Check the initial condition
One curve from the family
The equation \(dy/dx = 4 - y\) has many solutions. The faint segments show the slope that the equation gives at each point, and every curve \(y = 4 - Ke^{-x}\) runs along them, for any constant \(K\). Each curve is a candidate, which is a function that someone proposes as a solution.
A problem often adds an initial condition, \(y(0) = 1\). It says that the solution passes through the orange dot at \((0, 1)\).
Drag the slider for \(K\). Most curves pass above or below the dot. The orange dashed line reads the height of the curve at \(x = 0\), and the red segment from that height to the dot is the gap.
The height at \(x = 0\) is \(4 - K\), because \(e^0 = 1\). At \(K = 3\) the gap closes, because \(4 - 3 = 1\), and only this curve satisfies the condition.
A point that the curve must pass through
An initial condition is a point \((a, b)\). The number \(a\) is the starting input, and \(b\) is the starting output. Drag the orange dot, or move the second slider, to change the point.
The dashed line reads the height of the candidate above \(a\). The red segment is the gap between that height and \(b\), and the candidate passes through the point when the gap is 0.
The condition concerns one input only. To test it, you evaluate the candidate at that input, and you do not need the whole interval.
Notation
An initial condition \(y(a) = b\) says that the solution has the output \(b\) at the input \(a\).
Write the candidate as \(g\). Evaluate it at the starting input \(a\), without using the equation.
\[\cy{g(a)}\]Compare the result with the starting output. If \(g(a) = b\), the candidate passes through the point.
\[g(a) = b\]An initial value problem is a differential equation together with an initial condition. Its solution must pass two tests: it must satisfy the equation on the interval, and it must satisfy the condition at one input. The orange segments show that this candidate runs along the field, so it passes both tests. A candidate can also pass one test and fail the other.
The starting input is the number in the condition, and it need not be 0.
When it fails
The candidate \(y = \ln x\) has no value at \(x = 0\), so it cannot satisfy a condition at \(x = 0\).
The starting input must lie in the interval where the candidate is defined. If it does not, the candidate fails the condition, whatever the equation says.
Example 1: a parabola
The equation is \(\dfrac{dy}{dx} = 6x\). Does the candidate \(y = 3x^2 - 4\) satisfy the initial condition \(y(1) = -1\)?
The starting input is 1. Evaluate the candidate at \(x = 1\).
\[\cy{g(1)} = 3(1)^2 - 4 = -1\]The starting output is −1. The two values are equal, so the candidate passes through the point \((1, -1)\).
The derivative of the candidate is \(6x\), which matches the equation, so the candidate also solves the equation and solves the initial value problem. At the dot, the orange tangent has slope \(6(1) = 6\).
\[g'(x) = 6x\]
Example 2: a different starting input
The equation is \(\dfrac{dy}{dx} = y - 1\), and the candidate \(y = 1 + 2e^{x-1}\) solves it. Does the candidate satisfy \(y(1) = 3\)?
A tempting wrong move: Initial conditions are often given at \(x = 0\), so evaluate the candidate there. The result is not 3, so the condition fails.
\[g(0) = 1 + 2e^{-1} \approx 1.74\]The condition says \(y(1) = 3\), so the dot is above \(x = 1\). The input 0 is not the starting input, and the red dashed line at \(x = 0\) reads a point that the condition does not mention.
Evaluate at the starting input, \(x = 1\).
\[\cy{g(1)} = 1 + 2e^{0} = 3\]The result equals the starting output 3, so the candidate passes through the dot and satisfies the condition.
Practice
The candidate \(y = 5 - 4e^{-2t}\) solves \(\dfrac{dy}{dt} = 10 - 2y\). Does it satisfy \(y(0) = 2\)? If it does not, find the value of the candidate at \(t = 0\).
Evaluate the candidate at the starting input. Then compare the result with the output in the condition. The red gap shows the difference.
Solution
The starting input is 0. Evaluate the candidate there.
\[\cy{g(0)} = 5 - 4e^{0} = 1\]The condition asks for 2, and the candidate gives 1. The values differ, so the candidate does not satisfy \(y(0) = 2\). It passes through \((0, 1)\).
\[1 \neq 2\]The candidate still solves the equation. It solves the problem with \(y(0) = 1\), not the problem with \(y(0) = 2\), so the dot at \((0, 1)\) lies on the curve.
Questions
The graph shows the candidate \(g(x) = (x - 1)^2 + 1\). After you answer, an orange dot marks the initial condition.
The initial condition is y(3) = 5. Does the candidate satisfy it?
- Yes, the curve has height 5 above x = 3.
- No, the curve has height 2 above x = 0.
- It cannot be tested without the equation.
Answer
Yes, the curve has height 5 above x = 3.
The initial condition is y(1) = 2. What is the gap between the point (1, 2) and the curve?
- 1
- 0
- 2
Answer
1
Does the candidate \(y = 3t^2 + t\) satisfy the initial condition \(y(2) = 14\)?
- Yes, because the candidate equals 14 at t = 2.
- No, because the candidate equals 12 at t = 2.
- No, because the candidate equals 0 at t = 0.
Answer
Yes, because the candidate equals 14 at t = 2.
Summary
You evaluate the candidate at the starting input and compare the result with the starting output.
This works because the condition names one point that the solution must pass through, so one value decides it.
Use it after the equation test, because a solution of an initial value problem must pass both tests.
This is one of four skills in topic 7.2 of Which & Why: AP Calculus.