Differential Equations

Integrating Factors<!-- --> | Differential Equations

Integrating Factors

Sharan Sajiv Menon - October 12th, 2021

Sometimes, there are differential equations that arent exact, but they can be made exact by multiplying the function by something known as an integrating factor.

Example of what integrating factors can do

Here is an example of a function that isn't exact. (3x+2y2)dx+(2xy)dy=0(3x+2y^2)dx + (2xy)dy=0.

When you take the partial derivatives MyM_y and NxN_x, you get My=4yM_y = 4y and Nx=2yN_x=2y. And since My≠NxM_y \ne N_x, this is not a exact equation. That means the equation can't be solved.

However, by multiplying the equation by xx, yields (3x2+2xy2)dx+(2x2y)dy=0(3x^2 + 2xy^2)dx + (2x^2y)dy=0. Taking the partial derivatives we get My=Nx=4xyM_y = N_x = 4xy. This is an exact equation and can be easily solved.

A function μ(x,y)\mu(x, y) is an integrating factor if μ(x,y)M(x,y)dx+μ(x,y)N(x,y)dy=0\mu(x, y)M(x,y)dx + \mu(x, y)N(x,y)dy = 0 (eq 1)

Finding Integrating factors

By taking the concept discussed in the Exact Equations section, eq 1 is an exact equation and that means

∂∂y(μM)=∂∂x(μN)\frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N)

Solving this yields the following equation.

μyM+Myμ=μxN+Nxμ\mu_yM + M_y\mu = \mu_xN + N_x\mu

This can be rewritten as the following equation below

μ(My−Nx)=μxN−μyN\mu(M_y-N_x)=\mu_xN - \mu_yN

If My=NxM_y = N_x, then μ=1\mu=1, making the equation exact.

If μ(x,y)=P(x)Q(y)\mu(x, y)=P(x)Q(y), then μx(x,y)=P′(x)Q(y)\mu_x(x, y)=P'(x)Q(y) and μy(x,y)=P(x)Q′(y)\mu_y(x, y)=P(x)Q'(y). Replace this into the equation to get.

P(x)Q(y)My−Nx)=P′(x)Q(y)N−P(x)Q′(y)NP(x)Q(y)M_y-N_x)=P'(x)Q(y)N - P(x)Q'(y)N

This simplifies into the following equation below.

My−Nx=P′(x)P(x)N−Q′(y)Q(y)MM_y - N_x=\frac{P'(x)}{P(x)}N - \frac{Q'(y)}{Q(y)}M

Replace P′(x)P(x)=p(x)\frac{P'(x)}{P(x)}=p(x) and Q′(y)Q(y)=q(y)\frac{Q'(y)}{Q(y)}=q(y) and the equation now becomes the following equation down below.

My−Nx=p(x)N−q(y)MM_y - N_x=p(x)N - q(y)M

We can use this to obtain the following idea below.

Theorem If p(x)=My−NxNp(x)=\frac{M_y-N_x}{N}, then μ(x)=±e∫p(x)dx\mu(x)=\pm e^{\int p(x) dx} If q(y)=Nx−MyMq(y)=\frac{N_x - M_y}{M}, then μ(y)=±e∫q(y)dy\mu(y)=\pm e^{\int q(y) dy}

Examples

Example 1: Find the integrating factor and solve the following Differential equation

(2xy3−2x3y3−4xy2+2x)dx+(3x2y2+4y)dy=0(2xy^3-2x^3y^3-4xy^2+2x)dx + (3x^2y^2+4y)dy =0

Solution: We need to find a integrating factor to turn this into an easily solvable exact equation.

  • M(x,y)=2xy3−2x3y3−4xy2+2xM(x, y)=2xy^3-2x^3y^3-4xy^2+2x
  • N(x,y)=3x2y2+4yN(x, y)=3x^2y^2+4y
  • My=6xy2−6x3y2−8xyM_y = 6xy^2-6x^3y^2-8xy
  • Nx=6xy2N_x = 6xy^2
  • My−Nx=−6x3y2−8xyM_y - N_x = -6x^3y^2-8xy

You can see My≠NxM_y \ne N_x, therefore this is not an exact equation.

To find the integrating factor, we use the theorem mentioned above.

p(x)=My−NxN=−6x3y2−8xy3x2y2+4y=−2xp(x)=\frac{M_y-N_x}{N} = \frac{-6x^3y^2-8xy}{3x^2y^2+4y}=-2x

We can find μ\mu our integrating factor from p(x)p(x).

μ(x)=e∫−2xdx=e−x2\mu(x)=e^{\int -2xdx} = e^{-x^2}

Multiply this into our original equation to get an exact equation

e−x2(2xy3−2x3y3−4xy2+2x)dx+e−x2(3x2y2+4y)dy=0 e^{-x^2}(2xy^3-2x^3y^3-4xy^2+2x)dx + \\e^{-x^2} (3x^2y^2+4y)dy =0

To solve the equation, we had to find a function F(x,y)F(x,y) where it satisfied the following conditions

  1. Fx=e−x2(2xy3−2x3y3−4xy2+2x)F_x= e^{-x^2}(2xy^3-2x^3y^3-4xy^2+2x)
  2. Fy=e−x2(3x2y2+4y)F_y=e^{-x^2} (3x^2y^2+4y)

We can integrate either equation to find FF, we will integrate the FyF_y with respect to y since it is simpler

∫e−x2(3x2y2+4y)dy=e−x2∫(3x2y2+4y)dy=e−x2(x2y3+2y2)+Ψ(x)\int e^{-x^2} (3x^2y^2+4y) dy= e^{-x^2} \int(3x^2y^2+4y) dy \\= e^{-x^2}(x^2y^3 + 2y^2) + \Psi(x)

So, F(x,y)=−e−x2(x2y3+2y2)+Ψ(x)F(x, y) = -e^{-x^2}(x^2y^3 + 2y^2) + \Psi(x). To find Ψ(x)\Psi(x), we just take FxF_x

Fx=−e−x2(2xy3−2x3y3−4xy2)+Ψ′(x)F_x=-e^{-x^2}(2xy^3-2x^3y^3-4xy^2) + \Psi'(x). Setting this equal to e−x2(2xy3−2x3y3−4xy2+2x)e^{-x^2}(2xy^3-2x^3y^3-4xy^2+2x) we find that Ψ′(x)=2xe−x2\Psi'(x) = 2xe^{-x^2} and Ψ(x)=−e−x2\Psi(x) = -e^{-x^2}

Our final answer then is C=e−x2(x2y3+2y2−1)C= e^{-x^2}(x^2y^3 + 2y^2 -1)

Example 2: Coming soon... Solution: Coming soon...

Created by Sharan Sajiv Menon, © 2022