A) \[{{e}^{2x}}(1-x+y)=1+x-y\]
B) \[{{e}^{2x}}(1+x-y)=1-x+y\]
C) \[{{e}^{2x}}(1-x+y)+(1+x-y)=0\]
D) \[{{e}^{2x}}(1+x+y)=1-x+y\]
E) none of the above
Correct Answer: A
Solution :
The given differential equation is \[\frac{dy}{dx}={{(x-y)}^{2}}\] Let \[x-y=t\] \[\Rightarrow \] \[1-\frac{dy}{dx}=\frac{dt}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=1-\frac{dt}{dx}\] \[\therefore \] \[1-\frac{dt}{dx}={{t}^{2}}\] \[\Rightarrow \] \[(1-{{t}^{2}})=\frac{dt}{dx}\] On integrating both sides \[\Rightarrow \] \[\int{\frac{1}{1-{{t}^{2}}}}dt=\int{1\,}dx\] \[\Rightarrow \] \[\frac{1}{2}\log \left( \frac{1+t}{1-t} \right)=x+c\] \[\Rightarrow \] \[\left( \frac{1+t}{1-t} \right)={{e}^{2x+2c}}\] \[\Rightarrow \] \[\frac{1+x-y}{1-x+y}=A{{e}^{2x}}\] \[\because \]It passes through origin \[\therefore \] \[A=1\] \[\therefore \]Required curve is \[(1+x-y)={{e}^{2x}}(1-x+y)\]
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