btech m2 unit-1|non exact differential equ|methods of non exact|methods 2| y(1+xy)dx+x(1-xy)dy=0

Описание:    • btech m2 unit-1,2,3,4,5 imp Questions  


How do I solve the non-exact differential equation y (1+xy) dx+x (1-xy) dy=0?


The given first order differential equation is of the form
M(x,y)dx + N(x,y)dy = 0. . . . . (1)
Since both the terms contain (xy), the following result tests whether (1) has an integrating factor which is a function u(xy) of xy and also finds it.
Theorem: The equation (1) has an integrating function u(x,y) of (xy) alone, if and only if
(M_y -N_x)/(yN - xM) = f(xy) of the product xy alone. And if the above holds, then u(x,y) = exp(Int[f(t)dt]), (with t=xy), will be an integrating factor. (Here M_y denotes the partial derivative of M(x,y) with respect to y, etc.)
For the present question, M(x,y) = y(1+xy) and N(x,y) = x(1-xy). Therefore
(M_y - N_x)/(yN-xM) =[(1+2xy)-(1–2xy)]/[xy(1-xy)-xy(1+xy)] = 4xy/[-2(xy)^2] = (-2/xy), which is indeed a function (-2/t) of t=xy alone.
Hence we take u(t) = exp(Int[(-2/t)dt] = 1/t^2. Thus 1/(xy)^2 is an integrating factor of (1). Multiplying (1) by 1/(xy)^2, we get
[(1/y.x^2)+1/x]dx + [(1/x.y^2)-1/y]dy = 0, . . . . .(2). If we write it as P(x,y)dx+Q(x,y)dy = 0, then P_y = -1/x^2.y^2 = Q_x and so (2) is indeed now an exact equation.
To solve the exact equation (2), put Int[P(x,y)dx] = G(x,y), where the integration w.r.t. x, is carried out treating y as a constant.
Hence G(x,y) = -1/(xy) +ln(x). Next Q(x,y) - G_y = (1/x.y^2 -1/y) - (1/x.y^2) = (-1/y) = A(y) say. Put F(x,y) = G(x,y) + Int[A(y)dy] = -1/(xy) +ln(x) +Int[(-1/y)dy] = -1/(xy)+ln(x)-ln(y) = ln(x/y)-1/(xy).
Accordingly F(x,y) = c or ln(x/y) - 1/(xy) = c is a general solution of the given equation wherever x and y are both nonzero.
The method is a particular case of a general class of methods for finding integrating factors of the form t(x,y) of special types of functions like x, y, x^n + y^n, xy, y/x etc