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(A) Given differential equation: $(x-y)(dx+dy)=dx-dy$Rearranging the terms:$(x-y)dx + (x-y)dy = dx - dy$$(x-y+1)dy = (1-x+y)dx$$\frac{dy}{dx} = \frac{

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$\log |x-y|=x+y+1$

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(A) Given differential equation: $(x-y)(dx+dy)=dx-dy$Rearranging the terms:$(x-y)dx + (x-y)dy = dx - dy$$(x-y+1)dy = (1-x+y)dx$$\frac{dy}{dx} = \frac{1-(x-y)}{1+(x-y)}$ ............$(1)$Let $x-y=t$. Then $1-\frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = 1-\frac{dt}{dx}$.Substituting into $(1)$:$1-\frac{dt}{dx} = \frac{1-t}{1+t}$$\frac{dt}{dx} = 1 - \frac{1-t}{1+t} = \frac{1+t-1+t}{1+t} = \frac{2t}{1+t}$Separating variables:$\frac{1+t}{2t} dt = dx$$\frac{1}{2} (\frac{1}{t} + 1) dt = dx$Integrating both sides:$\frac{1}{2} (\log |t| + t) = x + C$$\log |t| + t = 2x + 2C$$\log |x-y| + x - y = 2x + C_1$$\log |x-y| = x + y + C_1$Given $y=-1$ when $x=0$:$\log |0 - (-1)| = 0 + (-1) + C_1$$\log 1 = -1 + C_1$$0 = -1 + C_1 \Rightarrow C_1 = 1$Thus,the particular solution is $\log |x-y| = x + y + 1$.

Find a particular solution of the differential equation $(x-y)(dx+dy)=dx-dy$ given that $y=-1$ when $x=0$. (Hint: put $x-y=t$)

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