InterviewSolution
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InterviewSolution
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The equation of the curve passing through the origin and satisfying the differential equation `((dy)/(dx))^(2)=(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)`D. `e^(2x)(1+x+y)=1-x+y` |
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Answer» Correct Answer - A We have, `(dy)/(dx)=pm(x-y)` `ul("CASE I")" When "(dy)/(dx)=(x-y)` In this case, we have `1-(dv)/(dx)=v," where "x-y=v` `rArr" "(dv)/(dx)=1-v` `rArr" "(1)/(1-v)dv=dx` `-log(1-v)=x+logC` `rArr" "(1-v)^(-1)=Ce^(x) rArr(1)/(1-x+y)=Ce^(x)` It passes through the origin. `therefore" "C=1` Hence, `(1)/(1-x+y)=e^(x)" ...(i)"` `ul("CASE II")" When "(dy)/(dx)=(-x-y)=y-x.` In this case, we have `(du)/(dx)+1=u," where"u=y-x` `rArr" "(du)/(u-1)=dx` `rArr" "log(u-1)=x+logC` `rArr" "u-1=Ce^(x)rArr y-x-1=Ce^(x)` It passes through the origin. `therefore" "C=-1` Hence, `y-x-1=-e^(x)or, x-y+1=e^(x)" ...(ii)"` From (i) and (ii), we obtain `(x-y+1)/(1-x+y)=e^(2x)rArr (x-y+1)=(1-x+y)e^(2x)` |
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