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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+logy+e^(x)cos^(2)x=2`B. `log(y+1)+e^(x)cos^(2)x=1`C. `y+logy=e^(x)cos^(2)x`D. `(y+1)+e^(x)cos^(2)x=2` |
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Answer» Correct Answer - a Given , ` e^(-x) (y+1) dy + (cos^(2) x - sin 2x ) ydx = 0 ` ` rArr (1+1/y ) dy = - e^(x) (cos ^(2)x - sin 2 x) dx ` On integrating both sides , we get ` y + lo y = -e^(x) cos^(2) x + int e^(x) sin 2x dx` ` -int e^(x) sin 2x dx + C` ` rArr y + log y = -e^(x) cos^(2) x + C ` At x = 0 , y = 1 ` 1+ 0 = - e^(0) cos 0 + C rArr C = 2` ` :. "Required solution is " y + log y = = -e^(x) cos^(2) x +2` |
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