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Exact ODE and Integrating Factors

Exact differential equation #

A differential equation of the form M(x,y)dx+N(x,y)dy=0 is said to be exact if its left hand member is the exact differential of some function Φ(x,y)

$$ \begin{align*} d\Phi=Mdx+Ndy\\ Also, d\Phi=\pd{\Phi}{x}d x+\pd{\Phi}{y}d y \end{align*} $$ So has to be the case that $\pd{\Phi}{x}=M$ and $\pd{\Phi}{y}=N$

Testing for exactness #

Theorem:

The necessary and sufficient condition for the differential equation Mdx+Ndy=0 to be exact is $$\pd{M}{y}=\pd{N}{x}$$

Proof: #

Necessary condition #

The necessary condition for the differential equation to be exact is: $\pd{M}{y}=\pd{N}{x}$

We need to prove that if differential equation is exact ⇒ ($\pd{M}{y}=\pd{N}{x}$)

$$ \begin{align*} Assume\\ Mdx+Ndy=0\\ is\\ exact\\ \exists \Phi(x,y)s.t. d\Phi=Mdx+Ndy=0\\ \therefore M=\pd{\Phi}{x}; N=\pd{\Phi}{y}\\ \Rightarrow \pd{M}{y}=\pd{N}{x}=\pd{^2\Phi}{xy} \end{align*} $$

Sufficient condition #

The sufficient condition for the differential condition to be exact is: $\pd{M}{y}=\pd{N}{x}$ We need to prove that if ($\pd{M}{y}=\pd{N}{x}$) ⇒ differential equation is exact

$$ \begin{align*} Let \int Mdx&= u\\ \pd{u}{x}&= M\\ Given:\pd{M}{y}&= \pd{N}{x}\\ \pd{^2u}{xy} &= \pd{N}{x}\\ N&= \pd{u}{dy}+f(y)\\ \\ -(1)\\ Now, Mdx+Ndy &= \pd{u}{x}dx+(\pd{u}{y}+f(y))dy\\ &= d\left(u+\int f(y)dy\right)=d\Phi \end{align*} $$ Shows that Mdx+Ndy=0 is an exact D.E.

Solution #

Find M and N, coeffs of dx, dy respectively

$$ \begin{align*} \Phi &= \int M dx+k(y)\\ \dd{\Phi}{y}&= \dd{\int M dx}{y}+k’(y)=N\\ k’(y)&= N-\dd{\int M dx}{y}\\ k(y)&= \int N-\dd{\int M dx}{y} dy \end{align*} $$

Quick solution #

Solution of the D.E. Mdx+Ndy=0 is

$$ \int M dx + \int (terms\\ of\\ N\\ not\\ containing\\ x)dy=0 $$

Finding Integrating factor #

Say we have a nonexact ODE $Mdx+Ndy=0$ where M and N are functions of x,y. Integrating factor of this ODE is a function F(x,y) where $MFdx+NFdy=0$ becomes an exact ODE.

Find I.F. by inspection #

$$ \begin{align*} xdy+ydx &= d(xy)\\ \frac{xdy-ydx}{x^{2}}&= d(\frac{y}{x})\\ \frac{xdy-ydx}{y^{2}} &= d(\frac{x}{y})\\ \frac{xdy-ydx}{xy}&= d(\ln(\frac{y}{x}))\\ \frac{xdy-ydx}{x^{2}+y^{2}} &= d(\arctan(\frac{y}{x})) \end{align*} $$

Exactness condition #

Say Mdx+Ndy=0 is non exact

$$ \begin{align*} \pd{MF}{y}=\pd{NF}{x}\\ M_yF+MF_y=N_xF+NF_x \end{align*} $$

If I.F. is a function of x #

$$ \begin{align*} M_yF=N_xF+NF_x\\ (M_y-N_x)F=NF_x\\ \frac{M_{y}-N_{x}}{N}=\frac{F_{x}}{F}=R\\ I.F.=F(x)=e^{\int R(x) dx} \end{align*} $$

If I.F. is a function of y #

$$ \begin{align*} M_yF+MF_y=N_xF\\ (M_y-N_x)F=-MF_y\\ \frac{M_{y}-N_{x}}{M}=\frac{-F_{y}}{F}=R\\ I.F.=F(y)=e^{-\int R(y)dy} \end{align*} $$

Special Cases #

If M, N are homogeneous functions of degree n #

$$ \begin{align*} I.F. &= \frac{1}{Mx+Ny}\\ if\\ denominator\\ =0: I.F.&= \frac{1}{xy}\\ or\\ \frac{1}{x^{2}}\\ or\\ \frac{1}{y^{2}} \end{align*} $$

If M=F₁(xy)y and N=F₂(xy)x #

$$ I.F.=\frac{1}{Mx-Ny} $$

If Mdx + Ndy can be put in the form $x^ay^b(mydx+nxdy)+x^cy^d(pydx+qxdy)=0$ #

$$ \begin{align*} I.F.=x^hy^k\\ obtained\\ by\\\ \frac{a+h+1}{m}=\frac{b+k+1}{n}\\ and\\ \frac{c+h+1}{p}=\frac{d+k+1}{q} \end{align*} $$