Particular Solutions

A general solution describes infinitely many solutions, and for the equations in this course an initial condition picks out exactly one of them. That solution may exist on only part of the number line. Finding it, and the interval where it lives, is the work of a separation question.

The steps in order

A particular solution by separation of variables takes four steps. Separate the variables, antidifferentiate both sides with a constant

C

, use the initial condition to find

C

, and solve for

y

. Finding

C

before solving for

y

usually keeps the algebra simple. A free-response answer earns its points in this order. An answer that drops

C

can't earn the points for the initial condition or the final solution.

Example 1

Find the particular solution

y = f(x)

\frac{dy}{dx} = 2x\tan y

with

f(0) = \frac{\pi}{6}

, and give its domain.

Near

y = \frac{\pi}{6}

\tan y \ne 0

, so separate the variables and antidifferentiate:

\begin{aligned} \int \frac{\cos y}{\sin y}\,dy &= \int 2x\,dx \\[4pt] \ln|\sin y| &= x^2 + C. \end{aligned}

Use

f(0) = \frac{\pi}{6}

\ln\frac{1}{2} = 0 + C

, so

C = -\ln 2

. Near

y = \frac{\pi}{6}

\sin y > 0

, so

|\sin y| = \sin y

. Then solve for

y

\begin{aligned} \sin y &= e^{x^2 - \ln 2} = \tfrac{1}{2}e^{x^2} \\[4pt] y &= \arcsin\left(\tfrac{1}{2}e^{x^2}\right). \end{aligned}

The arcsine gives the value near

\frac{\pi}{6}

, between

-\frac{\pi}{2}

and

\frac{\pi}{2}

. As

|x|

grows,

\frac{1}{2}e^{x^2}

grows, and

y

reaches

\frac{\pi}{2}

when

e^{x^2} = 2

, at

x = \pm\sqrt{\ln 2}

. There

\tan y

is undefined, so the equation itself breaks down. The domain is

-\sqrt{\ln 2} < x < \sqrt{\ln 2}

The particular solution y = arcsin(½e^(x²)) through (0, π/6). It rises to π/2 (dashed) at x = ±√(ln 2) ≈ ±0.83, where tan y is undefined, so the domain stops there.

Absolute values

Antiderivatives such as

\int \frac{1}{y}\,dy = \ln|y| + C

bring in absolute values. To drop one, use the sign that the quantity has near the initial point, and say why.

Example 2

Find the particular solution

y = f(x)

\frac{dy}{dx} = y \tan x

with

f(0) = -3

, and give its domain.

Since

f(0) \ne 0

, separate and antidifferentiate:

\begin{aligned} \int \frac{1}{y}\,dy &= \int \tan x\,dx \\[4pt] \ln|y| &= -\ln|\cos x| + C. \end{aligned}

Use

f(0) = -3

\ln 3 = -\ln 1 + C

, so

C = \ln 3

. Near

x = 0

\cos x > 0

, so

|\cos x| = \cos x

. The solution starts at

-3

, so

y < 0

near

x = 0

and

|y| = -y

. Then

\begin{aligned} \ln(-y) &= \ln 3 - \ln(\cos x) = \ln\frac{3}{\cos x} \\[4pt] y &= -\frac{3}{\cos x} = -3\sec x. \end{aligned}

The equation and the solution are undefined where

\cos x = 0

. The nearest such points to

0

are

\pm\frac{\pi}{2}

, so the domain is

-\frac{\pi}{2} < x < \frac{\pi}{2}

. The solution stays negative there, as the choice

|y| = -y

assumed.

The particular solution y = −3 sec x through (0, −3), which falls without bound toward x = ±π/2 (dashed).

Particular solutions in context

Example 3

A spherical gumball dissolves in water. Its volume

V

, in cubic centimeters, decreases at a rate proportional to its surface area, which is proportional to

V^{2/3}

. With

t

in minutes,

V

satisfies

\frac{dV}{dt} = -0.6\,V^{2/3}, \qquad V(0) = 27.

(a) Find

V(t)

(b) Find the time at which half of the gumball has dissolved.

(a) Separate and antidifferentiate:

\begin{aligned} \int V^{-2/3}\,dV &= \int -0.6\,dt \\[4pt] 3V^{1/3} &= -0.6t + C. \end{aligned}

t = 0

3(27)^{1/3} = 9 = C

. So

3V^{1/3} = 9 - 0.6t

, and

\begin{aligned} V^{1/3} &= 3 - 0.2t \\[4pt] V(t) &= (3 - 0.2t)^3. \end{aligned}

(b) Half of the gumball is left when

V = 13.5

. Since

13.5 = \frac{27}{2}

, its cube root is

\frac{3}{\sqrt[3]{2}}

, so

\begin{aligned} 3 - 0.2t &= \frac{3}{\sqrt[3]{2}} \\[4pt] t &= 15\left(1 - 2^{-1/3}\right) \approx 3.09. \end{aligned}

That's about

3.09

minutes after the start.

(c)

V = 0

when

3 - 0.2t = 0

, at

t = 15

minutes. The formula satisfies the equation for every

t

. The context ends the model at

t = 15

, since a volume can't be negative.

The volume V(t) = (3 − 0.2t)³ falls from 27 cm³ to 0 at t = 15. Half is gone at t ≈ 3.09 (blue).

When there's no antiderivative formula

By the Fundamental Theorem of Calculus,

F'(x) = f(x)

, and

F(a) = y_0 + 0 = y_0

. This form works even when

f

has no antiderivative that can be written with familiar functions.