1. Separable Differential Equations
Lesson 9-3
Separable Differential Equations

2. Solutions to Differential Equations
A separable first–order differential equation has the form
To solve the equation,
1. Rewrite it as h(y) dy = g(x) dx (cross multiply)
2. Integrate both sides ∫ h(y) dy = ∫ g(x) dx
3. Solve for y (remember to include C)
4. Use initial condition to solve for C
dy g(x)
----- =
dx h(y)

3. Example 1 Solve 3y² dy = x dx 3∫y² dy = ∫x dx y³ = ½x² + C y = ½x² + C
----- =
dx y² 3 3

4. Example 2 2y dy = (6x – x³) dx 2∫y dy = ∫(6x – x³) dx
----- =
dx y
Given initial condition: y(0) = 3
2y dy = (6x – x³) dx
2∫y dy = ∫(6x – x³) dx
y² = 3x² – ¼x4 + C
y = 3x² – ¼x4 + C
y = 3x² – ¼x4 + 9 4

5. Example 3
Find a function f whose graph passes through (1, 0) and has slope 1 – x. dy
----- = 1 - x dx
dy = (1 – x) dx
∫dy = ∫(1 – x) dx
y = x – ½x² + C
0 = 1 – ½(1)² + C  = ½ + C
y = x – ½x² - ½ 5

6. Example 4
Find the particular solution for the initial condition y(0) = 2 for the equation ex² yy’ + x = 0.
ex² yy’ + x = 0.
ex² yy’ = - x
y’ = (-xe-x²) / y
dy xe-x²
----- =
dx y
y dy = -xe-x² dx
u = -x²
du = -2xdx
∫y dy = -∫xe-x² dx
½y² = ½e-x² + C
y = e-x² + C  2 = 1 + C  C = 3
y = e-x² + 3 6

7. Example 5 dy x ----- = - y² dx
Solve , x > 0, with the initial condition x0 = 1, y0 = ⅓
dy y²
----- =
dx x
x dy = - y² dx
dy dx
----- =
- y² x dx
-y-2dy = x dx
-∫ y-2dy = ∫ x 1
---- = ln |x| + C y 1
y = ln |x| + C 1
⅓ = C 1
y = ln |x| + 3 7

8. Example 6
A point is moving along a line in such a way that its velocity is 4- 2t. When t = 0, the position is 0. Find the position at time t, and the acceleration. ds
----- = v(t) = 4 – 2t dt
a(t) = -2
ds = (4 – 2t) dt
∫ ds = ∫ (4 – 2t) dt
s = 4t – t² + C
0 = 4(0) – (0)² + C
0 = C
s = 4t – t² 8

9. Example 7
A formula for acceleration, a, of a point P moving on a line is a= 6t - 6, where s = 0, v = 2 and t = 0. Find the law of motion.
a(t ) = 6t – 6 dv
= a(t ) = 6t – 6 dt
∫dv = ∫(6t – 6) dt
v(t) = 3t² – 6t + v0
v(0) = 3(0)² – 6(0) + v0 = 2
v0 = 2 ds
= v(t) = 3t² – 6t + 2 dt
∫ds = ∫(3t² + 6t + 2) dt
s(t) = t³ + 3t² + 2t + s0
s(0) = 0³ + 3(0)² + 2(0) + s0 = 0
s0 = 0
s(t) = t³ + 3t² + 2t 9

10. Example 8
Find the x,y equation of the curve through (1,2) whose slope at any point is 4 times its x-coordinate dy
----- = 4x dx
dy = 4x dx
∫dy = 4∫x dx
y = 2x² + C
2 = 2(1)² + C
2 = 2 + C
C = 0
y = 2x² 10

11. Summary & Homework Summary: Homework:
Separation of variables involves solving two differential equations with integration
Initial conditions allow us to solve for C
Homework:
pg 607 – 609, Day One: 1-4, 11, Day Two: 7-9, 15, 20