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If `ye^(y) dx = (y^(3) + 2xe^(y))dy, y(0) = 1`, then the value of x when y = 0 isA. `-1`B. 0C. 1D. 2 |
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Answer» Correct Answer - B `ye^(y) dx = (y^(3)+2xe^(y))dy` `rArr" "(dx)/(dy)=(y^(2))/(e^(y))+(2x)/(y)` `rArr" "(dx)/(dy)-(2)/(y)x=(y^(2))/(e^(y))` I.F. `= e^(-2 Iny) = (1)/(y^(2))` General solution `x (1)/(y^(2))= int (y^(2))/(e^(y))-(1)/(y^(2)) dy + c` `(x)/(y^(2))=e^(-y) + c = -e^(-y) + c` `x = 0, y = 1 rArr c = e^(-1)` `rArr" "x = -y^(2)(e^(-y) - e^(-1))` If y = 0, then x = 0 |
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