Answer:
Slope of tangent to the curve is \[\frac{dy}{dx}\]and the square of the difference of the abscissa and the ordinate is\[{{(x-y)}^{2}}\]. According to the question, slope of the tangent to the curve is equal to the square of the difference of the abscissa and ordinate of the point. \[\therefore \] \[\frac{dy}{dx}={{(x-y)}^{2}}\] ?(i) Put \[z=x-y\] \[\therefore \] \[\frac{dz}{dx}=1-\frac{dy}{dx}\] [on differentiating w.r.t. x] \[\Rightarrow \] \[\frac{dy}{dx}=1-\frac{dz}{dx}\] Now, substituting\[\frac{dy}{dx}=1-\frac{dz}{dx}\]and \[z=x-y\]in Eq. (i), we get \[1-\frac{dz}{dx}={{z}^{2}}\Rightarrow \frac{dz}{dx}=1-{{z}^{2}}\] By separating the variables, we get \[\frac{dz}{1-{{z}^{2}}}=dx\] On integrating both sides, we get \[\int{\frac{dz}{1-{{z}^{2}}}=\int{dx\Rightarrow \frac{1}{2}\log }\left| \frac{1+z}{1-z} \right|\,\,=x+{{C}_{1}}}\] \[\left[ \because \int{\frac{dx}{{{a}^{2}}-{{x}^{2}}}=\frac{1}{2a}\log \,\,\left| \frac{a+x}{a-x} \right|+C} \right]\] \[\Rightarrow \] \[\log \,\,\left| \frac{1+z}{1-z} \right|\,\,=2\,(x+{{C}_{1}})\] Now, substituting, \[z=x-y,\] we get \[{{\log }_{e}}\left| \frac{1+(x-y)}{1-(x-y)} \right|\,\,=2x+{{C}_{2}}\] \[[\because {{C}_{2}}=2{{C}_{1}}]\] \[\Rightarrow \] \[\frac{1+(x-y)}{1-(x-y)}=\pm \,{{e}^{2x\,\,+\,\,{{C}^{2}}}}\] \[=\pm \,\,C{{e}^{2x}}\] \[[\text{here},\,\,C={{e}^{{{C}_{2}}}}]\] It is also given that curve passing through origin. \[\therefore \]\[\frac{1+(0-0)}{1-(0-0)}=\pm \,\,C{{e}^{0}}\Rightarrow C=\pm \,\,1\] \[[\because {{e}^{0}}=1]\] \[\therefore \] \[\frac{1+(x-y)}{1-(x-y)}=\pm \,\,{{e}^{2x}}\] \[\Rightarrow \] \[\pm \,\,{{e}^{2x}}(1-x+y)-1=x-y\] \[\Rightarrow \] \[(1+x-y)=\pm \,\,(1-x+y){{e}^{2x}}\]
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