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find the particular solution satisfying the given condition, for the following differential equation: `(x+1)(dy)/(dx)=2e^-y-1` given that `y=0` when `x=0` |
Answer» Given differential equation is `(x+1)(dy)/(dx)=2e^(-y)-1`……`(1)` `implies (dy)/(2e^(-y)-1)=(dx)/(x+1)implies(e^(y)dy)/(2-e^(y))=(dx)/(x+1)` On integration, `int(e^(y)dy)/(2-e^(y))=int(dx)/(x+1)`……`(2)` Let `2-e^(y)=timplies-e^(y)=(dt)/(dy)impliese^(y)dy=-dt` From equation `(2)`, `int-(dt)/(t)=int(dx)/(x+1)` `implies -log|t|=log|x+1|+logC` `implies -log|2-e^(y)|=log|C(x+1)|` `implies (1)/(2-e^(y))=C(x+1)` `implies 2-e^(y)=(1)/(C(x+1))` Now, at `x=0` and `y=0`, `2-1=(1)/(C)impliesC=1` Put the value of `C` in equation `(3)`, `2-e^(y)=(1)/((x+1))impliese^(y)=2-(1)/(x+1)` `impliese^(y)=(2x+2-1)/(x+1)impliese^(y)=(2x+1)/(x+1)` `implies y=log|(2x+1)/(x+1)|` |
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