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The solution of the differential equation `e^(-x) (y+1) dy +(cos^(2) x - sin 2x) y (dx) = 0` subjected to the condition that y = 1 when x = 0 isA. `(y+1) + e^(x) cos^(2)x = 2`B. `y + log y = e^(x) cos^(2) x`C. `log(y+1)+e^(x) cos^(2)x = 1`D. `y = log y + e^(x) cos^(2)x = 2` |
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Answer» Correct Answer - D `e^(-x)(y+1)dy+(cos^(2)x-sin 2x) y (dx)=0` `therefore" "(1+y^(-1))dy + e^(x)(cos^(2)x-sin 2x)dx = 0` Integration we get, `y+log_(e) y + e^(x) cos^(2) x = c` y = 1 when x = 0 `therefore" "1+0+1 = c` `therefore" "c = 2` `therefore" ""Curve is y" + log y + e^(x) cos^(2) x = 2` |
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