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Find the general solution of the differential equations:`(x+3y^2)(dx)/(dy)=y(y >0)` |
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Answer» Here, given equation is, `(x+3y^2)dy/dx = y` `=>dx/dy = (x+3y^2)/y` `=>dx/dy -x/y = 3y` Comparing it with `dx/dy +Px = Q` `P = -1/y and Q = 3y` Integrating factor, `I.F. = e^(intPdy)` `I.F. = e^(int-1/ydy) = e^-lny = y^-lne = y-1 = 1/y` Now, general solution will be, `x(I.F.) = int(I.F.)Qdy` `=>x/y = int(1/y)(3y)dy` `=>x/y = 3y+c` `=>x = 3y^2+cy` |
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