# Recitation 1

Definition An ordinary differential equation (ODE) relates a function with its derivatives and a single (independent) variable.

Examples (1) $f'(x) = 0$, (2) $\frac{dy}{dt}=y$, (3) $\frac{dy}{dx}=y$. In the 2nd example, $y$ is seen as a function of $t$, whereas in the 3rd example, $y$ is a function of $x$.

### Slope Diagram

In the last lecture, you have seen the slope diagram (also known as slope field) of the ODE $y'=ay+b$. Slope diagram helps us visualize the actual solution to the ODE.

Problem 1 Draw the slope field for the ODE $y' = \frac{-x}{y}$.

Solution See the excellent explanation on Khan Academy.

### Ordinary Differential Equation $y'=ay+b$

Problem 2 Solve $y' = 2y + 3$.

Solution Rewrite the original ODE as $\frac{dy}{dx} = 2y+3$. Transform it to $\frac{dy}{2y+3} = dx$. Integrate both sides: $\frac{1}{2}\ln |2y+3| = x + C$, and so $2y+3 = \pm e^{2x+2C}=\pm e^{2C}e^{2x}$. Denote $\pm e^{2C}$ by a new constant $A$, we obtain $2y+3=Ae^{2x}$, thus $y=\frac{Ae^{2x}-3}{2}$.

Remark The general formula for $y' = ay+b$, where $a, b$ are constants, is $y = \frac{Ae^{ax}-b}{a}$.

### Initial Conditions and Initial Value Problem

A differential equation often has a family of solutions (for example, in the solution to Problem 2, one has the freedom to choose $A$), so to specify a unique solution, we usually need other data in addition (known as the initial conditions) to the differential equation.

The problem of finding the unique solution, given initial conditions, is sometimes called an initial value problem (IVP).

Problem 3 Solve the initial value problem $y' = 3y+5, y(0) = -1$.

Solution Using the general formula in the remark above, $y=\frac{Ae^{3x}-5}{3}$. By the initial condition, $-1=y(0)=\frac{Ae^{3\cdot 0}-5}{3}=\frac{A-5}{3}$, so $A=2$. Therefore the solution to the IVP is $y=\frac{2e^{3x}-5}{3}$.

Warnings

1. The initial value doesn’t necessarily have to just be $y$-values. Higher-order equations might have an initial value for both $y$ and $y'$, for example.
2. An initial value problem doesn’t always have a unique solution. It’s possible for an initial value problem to have multiple solutions, or even no solution at all.