#math

Exact Differential

• Implying, that if we have a differential equation $$M(x,y)\dd x + N(x,y)\dd y = 0$$and $$\frac{\partial }{\partial y}M(x,y) = \frac{\partial}{\partial x}N(x,y)$$then the differential equation is exact.
• Solution of this type is obviously $F(x,y)=c$.

Integrating factors

• When the ODE is not exact, but can be made exact by multiplying with an arbitrary function $\mu (x,y)$, then $\mu$ is called the integrating factor of the differential equation
• We claim that the solution set of both these equations are "essentially same", but it might result in
  • Loss of one or more solutions of the original
  • Gain of one or more new solutions which are not solutions of the original
  • Or, both of these can happen simultaneously.

Finding integrating factors

• If $$\frac{1}{N}\left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial y} \right) = p(x)$$with $p(x)$ being a function of $x$ only, then the integrating factor is

$$e^{\int p(x)dx}$$

• A first order linear equation of the fashion $$y' +p(x)y = r(x)$$has an integrating factor of $e^{\int p(x)dx}$, thus the solution being $$e^{\int p(x)dx}\times y = \int^xr(s)e^{\int p(s)ds}ds + C$$
• In a similar fashion if $$-\frac{1}{M}\left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) = g(y)$$then integrating factor is

$$e^{\int g(y)dy}$$

• Once we find the integrating factor, we multiply and try to reduce to exact differential, or apply the method to solve exact differentials.

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