InterviewSolution
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InterviewSolution
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If `(dx)/(dy)=(e^(y)-x)`, where y(0)=0, then y is expressed explicitly asA. `(1)/(2)ln(1+x^(2))`B. `ln(1+x^(2))`C. `ln(x-sqrt(1+x^(2)))`D. `ln(x+sqrt(1-x^(2)))` |
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Answer» Correct Answer - C We have, `(dx)/(dy)=e^(y)-x rArr(dx)/(dy)+x=e^(y)" ...(i)"` This is a linear differential equation with I.F. `=e^(y)`. Multiplying both sides of (i) by `"I.F."=e^(y)` and integrating with respect to y, we obtain `xe^(y)=(1)/(2)e^(2y)+C` It is given y = 0 when x = 0. Putting these values in (ii), we obtain `C=-(1)/(2)`. Substituting `C=-(1)/(2)` in (ii), we obtain `xe^(y)=(1)/(2)e^(2y)-(1)/(2)` `rArr" "e^(2y)-2xe^(y)-1=0` `rArr" "e^(y)=x+sqrt(x^(2)+1)rArry=ln(x+sqrt(x^(2)+1))` |
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