question_answer

                        Solve the differential equation \[(x-y)(dx+dy)=dx-dy,\] \[y(0)=-1.\]

Answer:

The given differential equation is                         \[(x-y)(dx+dy)=dx-dy\]             \[\Rightarrow \]   \[(x-y-1)\,dx=-\,(x-y+1)\,dy\]             \[\Rightarrow \]   \[\frac{dy}{dx}=-\frac{x-y-1}{x-y+1}\]                           ? (i) Let \[x-y=v.\] Then, \[1-\frac{dy}{dx}=\frac{dv}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=1-\frac{dv}{dx}\] Putting \[x-y=v.\] and  \[\frac{dy}{dx}=1-\frac{dv}{dx}\] in Eq. (i) we get \[1-\frac{dv}{dx}=-\,\frac{v-1}{v+1}\] \[\Rightarrow \]   \[\frac{dv}{dx}=\frac{v-1}{v+1}+1\]     \[\Rightarrow \]   \[\frac{dv}{dx}=\frac{2v}{v+1}\] \[\Rightarrow \]   \[\frac{v+1}{v}dv=2\,dx\]             \[\Rightarrow \]   \[\left( 1+\frac{1}{v} \right)dv=2\,dx\]             \[\Rightarrow \]   \[\int{\left( 1+\frac{1}{v} \right)dv}=2\int{dx}\]             \[\Rightarrow \]   \[v+\log |v|\,\,=2x+C\]             \[\Rightarrow \]   \[x-y+\log |x-y|\,\,=2x+C\]             \[\Rightarrow \]   \[\log |x-y|\,\,=x+y+C\] It is given that \[y(0)=-\,1,\] i.e. when x = 0, \[y=-\,1.\] Putting x = 0 and \[y=-\,1.\] in Eq. (ii) we get \[\log \,1=-1+C\]            \[\Rightarrow \] C = 1 Putting C = 1 in Eq. (ii), we get \[\log |x-y|\,\,=x+y+1\] \[\Rightarrow \] \[|x-y|={{e}^{x+y+1}}\] \[\Rightarrow \]   \[x-y=\pm \,{{e}^{x+y+1}}\]