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An ordinary differential equation (ODE) contains one or more derivatives of an unknown function of one variable.

y=cosxy+9y=0x2yy+2exy4=(x2+2)y2\begin{align} &y' = \cos x \\ &y'' + 9y = 0 \\ &x^2 y''' y' + 2e^x y^4 = (x^2 + 2)y^2 \end{align}

The order of an ODE is its highest derivative. First-order ODEs in implicit form look like:

F(x,y,y)=0F(x, y, y') = 0

and in explicit form look like:

y=f(x,y)y' = f(x, y)

When do we see first-order ODEs in chemical engineering?

Direction fields

ODEs have families of solutions due to integration constants. The direction field (or slope field) is obtained by plotting the slope at (x, y) from y=f(x,y)y' = f(x, y), and it shows how a point will evolve.

Example: Direction field

Given the first-order ODE

y=xyy(0)=1y' = xy \quad y(0) = 1

Draw the slope field, then verify that

y=cex2/2y = c e^{x^2 / 2}

is a general solution.


The direction field is

Directional field diagram

Now, let’s verify the proposed solution. The first derivative of y is

y=cxex2/2y' = c x e^{x^2/2}

Substituting y and yy' in the differential equation

y=xycxex2/2=x(cex2/2)\begin{align} y' &= x y \\ c x e^{x^2 / 2} &= x (c e^{x^2 / 2}) \end{align}

shows this is indeed a solution.

Initial value problems

To obtain a specific solution, an initial condition specifying a point y(x0)=y0y(x_0) = y_0 is needed. A first-order ODE with an initial condition is called an initial value problem.

Example: Initial value problem

For the first-order ODE above, find the solution when y(0)=1y(0) = 1.


Substitute x=0x = 0 and y=1y = 1 into the general solution y=cex2/2y = c e^{-x^2/2} to determine cc:

1=y(0)=ce0=c1 = y(0) = c e^0 = c

This gives the solution

y=ex2/2y = e^{x^2/2}