2.6 Exact DE

What is an exact DE?

• Exact DEs are DEs that can be written as a derivitive of a function

• Definition

  • $M(x,y)+N(x,y)\frac{dy}{dx}=0\text{ OR }M(x,y)dx+n(x,y)dy=0$

    • is called exact if there is a funciton $\psi (x,y)$ in the variables x and y such that
      • $\frac{\partial \psi }{\partial x}=M(x,y)$
      • $\frac{\partial \psi }{\partial y}=N(x,y)$

• Solving exact DEs

  • start with the function $M(x,y)$. since $\frac{\partial \psi }{\partial x}=M(x,y)$ intigrate $M(x,y)$ with repsect to x to obtian $\psi (x,y)$
  • the constant of integration in the previous step will be an unknow function of y. write this constant as h(y)
  • the equation $\frac{\partial \psi }{\partial y}=N(x,y)$ will allow us to compute the function h(y).
  • then

• Determining exactness

  • Let the function M,N my, and nx be continuous in rectangular region $\alpha <x<\beta$and $\gamma <y<\delta$. then there iextists a function $\psi (x,y)$ satisfying

    • $M_y(x,y)=N_x(x,y)$

• Converting non exact to exact

  • When a DE $M(x,y)+N(x,y)$

• Important note Mx is the function that isnt being multiplied by $\frac{dy}{dx}$ Nx is the one being multiplied by $\frac{dy}{dx}$

• criteara for the integrating factor to depend only on x or y:

  • 1. if $\frac{M_y-N_x}{N}=Q(x)$ depends only on x then the equation $M(x,y)+N(x,y)\frac{dy}{dx}$ becomes exact after multiplynig the equation by the integrating factor $\mu (n)=e^{\int Q(x)}dx$

Examples

Find the general solution for $2xy+(x^2+y^2)\frac{dy}{dx}=0$

• This is not seprable!!! so our previous methods will not work but we can see that the left hand side is the derivitive of

  • $\frac{d}{dx}\left(x^{2}y+\frac{y^3}{3}\right)=0$

    • how we found this ??

    • How did we find that this is an exact DE?

      • \begin{align} \frac{d}{dx}(\psi(x,y))=0\end{align}
        • $\frac{d\psi }{dx}=\frac{\partial \psi }{\partial x}+\frac{\partial \psi }{\partial y}\cdot \frac{dy}{dx}$
        • this is found from the chain rule for funcitons in multiple varibles
          • we know that y is a function of x, so the composite function thas both y and x, is just a function of x

    • How did we find that this is the anti derivitive of the function

  • this is just the general solution because we can cancel out the d/dx
    • $x^{2}y+\frac{y^3}{3}=C$

Find the general solution for the DE: $1+\ln x+2x\ln y+\left(\frac{x^2}{y}-2y\right)\frac{dy}{dx}$

• First is the DE is exact?

  • yes My=Nx (2x=2x)
• apply thereom

Show that the following ODE is not exact $\frac{1}{3}{y}^3+e^{-x}+y^2\frac{dy}{dx}=0$

• My=y^2 Nx=0

• find the integrating factor that makes the equation ecat
  • assume $\mu (x,y)=\mu (n)$ depends only on x.

\mu(x)\left( \frac{1}{3}y^{3}+e^{-x} \right)+(\mu)(x)y^{2}{\frac{dy}{dx}}=0 \\ My=Nx\to \mu(x)(y^{2})=\mu'(x)(y^{2})\implies \mu(x)=\mu'(x) \therefore \mu(x)=e^{x} \end{align}$$