Separable Differential Equations

1
Lesson 9-3

• Separable Differential Equations

2
Solutions to Differential Equations

• A separable firstorder 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
dy x ----- ------ dx 3y²
3
4
Example 2
dy 6x - x³ ----- ----------- dx
2y
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
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 ? 0 ½ C
y x ½x² - ½
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.
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 3
7
Example 5
dy x ----- - y² dx
Solve , x gt 0, with the
initial condition x0 1, y0 ?
dy - y² ----- ------ dx
x
dx -? y-2dy ? ------
x
1 ---- ln x
C y
1 y
------------- ln x C

1 ? ---------
0 C
1 y
------------- ln x 3

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
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
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
11
Summary Homework

• Summary
• 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, 19
  Day Two 7-9, 15, 20