Use and verify the particular solution

A curve that answers and passes two tests

The equation \(\frac{dy}{dx} = \frac{3x^2}{2y}\) with \(y(1) = 2\) has the particular solution \(y = \sqrt{x^3 + 3}\). The chalk curve is its graph, drawn over the faint slope field, and the orange point is the start \((1, 2)\).

A question asks for the value \(y(2)\). Move \(x\) to 2. The curve has height \(\sqrt{2^3 + 3} = \sqrt{11}\) there.

Test the curve twice before you trust the answer. It must pass through the starting point, and its slope must match the slope field at every input.

The orange line is the tangent of the curve, and the dashed chalk line has the slope that the equation asks for. The lower graph shows these two slopes at every \(x\) in the same colors. Both pairs lie on top of each other, so the curve follows the field.

Two tests, one answer

Let \(y = g(x)\) be a proposed particular solution. Two tests decide whether it is correct.

The first test is the equation: the slope of \(g\) must equal the right side of the equation along \(g\). In the lower graph, a red band fills the gap wherever they differ.

The second test is the starting value \((x_0, y_0)\). If \(g(x_0) \ne y_0\), a red segment joins the curve to the orange point.

Press the three buttons. Only curve 1 has no red band and no red segment, and a curve that passes only one test is not the particular solution.

Notation

1. Read what the question asks: an output for a given input, or an input for a given output. Note the unit.

2. For an output, substitute the input into the particular solution and simplify.

   \[y(2) = \sqrt{2^3 + 3} = \sqrt{11}\]

3. Test the equation: the derivative of the solution must equal the right side.

   \[\begin{aligned} \frac{d}{dx}\sqrt{x^3 + 3} &= \frac{3x^2}{2\sqrt{x^3 + 3}} \\ &= \frac{3x^2}{2y} \end{aligned}\]

4. Test the starting value. The solution at the initial input must give the initial output.

   \[y(1) = \sqrt{1 + 3} = 2\]

5. Write the answer with its unit. The input must lie in the interval of the solution, the orange bar on the x-axis.

When it fails

The curve \(y = \sqrt{x^3 + 5}\) also satisfies \(\frac{dy}{dx} = \frac{3x^2}{2y}\), because its derivative is \(\frac{3x^2}{2\sqrt{x^3 + 5}}\). No red band opens in the lower graph.

This curve has the wrong constant. At \(x = 1\) its height is \(\sqrt6 \approx 2.45\), not 2, and the red segment shows the gap. The equation alone does not find the particular solution.

Example 1: A draining tank

Water drains from a tank. The volume \(V\) in liters at time \(t\) minutes satisfies \(\frac{dV}{dt} = -2\sqrt V\) with \(V(0) = 36\). The particular solution is \(V = (6 - t)^2\). Find the volume at \(t = 4\) and the time when the tank is empty. Then verify the solution.

1. Substitute the input \(t = 4\). The tank holds 4 liters after 4 minutes, and the red arrow shows the water leaving.

   \[V(4) = (6 - 4)^2 = 4\]

2. The tank is empty when \(V = 0\). Solve for \(t\): the tank is empty after 6 minutes.

   \[(6 - t)^2 = 0 \ \Rightarrow\ t = 6\]

3. Test the equation. The derivative of \(V\) is \(-2(6 - t)\), the slope of the orange tangent. For \(t \le 6\), \(6 - t\) is not negative, so \(\sqrt V = 6 - t\) and \(-2\sqrt V\) is the same number.

   \[\begin{aligned} \frac{dV}{dt} &= -2(6 - t) \\ &= -2\sqrt{(6 - t)^2} = -2\sqrt V \end{aligned}\]

4. Test the starting value: the result is the given 36 liters.

   \[V(0) = 6^2 = 36\]

Example 2: A claim to test

A student claims that \(y = 1 - x + \frac{x^2}{2}\) solves \(\frac{dy}{dx} = x - y\) with \(y(0) = 1\). Decide whether the claim is correct.

1. Differentiate the claimed solution.

   \[y' = -1 + x\]

2. A tempting wrong move: Substitute \(x = 0\). Both sides equal \(-1\), so the claim is correct.

   \[\begin{aligned} y'(0) &= -1 \\ x - y &= 0 - 1 = -1 \end{aligned}\]

   One input is not enough. At \(x = 1\), the curve has slope 0, but the equation asks for slope \(\frac12\). The orange tangent and the dashed chalk line separate there.

3. Substitute the claimed \(y\) into the right side, to compare it with \(y'\) at every \(x\).

   \[\begin{aligned} x - y &= x - \left(1 - x + \frac{x^2}{2}\right) \\ &= 2x - 1 - \frac{x^2}{2} \end{aligned}\]

4. Compare the two sides. They are equal only where \(x - 1 = 2x - 1 - \frac{x^2}{2}\).

   \[x - \frac{x^2}{2} = 0 \ \Rightarrow\ x = 0 \ \text{or}\ x = 2\]

5. The sides differ at every other input, so the claim is not correct. The starting value \(y(0) = 1\) holds, but this alone does not make a solution.

Practice

Water drains from a tank. The volume \(V\) in liters satisfies \(\frac{dV}{dt} = -3\sqrt V\) with \(V(0) = 16\). The particular solution is \(V = \left(4 - \frac{3t}{2}\right)^2\). Find the volume at \(t = 2\) minutes. Then verify the solution.

Slide \(t\) to 2 and read the volume under the figure. Then compare the slope of the curve with the value of \(-3\sqrt V\).

Questions

Questions 1 and 2 use the equation \(\frac{dy}{dx} = y - x\) with \(y(0) = 1\). The orange point is the starting value.

1. Which curve is the particular solution of \(\frac{dy}{dx} = y - x\) with \(y(0) = 1\)?

   • The chalk line
   • The faint curve
   • The red dashed curve
   • None of them
   Answer

   The chalk line

2. At \(x = 1\) the red dashed curve is at height 3, and its tangent has slope 3. What slope does the equation ask for at \((1, 3)\)?

   • 2
   • 3
   • 1
   • 4
   Answer

   2

3. The general solution of \(\frac{dy}{dx} = \frac{4x}{y}\) is \(y^2 = 4x^2 + C\). With \(y(1) = 3\) and \(y > 0\), what is \(y(2)\)?

   • \(\sqrt{21}\)
   • 21
   • 5
   • 4
   Answer

   \(\sqrt{21}\)

Summary

You substitute the requested input into the particular solution, or solve for the requested input, and you give the answer with its unit.

This works because the particular solution satisfies the equation on its interval and passes through the starting point, and you can test both by substitution.

Use it when a problem asks for a value after the constant is known.

This is one of four skills in topic 7.7 of Which & Why: AP Calculus.