text{v}=\frac{\text{dv}}{\text{dx}}$
$\text{dx}=\frac{\text{dv}}{1-\sec\text{v}}$
$\text{dx}=\frac{\cos\text{v}}{1-\cos\text{v}}\text{dv}$
$\int\text{dx}=\int\frac{\cos^{2}\frac{\text{v}}{2}-\sin^{2}\frac{\text{v}}{2}}{2\sin^{2}\frac{\text{v}}{2}}\text{dv}$
$\int\text{dx}=\int\frac{1}{2}\cot\big(\frac{\text{v}}{2}\big)\text{dv}-\frac{1}{2}\text{dv}$
$2\int\text{dx}=\int\cot^{2}\big(\frac{\text{v}}{2}\big)-\int\text{dv}$
$2\int\text{dx}=\int\Big(\text{cosec}^{2}\frac{\text{v}}{2}-1\Big)\text{dv}-\int\text{dv}$
$2\text{x}=-2\cot\big(\frac{\text{v}}{2}\big)\text{dv}-\text{v}-\text{v}+\text{C}_{1}$
$2(\text{x}+\text{v})=-2\cot\frac{\text{v}}{2}+\text{C}_{1}$
$\text{x}+\text{x}-\text{y}=-\cot\Big(\frac{\text{x}-\text{y}}{2}\Big)+\text{C}$
$\text{C}+\text{y}=\cot\Big(\frac{\text{x}-\text{y}}{2}\Big)$

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