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If `(dy)/(dx)-y log_(e) 2 = 2^(sin x)(cos x -1) log_(e) 2`, then y =A. `2^(sin x) + c2^(x)`B. `2^(cos x)+c2^(x)`C. `2^(sin x) + c2^(-x)`D. `2^(cos x) + c2^(-x)` |
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Answer» Correct Answer - A `(dy)/(dx)-y log_(e) 2 = 2^(sin x) (cos x - 1) log_(e) 2` This is linear differential equation I.F. `=e^(-log_(e) 2int dx) = e^(-x log_(e) 2) = 2^(-x)` Solution is `y 2^(-x) = int 2^(-x) 2^(sin x) (cos x - 1) log_(e) 2 dx` put `sin x - x = t rArr (cos x -1) dx = dt` `therefore" "y 2^(-x) = log_(e) 2 int 2^(t) dt` `therefore" "y 2^(-x) = 2^(t) + c` `therefore" "y = 2^(x+t) + c 2^(x)` `therefore" "y = 2^(sin x) + c2^(x)` |
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