InterviewSolution
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InterviewSolution
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The equation of a curve passing through the origin and satisfying the differential equation `(dy)/(dx)=(x-y)^(2)`, isA. `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` |
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Answer» Correct Answer - A We have, `(dy)/(dx)=(x-y)^(2)` Let `(x-y)=v`. Then, `1-(dy)/(dx)=(dv)/(dx)rArr (dy)/(dx)=1-(dv)/(dx)` `therefore" "(dy)/(dx)=(x-y)^(2)` `rArr" "1-(dv)/(dx)=v^(2)` `rArr" "1-v^(2)=(dv)/(dx)` `rArr" "dx=(1)/(1-v^(2))dv` `rArr" "2intdx=2int(1)/(1-v^(2))dv` `rArr" "2x=log((1+v)/(1-v))+logC` `rArr" "C((1+v)/(1-v))=e^(2x)` `rArr" "C((x-y+1)/(y-x+1))=e^(2x)rArrC(x-y+1)=e^(2x)(y-x+1)` Taking C = 1, we find that option (a) is correct. |
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