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$$\text { Solve } \quad \frac{d y}{d x}=1+\frac{e^{(x-y)^{2}} \tan x}{x-y}$$

$$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$$ parallel

let $$t=x-y \quad \frac{d t}{d x}=1-\frac{d y}{d x}$$

$$\frac{d y}{d x}=1-\frac{d t}{d x} \cdots-(2)$$


$$1-\frac{d t}{d x}=1+\frac{e^{(x-y)^{2}} \tan x}{x-y}$$

$$d x\frac{d t}{d x}=\frac{e^{(x-y)^{2}} \tan x}{x-y} d x$$

$$-d t=\frac{e^{t^2} \tan x d x}{t}$$

(1) Separate the variables

(2) Integration

$$* \frac{t}{e^{t^2}}=t \cdot e^{-t^{2}}$$

$$-t e^{-t^{2}} d t=\tan x d x$$ Integrate

$$\int-t e^{-t^{2}} d t=\int \tan x d x$$

$$\int {-\frac{2t}{2}}e^{-t^{2}} d t=\int \tan x d x$$

$$\frac{1}{2} e^{-t^{2}}=\ln |\sec x|+c$$


$$\frac{1}{2} e^{-(x-y)^{2}}=\ln |\sec x|+c$$

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