An equation with a family of answers becomes a question with one answer as soon as a single measurement is added: the value of the solution at one point. Geometrically the condition names a point of the plane, and the answer is the one curve of the family through it. The order in which the work is done matters — find the family first, with its constant, and use the condition last.

The equation is the rule of the road and the initial condition is where the car is parked at noon. Neither alone tells you where the car will be at one o’clock; together they do, and there is nothing left to choose.

The family $y=Ce^{x}$ in grey, and the one member through $(0,2)$ in green. The condition $y(0)=2$ does not change the equation; it chooses.

An initial value problem is a differential equation together with enough conditions at a single point to fix every constant. A first-order equation needs one condition, $y(x_{0})=y_{0}$; an $n$th-order equation needs , giving the value and the first derivatives at . The conditions are all imposed at the same point — conditions at two different points make a boundary value problem, which is a different subject.

• Find the general solution first, keeping the constant, then substitute the condition. Imposing the condition part-way through is the mistake that produces an answer with no constant left to fix.
• For a second-order equation the two conditions are $y(x_{0})=a$ and $y'(x_{0})=b$: a position and a velocity, both at the same instant.

The number of conditions needed matches the order, because that is how many constants there are to fix.

Solve the equation with its constant still in place. Only then substitute the condition and solve for the constant.

Great work! You now know how to:

• count the conditions an equation of a given order needs
• solve a simple initial value problem in the right order: family first, condition last
• read an initial condition as a point the solution curve must pass through
• tell an initial value problem from a boundary value problem