An initial value problem (IVP) is the problem of solving an ODE subject to some initial value problem initial conditions of the form $y(t_0)=a,$ $y'(t_0)=b,$ etc.

The solution to an initial value problem no longer contains arbitrary constants from the general solution to the ODE - these are determined by the initial conditions of the problem at hand.

Example: A flow-meter in a pipeline measures flow-through as $2+\sin t $ litres/second. How much fluid passes through the pipeline from time zero up to time $T$?

The order of the highest-order derivative in an ODE defines the order of the ODE.

Exercise: What is the order of each of these ODEs?

• $\ds \dif{P}{t} = kP $                                - 1st order
• $ m\ddot{x} = -kx $                               - 2nd order
• $ x(y'')^2+y'y'''+4y^5 = yy' $   - 3rd order
• $ y' = xy $                                    - 1st order

Find the position of your bike (at time $t$) if you are travelling at a constant speed of 60 km/h along a perfectly straight road.

If $x=x(t)$ is the distance you have travelled at time $t$ then the corresponding ODE is \[ \dif{x}{t}=60. \qquad \qquad (*) \]

We can directly integrate to find $x(t):$

We can directly integrate to find $x(t):$

This yields the general solution to the ODE \[ x(t)=60t+C. \] To determine the constant $C$ we need more information, such as your initial position at time $0$, in which case we would be solving an IVP.

For different values of $C$ we get different solutions, and below we graph some of these. If $C=0$ then $x=60t.$ All the other solutions are parallel to this line.

The curves $x=60t+C$ are called solution curves to the ODE $\ds\dif{x}{t}=60$.

Note that in this particular case all curves are straight lines with slope 60.

Example: Consider an apple falling under gravity. Find an expression for the height $y$ of the apple at time $t.$

Example: Consider an apple falling under gravity. Find an expression for the height $y$ of the apple at time $t$.

Example: Now suppose you are throwing apples over your fence to your neighbour. Find an expression for the position $(x(t),y(t))$ of the apple if you assume the initial position is $x(0)= 0, y(0)=0$ and the initial velocity is $\dot{x}(0)=u,$ $\dot{y}(0)=v$.

Example: Now suppose you are throwing apples over your fence to your neighbour. Find an expression for the position $(x(t),y(t))$ of the apple if you assume the initial position is $x(0)= 0, y(0)=0$ and the initial velocity is $\dot{x}(0)=u,$ $\dot{y}(0)=v$.

Example: Now suppose you are throwing apples over your fence to your neighbour. Find an expression for the position $(x(t),y(t))$ of the apple if you assume the initial position is $x(0)= 0, y(0)=0$ and the initial velocity is $\dot{x}(0)=u$, $\dot{y}(0)=v$.

To solve an ODE (or IVP) analytically means to give a solution curve in terms of continuous functions defined over a specified interval, where the solution is obtained exactly by analytic means (e.g., by integration). The solution satisfies the ODE (and initial conditions) on direct substitution.

To solve an ODE (or IVP) numerically means to use an algorithm to generate a sequence of points which approximates a solution curve.

Important remark: As we have already seen, the ODE $$\dif{y}{t}=f(t)$$ can be solved analytically. The solution simply is \[ y=\int f(t) \dup t. \]

Important remark: As we have already seen, the ODE $\ds \dif{y}{t}=f(t)$ can be solved analytically. The solution simply is $ \ds y=\int f(t) \dup t. $

Now this may seem like a cop-out because all we are saying is that the solution is given by the anti-derivative of $f(t),$ and the above solution is as informative as the actual ODE. In practice one hopes to be able perform the above integral to get a more explicit form of the solution. Depending on the form of $f$ this may either be done exactly or by numerical means.