Particular Solutions from Initial… — AP® Calculus AB Notes

Separation of variables hands you a general solution: a family of curves with a constant $C$ in it. A real question usually wants one curve. The coffee was 90 °C at noon; the population was 200 at the start; the curve passes through $(1, -3)$ . That one fact is an initial condition, and the single curve it selects is a particular solution. A differential equation together with an initial condition is called an initial value problem.

The procedure

Separate and integrate, adding $+ C$ (exactly as before).
Substitute the initial condition immediately. An initial condition $y(x_0) = y_0$ means: replace $x$ by $x_0$ and $y$ by $y_0$ . Solve for $C$ .
Put the value of $C$ back and solve for $y$ . If a square root or an absolute value appears, keep the sign that matches $y_0$ .
State the domain of the solution (below).

Why substitute before solving for $y$ ? Because the equation right after integrating is the simplest place to find $C$ . Solving for $y$ first means exponentiating or taking roots with $C$ still unknown, where most sign and algebra slips happen.

Worked example 1. Solve $\frac{dy}{dx} = \frac{2x}{y}$ with $y(1) = -3$ .

Separate and integrate: $y\,dy = 2x\,dx$ , so $\frac{y^2}{2} = x^2 + C$ .

Use the condition now: $\frac{(-3)^2}{2} = 1^2 + C$ , so $C = \frac{9}{2} - 1 = \frac{7}{2}$ .

Solve: $y^2 = 2x^2 + 7$ , so $y = \pm\sqrt{2x^2 + 7}$ . The point $(1, -3)$ has a negative $y$ , so take the negative root: $y = -\sqrt{2x^2 + 7}$ .

Check: at $x = 1$ , $y = -\sqrt{9} = -3$ . Good.

Worked example 2. Coffee cools by $\frac{dT}{dt} = -0.05(T - 20)$ , with $T(0) = 90$ . Find $T(t)$ .

Separate and integrate: $\frac{dT}{T - 20} = -0.05\,dt$ , so $\ln|T - 20| = -0.05t + C$ .

Use the condition: $\ln|90 - 20| = 0 + C$ , so $C = \ln 70$ .

Solve: $|T - 20|$ $= e^{-0.05t + \ln 70}$ $= 70e^{-0.05t}$ . At $t = 0$ , $T - 20 = 70 > 0$ , so $T - 20 = 70e^{-0.05t}$ and

$T(t) = 20 + 70e^{-0.05t}.$

The domain of a particular solution

A solution must be differentiable at every point where you claim it works, and it must be one connected piece. So the domain of a particular solution is the largest open interval that contains $x_0$ and on which the formula is defined and differentiable (and makes the differential equation meaningful).

Worked example 3. Solve $\frac{dy}{dx} = y^2$ with $y(0) = 1$ , and give the domain.

Separate and integrate: $y^{-2}\,dy = dx$ , so $-\frac{1}{y} = x + C$ .

Condition: $-1 = 0 + C$ , so $C = -1$ . Then $-\frac{1}{y} = x - 1$ , and $y = \frac{1}{1 - x}$ .

The formula breaks at $x = 1$ . The interval containing $x_0 = 0$ is $x < 1$ , so the particular solution is $y = \frac{1}{1 - x}$ for $x < 1$ .

The slope field of dy/dx = y² with the curve y = 1/(1 − x). The crimson branch through the closed point (0, 1) rises without bound as x approaches the dashed vertical asymptote x = 1; it is the solution, with domain x < 1. The gray dashed branch for x > 1, below the x-axis, also fits the equation but is not connected to (0, 1), so it is not part of this solution. — **Figure 1.** The particular solution of worked example 3 is the crimson branch. The gray branch for $x > 1$ fits the differential equation too, but it is a different solution, not connected to $(0, 1)$ . LevlPrep original diagram.

Notice what happened. The equation $\frac{dy}{dx} = y^2$ looks harmless, yet the solution through $(0, 1)$ shoots to infinity at $x = 1$ . You could not have seen that from the equation alone; the formula shows it.

Watch for three ways a formula breaks: a division by 0, an even root of a negative number, and a logarithm of a number that is not positive. Also look at the differential equation itself: for $\frac{dy}{dx} = -\frac{x}{y}$ with $y(0) = 3$ , the solution $y = \sqrt{9 - x^2}$ reaches $y = 0$ at $x = \pm 3$ , where $-\frac{x}{y}$ divides by 0. So the domain is the open interval $-3 < x < 3$ .

Common slips

Adding the constant after solving: from $\ln|y| = \sin x + C$ , writing $y = e^{\sin x} + C$ . Exponentiating gives $y = Ae^{\sin x}$ .
Substituting the condition the wrong way round: $y(0) = 2$ means $x = 0$ and $y = 2$ .
Keeping the wrong sign after a square root.
Giving the formula without its domain, or a domain that does not contain $x_0$ .