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Solve thedifferential equation `(dy)/(dx)+ycotx=2cosx ,`given that `y=0,`when `x=pi/2dot` |
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Answer» Comparing the given equation with first order differential equation, `dy/dx+Py = Q(x)`, we get,`P = cotx and Q(x) = 2cosx` So, Integrating factor `(I.F) = e^(intcotxdx)` `I.F.= e^(ln|sinx|) = sinx` we know, solution of differential equation, `y(I.F.) = intQ(I.F.)dx` `:.`Our solution will be, `ysinx = int sinx(2cosx)dx` `=>ysinx = int sin2xdx` `=>ysinx = -cos(2x)/2+c` At `y = 0 and x = pi/2`, equation becomes `0 = -cospi/2 +c => c = -1/2` So, solution will be, `ysinx = -(cos2x)/2-1/2` `=>2ysinx+cos2x+1 = 0` |
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