Integrating Factors

On may occasions,

$$M(x, y) + N(x, y) \frac{dy}{dx} = 0, \; {\mbox{ or equivalently }} \; M(x, y) dx + N(x, y) dy = 0$$

may not be exact. But the above equation may become exact, if we multiply it by a proper factor. For example, the equation

$$y dx - dy = 0$$

is not exact. But, if we multiply it with $e^{-x},$ then the equation reduces to

$$e^{-x} y dx - e^{-x} d y = 0, \; {\mbox{ or equivalently }} \; d\left( e^{-x} y \right) = 0,$$

an exact equation. Such a factor (in this case, $e^{-x}$ ) is called an INTEGRATING FACTOR for the given equation. Formally

DEFINITION 7.3.6 (Integrating Factor)   A function $Q(x,y)$ is called an integrating factor for the (7.3.1), if the equation

$$Q(x,y) M(x,y) dx + Q(x,y) N(x,y) dy = 0$$

is exact.

EXAMPLE 7.3.7  

1. Solve the equation $y dx - x dy = 0, \;\; x, y > 0.$
   Solution: It can be easily verified that the given equation is not exact. Multiplying by $\frac{1}{xy},$ the equation reduces to
   $$\frac{1}{x y} y dx - \frac{1}{x y} x dy = 0, \; {\mbox{ or equivalently }} \; d \left(\ln x - \ln y \right) = 0.$$
   Thus, by definition, $$\frac{1}{x y}$$ is an integrating factor. Hence, a general solution of the given equation is $G(x, y ) = \displaystyle\frac{1}{ x y } = c,$ for some constant $c \in {\mathbb{R}}.$ That is,
   $$y = c x, \; {\mbox{ for some constant }} c \in {\mathbb{R}}.$$

2. Find a general solution of the differential equation
   $$\bigl(4 y^2 + 3 x y \bigr) dx - \bigl( 3 x y + 2 x^2 \bigr) d y = 0.$$
   
   Solution: It can be easily verified that the given equation is not exact.

   METHOD 1: Here the terms $M = 4 y^2 + 3 x y$ and $N = - ( 3 x y + 2 x^2)$ are homogeneous functions of degree $2.$ It may be checked that an integrating factor for the given differential equation is

   $$\frac{1}{M x + N y} = \frac{1}{ x y \bigl( x + y \bigr)}.$$
   Hence, we need to solve the partial differential equations
   

   $$\frac{\partial G(x,y)}{\partial x} = \frac{y \bigl( 4 y + 3 x \bigr)}{xy \bigl( x+y \bigr)} = \frac{4}{x} - \frac{1}{x+y} \; {\mbox{ and }}$$ (7.3.5)

   $$\frac{\partial G(x,y)}{\partial y} = \frac{ - x( 3 y + 2 x)}{xy \bigl(x+y \bigr)} = -\frac{2}{y} - \frac{1}{x+y}.$$ (7.3.6)

   
   Integrating (keeping $y$ constant) (7.3.5), we have

   $$G(x,y) = 4 \ln\vert x\vert - \ln \vert x+y\vert + h(y)$$ (7.3.7)

   
   
   and integrating (keeping $x$ constant) (7.3.6), we get

   $$G(x,y) = -2 \ln\vert y\vert - \ln \vert x+y\vert + g(x).$$ (7.3.8)

   
   
   Comparing (7.3.7) and (7.3.8), the required solution is
   $$G(x,y) = 4 \ln\vert x\vert - \ln\vert x+y\vert - 2 \ln\vert y\vert = \ln c$$
   for some real constant $c.$ Or equivalently, the solution is
   $$x^4 = c \bigl(x+y \bigr) y^2.$$

   METHOD 2: Here the terms $M = 4 y^2 + 3 x y$ and $N = - ( 3 x y + 2 x^2)$ are polynomial in $x$ and $y.$ Therefore, we suppose that $x^{\alpha} y^{\beta}$ is an integrating factor for some $\alpha, \beta \in {\mathbb{R}}.$ We try to find this $\alpha$ and $\beta.$

