InterviewSolution
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InterviewSolution
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The solution of the differential equation `dy/dx=y tanx-2 sinx` is |
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Answer» `dy/dx = ytanx-2sinx` `=>dy/dx - ytanx = -2sinx` Comparing the given equation with first order differential equation, `dy/dx+Py = Q(x)`, we get,`P = -tanx and Q(x) = -2sinx` So, Integrating factor `(I.F) = e^(int -tandx)` `I.F.= e^(-ln|secx|) = 1/secx = cosx` we know, solution of differential equation, `y(I.F.) = intQ(I.F.)dx` `:.`Our solution will be, `ycosx = int cosx(-2sinx)dx` `=>ycosx = int -sin2xdx` `=>ycosx = cos(2x)/2+c` `=>y = cos(2x)/2cosx + csecx` `=>y = (2cos^2x-1)/2cosx + csecx` `=>y = cosx - secx/2+csecx` `=>y = cosx + secx(c-1/2)` `=>y = cosx + Csecx`, where `C = (c-1/2)` is a constant. |
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