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Solve `(x^(2) + 2xy + y^(2) + 1)(dy)/(dx) = 2(x+y)` |
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Answer» Observe that `x^(2) + 2xy + y^(2) = (x+y)^(2)` The equation becomes, `{(x+y)^(2)+1}(dy)/(dx) = 2(x+y)` Set x+ y = t, so that `1+(dy)/(dx) = (dt)/(dx)` (on differentiation w.r.t. x) Now, `(t^(2)+1)((dt)/(dx) - 1) = 2t` `rArr (dt)/(dx) = (2t)/(t^(2) + 1) +1 = (t^(2) + 2t +1)/(t^(2)+1) = (t^(2) + 2t + 1)/(t^(2) + 1) = ((t + 1)^(2))/(t^(2)+1)` `rArr (t^(2) + 1)/((t + 1)^(2))dt = dx` `rArr ((t+1)^(2)-2(t+1)+2)/((t+1)^(2)) dt = dx` `rArr dt - (2)/(t+1)dt + 2(dt)/((1+t)^(2)) = dx` Integrating, we obtain, `t-2ln |t+1| - (2)/(t-1) = x + k` `rArr x+y - 2ln|x+y+1| - (1)/(x+y+1) = x+k` `:. y- 2ln|x+y+1| - (2)/(x+y+1) = k`, k being the constant of integration |
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