MA 3972-MA-Book April 11, 2018 16:5
More Applications of Definite Integrals 327
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 Steps:
- Separate the variables:g(y)dy=f(x)dx.
- Integrate both sides:
∫
g(y)dy=
∫
f(x)dx.
- Solve foryto get a general solution.
- Substitute given conditions to get a particular solution.
- 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 thaty=
− 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.