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Solve the differential equation: `dy/dx + y cotx = 4x cosec x`, given that y=0 at x = `pi/2` |
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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) = 4xcosecx` 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 4xcosecxsinxdx` As `sinx cosecx = 1`, `=>ysinx = int 4xdx` `=>ysinx = 2x^2+c` At `y = 0 and x = pi/2`, equation becomes `0 = 2(pi/2)^2 +c => c = -pi^2/2` So, solution will be, `ysinx = 2x^2-(pi)^2/2` |
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