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If `dy/dx=e^(-2y) and y=0` when `x=5,` then the value of x for `y=3` is |
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Answer» Given that, `(dy)/(dx)=e^(-2 y)Rightarrow (dy)/(e^(-2y))=dx` `Rightarrow inte^(2y)=intdx Rightarrow (e^(2y))/(2)=x+C ....(i)` When x=5 and y=0, then substituting these values in Eq. (i), we get `(e^(0))/(2)=5+C` `Rightarrow (1)/(2)=5+C Rightarrow (1)/(2)-5=(9)/(2)` `"Eq. = (i) becomes e"^(2y)=2x-9` When, y=3, then `e^(6)=2x-9 Rightarrow 2x=e^(6)+9` `x=((e^(6))+9)/(2)` |
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