# Can any one solve This with intregrated factor method

• Oct 16th 2007, 06:58 AM
king imran
Can any one solve This with intregrated factor method
Please solve this with Intregrating factor method
• Oct 16th 2007, 07:55 AM
Krizalid
I assume that you already have taken both partial derivatives.

When we want to find a suitable integrating factor, sometimes it's not so easy, so we may introduce the following cases.

CASE I. Integrating factor dependent of $\displaystyle x.$ Suppose that

$\displaystyle \frac{{\dfrac{{\partial M}}{{\partial y}} - \dfrac{{\partial N}} {{\partial x}}}} {N}$

it's a function that depends only of $\displaystyle x,$ which we'll denote it by $\displaystyle g(x).$ Then, an integrating factor for the given equation is $\displaystyle \mu(x)=\exp\int g(x)\,dx.$

CASE II. Integrating factor dependent of $\displaystyle y.$ If we have that

$\displaystyle \frac{{\dfrac{{\partial N}}{{\partial x}} - \dfrac{{\partial M}}{{\partial y}}}}{M}$

it's only a function of $\displaystyle y,$ denoted by $\displaystyle h(y),$ then $\displaystyle \mu(y)=\exp\int h(y)\,dy$ is an integrating factor for the differential equation on the form $\displaystyle M(x,y)\,dx+N(x,y)\,dy=0.$

CASE III. Integrating factors of the form $\displaystyle x^my^n.$ If they exist $\displaystyle m$ and $\displaystyle n$ such that

$\displaystyle \frac{{\partial M}}{{\partial y}} - \frac{{\partial N}}{{\partial x}} = m\frac{N}{x} - n\frac{M}{y},$

then $\displaystyle \mu(x,y)=x^my^n$ is an integrating factor for the differential equation mentioned previously.

CASE IV. If they exist functions $\displaystyle P(x)$ y $\displaystyle Q(y)$ which satisfy

$\displaystyle \frac{{\partial M}}{{\partial y}} - \frac{{\partial N}}{{\partial x}} = N(x,y)P(x) - M(x,y)Q(y),$

then an integrating factor for the differential equation is $\displaystyle \mu(x,y)=\exp\int P(x)\,dx\cdot\exp\int Q(y)\,dy.$

Note that the case IV includes to the cases I, II y III if we take $\displaystyle Q(y)=0,\,P(x)=0$ and $\displaystyle P(x)=\frac mx,\,Q(y)=\frac ny;$ respectively.