   Multiplying the terms $M(x,y)$ and $N(x,y)$ with $x^{\alpha} y^{\beta},$ we get

   $$M(x,y) = x^{\alpha} y^{\beta} \bigl(4 y^2 + 3 x y \bigr), \; {\mbox{ and }} N(x, y) = - x^{\alpha} y^{\beta} (3 x y + 2 x^2).$$
   For the new equation to be exact, we need $$\frac{\partial M(x, y)}{\partial y} = \frac{\partial N(x, y)}{\partial x}.$$ That is, the terms
   $$4(2 + \beta) x^{\alpha} y^{1 + \beta} + 3 (1 + \beta) x^{1 + \alpha} y^{\beta}$$
   and
   $$-3(1+ \alpha) x^{\alpha} y^{1 + \beta} - 2 (2 + \alpha) x^{1 + \alpha} y^{\beta}$$
   must be equal. Solving for $\alpha$ and $\beta,$ we get $\alpha = -5$ and $\beta = 1.$ That is, the expression $$\frac{y}{x^5}$$ is also an integrating factor for the given differential equation. This integrating factor leads to
   $$G(x,y) = - \frac{y^3}{x^4} - \frac{y^2}{x^3} + h(y)$$
   and
   $$G(x,y) = - \frac{y^3}{x^4} - \frac{y^2}{x^3} + g(x).$$
   Thus, we need $h(y) = g(x) = c,$ for some constant $c \in {\mathbb{R}}.$ Hence, the required solution by this method is
   $$y^2 \bigl( y + x \bigr) = c x^4.$$

Remark 7.3.8  

1. If (7.3.1) has a general solution, then it can be shown that (7.3.1) admits an integrating factor.
2. If (7.3.1) has an integrating factor, then it has many (in fact infinitely many) integrating factors.
3. Given (7.3.1), whether or not it has an integrating factor, is a tough question to settle.
4. In some cases, we use the following rules to find the integrating factors.
   1. Consider a homogeneous equation $M(x,y) dx + N(x,y) dy = 0.$ If
      $$M x + N y \neq 0,\;\; {\mbox{ then }} \;\; \frac{1}{M x + N y}$$
      is an Integrating Factor.
   2. If the functions $M(x,y)$ and $N(x,y)$ are polynomial functions in $x, y;$ then $x^{\alpha} y^{\beta}$ works as an integrating factor for some appropriate values of $\alpha$ and $\beta.$
   3. The equation $M(x,y) dx + N(x,y) dy = 0$ has $e^{\int f(x) dx}$ as an integrating factor, if $f(x) = \displaystyle \frac{1}{N}\left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right)$ is a function of $x$ alone.
   4. The equation $M(x,y) dx + N(x,y) dy = 0$ has $e^{-\int g(y) dy}$ as an integrating factor, if $g(y) = \displaystyle \frac{1}{M}\left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right)$ is a function of $y$ alone.
   5. For the equation
      $$y M_1(x y) dx + x N_1(x y) dy = 0$$
      with $M x - N y \neq 0,$ the function $$\frac{1}{M x - N y}$$ is an integrating factor.

EXERCISE 7.3.9  

1. Show that the following equations are exact and hence solve them.
   1. $(r + \sin \theta + \cos \theta) \displaystyle\frac{dr}{d\theta} + r( \cos \theta - \sin \theta) = 0.$
   2. $(e^{-x} - \ln y + \displaystyle\frac{y}{x}) + ( -\frac{x}{y} + \ln x + \cos y) \frac{dy}{dx} = 0.$
2. Find conditions on the function $g(x,y)$ so that the equation
   $$(x^2 + x y^2) + \{ a x^2 y^2 + g(x,y) \} \frac{dy}{dx} = 0$$
   is exact.
3. What are the conditions on $f(x),\; g(y), \;\phi(x),$ and $\psi(y)$ so that the equation
   $$(\phi(x) + \psi(y) ) + (f(x) + g(y))\frac{dy}{dx} = 0$$
   is exact.
4. Verify that the following equations are not exact. Further find suitable integrating factors to solve them.
   1. $y + (x + x^3 y^2 ) \displaystyle\frac{dy}{dx} = 0.$
   2. $y^2 + (3 x y + y^2 -1 ) \displaystyle\frac{dy}{dx} = 0.$
   3. $y + (x + x^3 y^2 ) \displaystyle\frac{dy}{dx} = 0.$
   4. $y^2 + (3 x y + y^2 -1 ) \displaystyle\frac{dy}{dx} = 0.$
5. Find the solution of
   1. $(x^2 y + 2 x y^2) + 2 (x^3 + 3 x^2 y) \displaystyle\frac{dy}{dx} = 0$ with $y(1) = 0.$
   2. $y(x y + 2 x^2 y^2) + x (x y - x^2 y^2) \displaystyle\frac{dy}{dx} = 0$ with $y(1) = 1.$