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Q)

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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A)
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• A linear differential equation of the form $\large\frac{dy}{dx}$$=F(x,y) where F(x.y) is of the form g(x).h(y) where g(x) is a function of x and h(y) is a function of y,then the equation is variable separable type.Such equations can be solved by separating the variables. Step 1: Given :(x-y)(dx+dy)=dx-dy On expanding and rearranging we get, x(dx-dy)-y(dx-dy)=dx-dy (x-y+1)dy=(1-x+y)dx \large\frac{dy}{dx}=\frac{(1-x+y)}{(x-y+1)} Step 2: Let (x-y)=t then 1-\large\frac{dy}{dx}=\frac{dt}{dx} or \large\frac{dy}{dx}=$$1-\large\frac{dt}{dx}$
$1-\large\frac{dt}{dx}=\frac{(1-t)}{(1+t)}$
$-\large\frac{dt}{dx}=\frac{(1-t)}{(1+t)}$$-1 -\large\frac{dt}{dx}=\frac{-2t}{(1+t)} \large\frac{dt}{dx}=\frac{2t}{(1+t)} Separating the variables \large\frac{(1+t)dt}{2t}$$=dx$
$\large\frac{dt}{2t}+\frac{dt}{2}$$=dt Step 3: Integrating on both sides we get, \large\frac{1}{2}$$\log t+\large\frac{1}{2}$$t=t+C substituting for t we get, \large\frac{1}{2}$$\log (x-y)+\large\frac{1}{2}$$(x-y)=(x-y)+C \large\frac{1}{2}$$\log (x-y)=\large\frac{(x+y)}{2}$$+C$
$\log(x-y)=(x+y)+2C$
Step 4:
To evaluate for $C$ let us substitute the given values of $x=0$ and $y=-1$
$\log(1)=-1+k$(where $2C=k$)
$k=1$
$\log(x-y)=(x+y)+1$
$\log|x-y|=x+y+1$
This is the required solution.