Integrating Factors

Integrating factors

Few equations that are not exact may be multiplied by some factor ,a function u(x,y),to make them exact.

When this function u(x,y)exists it is called an integrating factor.It will make true the following expression

$$\frac{\partial\;(u,N(x,y))}{\partial\;x}=\frac{\partial\;(u,M(x,y))}{\partial\;y}$$

There are some important cases:

1. $$u(x,y)=x^my^n$$
2. $$u(x,y)=u(x)\;\text{(That is, u is a function only of x)}$$
3. $$u(x,y)=u(y)\;\text{(That is, u is a function only of y)}$$

Let's move forward to these cases:

1. $$u(x,y)\;=x^my^n$$

   Example

   $$(y^2+3xy^3)dx+(1-xy)dy=0$$

   Solution

   $$M=(y^3+3xy^3)$$

   $$\frac{\partial\;M}{\partial\;y}=2y-9xy^2$$

   $$N=(1-xy)$$

   $$\frac{\partial\;M}{\partial\;y}=-y$$

   $$\frac{\partial\;M}{\partial\;y}\;\neq\;\frac{\partial\;N}{\partial\;x}$$

   Fact:bSo,it is not exact we try to make it exact by taking product of each part of equation by

   So,we have:

   $$(x^my^ny^2+x^my^n3xy^3)dx+(x^my^n-x^my^nxy)dy=0$$

   Which is solved as:

   $$(x^my^{n+2}+3x^{m+1}y^{n+3})dx+(x^my^n-x^{m+1}y^{n+1}x)dy=0$$

   $$M=x^my^{n+2}+3x^{m+1}y^{n+1}$$

   $$N=x^my^n-x^{m+1}y^{n+1}$$

   $$\frac{\partial\;M}{\partial\;y}=(n+2)x^my^{n+1}+3(n+3)x^{m+1}y^{n+2}$$

   $$\frac{\partial\;N}{\partial\;x}=mx^{m-1}y^n-(m+1)x^my^{n+1}$$

   Need:

   $$\frac{\partial\;M}{\partial\;y}=\frac{\partial\;N}{\partial\;x}$$

   Let us choose specific value of M and N to make given equation exact:

   Write as:

   $$(n+2)x^my^{n+1}+3(n+3)x^{m+1}y^{n+2}=mx^{m-1}y^n-(m+1)x^my^{n+1}$$

   Rearrange and solve:

   $$[(m+1)+(n+2)]x^my^{n+1}+3(n+2)x^{m+1}y^{n+2}-mx^{m-1}y^n=0$$

   Now,for it to be zero then there must be every coefficient of equation equals to 0:

   • (m+1)+(n+2)=
   • 3(n+3)=0
   • m=0

   The last point m=0 is very helpful after putting m=0 and then we have n=-3 Then,

   $$x^my^n=y^{-3}$$

   Multiply our real differential equation by y^-3:

   $$u(x)=e^{\int\;z(x)dx}$$

   and

   $$M=y^{-1}+3x$$

   $$N=y^{-3}-xy^{-2}$$

   $$\frac{\partial\;M}{\partial\;y}=\frac{\partial\;N}{\partial\;x}=-y^{-2}$$

   Thus,it is exact further you can solve it...

2. $$u(x,y)=u(x)$$

   For

   $$u(x,y)=u(x)$$

   We must check for important condition:

   Expression:

   $$Z(x)=\frac{[M_y-N_x]}{N}$$

   Fact: It must not contain y term,then the integrating factor is a function of x only.

   If above condition holds then our integrating factor

   $$u(x)=e^{\int\;z(x)dx}$$

   Example:

   $$(3xy-y^2)dx+x(x-y)dy=0$$

   Solution:

   You can check it and it is not exact as :

   $$\frac{\partial\;M}{\partial\;y}\neq\frac{\partial\;N}{\partial\;x}$$

   Lets try for Z(x):

   $$z(x)=\frac{[M_y-N_x]}{N}$$

   $$\frac{1}{N}[3x-2y-2x-(2x-y)]$$

   $$\frac{x-y}{x(x-y)}$$

   $$\frac{1}{x}$$

   Thus,Z(x) is a function of x:

   So, our integrating factors:

   $$u(x)=e^{\int\;z(x)dx}$$

   $$u(x)=e^{\int(\frac{1}{x})dx}$$

   $$u(x)=e^{\int\;lnx}$$

   $$u(x)=x$$

   So,we have found our integrating factor then multiply our differential equation by “x” We have:

   $$(3x^2y-y^2)dx+(x^3-x^2y)dy=0$$

   Now, you can check it is exact as:

   $$\frac{\partial\;M}{\partial\;y}\neq\frac{\partial\;N}{\partial\;x}$$

   Further you can solve it...

3. $$u(x,y)=u(y)$$

   For,

   $$u(x,y)=u(y)$$

   We must check for important condition:

   Expression:

   $$z(y)=\frac{[N_x-M_y]}{M}$$

   Fact: If above condition holds then our integrating factor:

   $$u(y)=e^{\int\;z(y)dy}$$

   Then,I hope that you can solve any type of example related to this case...