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The differential equations, find a particular solution satisfying the given condition: `(dx)/(dy)+2ytanx=sinx ; y=0`when `x=pi/3` |
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Answer» Comparing the given equation with first order differential equation, `dy/dx+Py = Q(x)`, we get,`P = 2tanx and Q(x) = sinx` So, Integrating factor `(I.F) = e^(int2tanxdx)` `I.F.= e^(2ln secx) = e^(ln sec^2x) = sec^2x` We know, solution of differential equation, `y(I.F.) = intQ(I.F.)dx` `:.`Our solution will be, `ysec^2x = int sinx(sec^2x)dx` `=>ysec^2x = int sinx/cos^2xdx` `=>ysec^2x = int tanx secx dx` `=>ysec^2x =secx+c` `=>y = cosx +c cos^2x` At `x = pi/3, y=0` `=>0 = 1/2 +c/4` `=> -2 = c` So, our solution will be, `=>y = cosx - 2cos^2x` `=>y = cosx(1-2cosx)`, which is the required solution. |
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