5 Steps to a 5 AP Calculus BC 2019

More Applications of Definite Integrals 321

Step 2. y(t)=y 0 ekt

y(t)= 60 e

⎡

⎢⎣−ln2

5750

⎤

⎥⎦

y(t)= 60 e

⎡

⎢⎣−ln2

5750

⎤

⎥⎦(3000)

y(3000)≈ 41. 7919

Thus, there will be approximately 41.792 grams of carbon-14 after 3000 years.

Separable Differential Equations

General Procedure

STRATEGY 1. Separate the variables:g(y)dy=f(x)dx.

2. Integrate both sides:

∫

g(y)dy=

∫

f(x)dx.

3. Solve foryto get a general solution.


4. Substitute given conditions to get a particular solution.


5. Verify your result by differentiating.

Example 1

Given

dy

dx

= 4 x^3 y^2 andy(1)=−

1

2

, solve the differential equation.

Step 1. Separate the variables:

1

y^2

dy= 4 x^3 dx.

Step 2. Integrate both sides:

∫

1

y^2

dy=

∫

4 x^3 dx;−

1

y

=x^4 +C.

Step 3. General solution:y=

− 1

x^4 +C

.

Step 4. Particular solution:−

1

2

=

− 1

1 +C

⇒c=1;y=

− 1

x^4 + 1

.

Step 5. Verify the result by differentiating.

y=

− 1

x^4 + 1

=(−1) (x^4 +1)−^1

dy

dx

=(−1) (−1) (x^4 +1)−^2 (4x^3 )=

4 x^3

(x^4 +1)^2

.

Note:y=

− 1

x^4 + 1

impliesy^2 =

1

(x^4 +1)^2

.

Thus,

dy

dx

=

4 x^3

(x^4 +1)^2

= 4 x^3 y^2.