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Solve `(dy)/(dx) = (x+2y - 3)/(2x + y - 3)` |
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Answer» Here `a_(1) = 1, b_(1) = 2 " " :. (a_(1))/(b_(1)) = (1)/(2)` `a_(2) = 2, b_(2) = 1 " " :. (a_(2))/(b_(2)) = 2` Now, `(a_(1))/(b_(1)) != (a_(2))/(b_(2))` Thus case - I applies, we have setting, `{:(x=u+h),(y=v+k):}}` `(dv)/(du) = ((u+2v) + (h+2k-3))/((2u + v) + (2h + k-3))0` Choose h, k such that `{:(h+2k-3=0),(2h + k - 3 =0):}}` The solution is h = 1, k = 1 `rArr (dv)/(du) = (u+2v)/(2u + v)` Set `v = tu, rArr (dv)/(du) = t + (dt)/(du)` Our equation reduces to `t + u(dt)/(du) = (u+2tu)/(2u + tu) = (1+2t)/(2+t)` `rArr u(dt)/(du) = (1+2t)/(2+t) - t = (1+2t-2t-t^(2))/(2+t) = (1-t^(2))/(2+t)` `rArr (2+t)/(1-t^(2)) dt = (du)/(2)` `rArr (2)/(1-t^(2))dt+(1)/(2).2(t)/(1-t^(2)) dt = (du)/(u)` On integrating, we get `ln|(1+t)/(1-t)| - (1)/(2)ln|1-t^(2)| = ln|u|+ ln|k|`, k being the constant of integration. `rArr ln|((1+(v)/(u))/(1-(v)/(u)))|-(1)/(2)ln(1-(v^(2))/(u^(2)))| = ln(uk)|` `rArr |((u+v)/(u-v))| - ln|(sqrt(u^(2)-v^(2)))/(u)| = ln|(uk)|` `rArr ln|((u+v)/(u-v).(u)/(sqrt(u^(2)-v^(2))))| = ln|uk|` `rArr (usqrt(u+v))/((u-v)^((3)/(2))) = +- uk` `rArr k(u-v)^((3)/(2)) = +-sqrt(u+v)` `rArr k(x-1-y+1)^((3)/(2)) = sqrt(x-1+y-1)` `rArr k(x-y)^((3)/(2)) = sqrt(x+y-2)` `:. k^(2)(x-y)^(3) = (x+y-2)` `rArr (x-y)^(3) = lambda(x+y-2), lambda` being another constant. |
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