Exact differential equations

Exact differential equations

Basic concepts: Implicit differentiation,


Recall Partial Derivatives:


dFdx=Fx\frac{dF}{dx}=F_x or

Exact Differential Equations:
Sometimes a differential equation is not solvable via the separable equations technique.

Let’s suppose the differential equation we want to solve is of the form:

M(x,y)+N(x,y)dydx=0M(x,y)+N(x,y) \frac{dy}{dx}=0

Now let’s imagine we have another function of xx and yy, denoted as Ψ(x,y)\Psi (x,y) “psi”.

It is demonstrated throughout the video that: ddxΨ(x,y)=Ψx+Ψydydx\frac{d}{dx} \Psi(x,y)=\frac{\partial\Psi}{\partial x}+\frac{\partial\Psi}{\partial y}\frac{dy}{dx}

So if we can find a Ψ(x,y)\Psi(x,y) such that Ψx=M(x,y)\Psi_x=M(x,y) and Ψy=N(x,y)\Psi_y=N(x,y), then

ddxΨ(x,y)=Ψx+Ψydydx=\frac{d}{dx} \Psi(x,y)=\frac{\partial\Psi}{\partial x}+\frac{\partial\Psi}{\partial y}\frac{dy}{dx}= Ψx+Ψydydx=\Psi_x+\Psi_y \frac{dy}{dx}=M(x,y)+N(x,y)dydx=0M(x,y)+N(x,y) \frac{dy}{dx}=0

Given that ΨxdΨdx\frac{\partial \Psi}{\partial x} \neq \frac{d \Psi}{dx}

So our solution to solving the differential equation is just

Ψ(x,y)=c\Psi(x, y)=c

Finding this Ψ(x,y)\Psi(x, y):

Ok, so provided that Ψ(x,y)\Psi(x, y) is continuous and it’s first derivative is continuous as well, we should have the following equality hold


(Remember this from Calculus 3, if not just trust me!)


Ψxy=(Ψx)y=(M)y=My\Psi_{xy}= (\Psi_x)_y=(M)_y=M_y
Ψyx=(Ψy)x=(N)x=Nx\Psi_{yx}= (\Psi_y)_x=(N)_x=N_x
My=Nx\Longrightarrow M_y=N_x

So a differential equation will be exact if and only if My=NxM_y=N_x

Then we can solve for our Ψ(x,y)\Psi(x, y) by integrating MM with respect to xx, or NN with respect to yy.

And once we have Ψ(x,y)\Psi(x, y) we have solved the differential equation just equate:
Ψ(x,y)=c\Psi(x, y)=c and that will be the answer
  • Introduction
    Partial Differentiation Review

    What are Exact Differential Equations and how do we solve them?

  • 1.
    Partial Differentiation
    Find the first order partial derivative of every variable for the following functions:
    f(x,y)=3x23y+2 f(x,y)=3x^2-3y+2

    N(y,z)=2y2z N(y,z)=2y^2 z

  • 2.
    Determining what is an Exact Differential Equation
    Solve the following differential equation

    2xy2y2+(2x2y2xy)dydx=02xy^2-y^2+(2x^2 y-2xy) \frac{dy}{dx}=0

    Using the equation: Ψ(x,y)=x2y2xy2=2\Psi(x,y)=x^2 y^2-xy^2=2

  • 3.
    Solving Exact Equations
    Solve the following differential equation

    3exy12y2+(3exxy)dydx=03e^x y- \frac{1}{2} y^2+(3e^x-xy) \frac{dy}{dx}=0

    With initial conditions y(0)=2y(0)=2

  • 4.
    Solve the following differential equation

    3x2y+2y2=3x+1+y(23y4xyx3)3x^2 y+2y^2=3x+1+y' (2-3y-4xy-x^3)

    With initial conditions y=1,x=1y=1, x=1

    Verify that the solution you found is in fact the solution to the above differential equation (may be useful to do on a test).