#### Lessons

__Recall Partial Derivatives:__

Notation:

$\frac{dF}{dx}=F_x$ or

$\frac{dF}{dx}=F_y$

__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:

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

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

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

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

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

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

So our solution to solving the differential equation is just

$\Psi(x, y)=c$

Finding this $\Psi(x, y)$:

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

$\Psi_{xy}=\Psi_{yx}$

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

So,

$\Psi_{xy}= (\Psi_x)_y=(M)_y=M_y$
$\Psi_{yx}= (\Psi_y)_x=(N)_x=N_x$
$\Longrightarrow M_y=N_x$

So a differential equation will be exact if and only if $M_y=N_x$

Then we can solve for our $\Psi(x, y)$ by integrating $M$ with respect to $x$, or $N$ with respect to $y$.

And once we have $\Psi(x, y)$ we have solved the differential equation just equate:

$\Psi(x, y)=c$ and that will be the answerIntroduction

a)

Partial Differentiation Review

b)

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:

2.

**Determining what is an Exact Differential Equation**

Solve the following differential equation

$2xy^2-y^2+(2x^2 y-2xy) \frac{dy}{dx}=0$

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

3.

**Solving Exact Equations**

Solve the following differential equation

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

With initial conditions $y(0)=2$

4.

a)

Solve the following differential equation

$3x^2 y+2y^2=3x+1+y' (2-3y-4xy-x^3)$

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

b)

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