InterviewSolution
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InterviewSolution
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Consider the differential equation, `y^(2)dx+(x-1/y)dy=0.` If value of y is 1 when x = 1, then the value of x for which y = 2, is(a) `5/2+1/sqrt e`(b) `3/2+1/sqrt e`(c) `1/2+1/sqrt e`(d) `3/2+sqrt e`A. (a) `5/2+1/sqrt e`B. (b) `3/2+1/sqrt e`C. (c) `1/2+1/sqrt e`D. (d) `3/2+sqrt e` |
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Answer» Correct Answer - (b) Given differential equation is `y^(2) dx+(x-1/y)dy =0` `rArr dx/dy+1/y^(2)x=1/y^(3),` which is the linear differential equation of the form `dx/dy+ Px = Q.` Here, `P=1/y^(2) and Q=1/y^(3)` Now, `IF=e^(int1/y^(2)dy)=e^(-1/y)` `therefore` The solution of linear differential equation is `x.(IF)intQ(IF)dy+c` `rArr x e^(-1//y)=int 1/y^(3) e^(-1//y)dy+C` `therefore x e^(-1//y) =int(-t) e^(t) dt+C [therefore let -1/y=trArr+1/y^(2)dy=dt]` `=-te^(t)+int e^(t)dt+C` [integration by parts] `=-te^(t)+ e^(t)+C` `rArr x c^(-1//y)=1/ye^(-1//y)+e^(-1//y)+C` Now, at y = 1, the value of x = 1, so `1cdote^(1)=e^(-1) +e^(-1)+C rArrC=-1/e` On putting the value of C, in Eq. (i), we get `x=1/y+1-e^(1//y)/e` So, at Y= 2, the value of `x=1/2+1-e^(1//2)/e= 3/2 - 1/sqrte` |
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