InterviewSolution
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InterviewSolution
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Show that `y^(2)dx + (xy + x^(2))dy = 0` is a homogeneous differential equation. Also find its general solution. |
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Answer» `(dy)/(dx) = -(y^(2))/(xy + x^(2))` Let `f(x,y) = -(y^(2))/(xy + x^(2))` Now `f(lambda x, lambda y) = -(lambda^(2) y^(2))/(lambda x.lambda y + lambda^(2) x^(2))` `= - lambda^(0)(y^(2))/(xy + x^(2))` `= lambda^(0)f(x, y)` `:.` The given differential equation is the homogeneous differential equation. Putting y = vx `rArr (dy)/(dx) = v + x(dv)/(dx)`, we get `v + x(dv)/(dx) = - (v^(2)x^(2))/(x.vx + x^(2))` `rArr x(dv)/(dx) = (-v^(2) - v^(2) - v)/(v+1)` `rArr (v+1)/(v(1+2v)) dv = -(dx)/(x)` Integrating both sides, we get `int ((1)/(v)-(1)/(1+2v))dv = -int (dx)/(x)` `rArr ln |v| - (1)/(2)ln|(1+2v)| = -ln|x| + lnc` `rArr (|vx|)/(sqrt(1+2v)) = c` `rArr xy^(2) = c^(2)(x+2y)` or `xy^(2) = k(x+2y)` |
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