InterviewSolution
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InterviewSolution
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Solve the following differential equations:`(x+y)(dx-dy)=dx+dy` |
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Answer» Given differential equation is `" "(x+y)(dx-dy)=dx+dy` `rArr" "(x+y)(1-(dy)/(dx))=1+(dy)/(dx)" "...(i)` Put `" "x+y=z` `rArr" "1+(dy)/(dx)=(dz)/(dx)` On substituting these values in Eq. (i), we get `" "z(1-(dz)/(dx)+1)=(dz)/(dx)` `rArr" "z(2-(dz)/(dx))=(dz)/(dx)` `rArr" "2z-z(dz)/(dx)-(dz)/(dx)=0` `rArr" "2z-(z+1)(dz)/(dx)=0` `rArr" "(dz)/(dx)=(2z)/(z+1)` `rArr" "((z+1)/(z))dz=2dx` On integrating both sides, we get `" "int(1+(1)/(z))dz=2intdx` `rArr" "z+logz=2x-logC` `rArr" "(x+y)+log(x+y)=2x-logC" "[because z=x+y]` `rArr" "2x-x-y=logC+log(x+y)` `rArr" "x-y=log|C(x+y)|` `rArr" "e^(x-y)=C(x+y)` `rArr" "(x+y)=(1)/(C)e^(x-y)` `rArr" "x+y=Ke^(x-y)" "[becauseK=(1)/(2)]` |
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