Bernoulli Differential
Equations
AP Calculus BC
Bernoulli Differential Equations
A Bernoulli differential equation can is of the form
dy
n
 P( x) y  Q( x) y
dx
where P and Q are continuous functions on a given interval.
Note that if n = 0 or 1, then we have a linear differential equation.
But what if n > 1?
Our goal is to take this non-linear differential equation and turn it into a
linear differential equation, so we can solve it.
Reduction
 Divide by
yn
y
n
dy
 y1n P ( x)  Q ( x)
dx
 We’re going to use u-substitution, so let u =
du
 n dy
dx
 So
y
(1  n)
dx
du
y1–n
 Substitute back into original equation 
du
 n dy
 (1  n) y

dx
dx
dx  u ( P ( x))  Q( x)
(1  n)
du
 (1  n) P ( x)u  (1  n)Q( x)
 Multiply by (1 – n) 
dx
 The differential equation is now linear (ugly, but linear)!
Example 1
dy 1
 y  xy 2
Find the general solution to the differential equation
dx x
1) n = 2, so u = y1–2, or u = y–1.
2) Divide equation by
y2
dy
1  1 
y  x
to get RHS to equal x  y
dx
x
du
2 dy
 y
3) Find du/dx 
dx
dx
dy
2 du
 y
4) Solve for dy/dx 
dx
dx
2
Example 1 (cont.)
du
1
5) Substitute into equation in #2  y ( y )  u    x
dx
x
du  1 
   u  x
dx  x 
2
2
du  1 
  u  x
dx  x 
6) Now this is a linear differential equation, so solve using integrating
1
factor.

 x dx
 ln x
ln x 1
I ( x)  e
e
e
 x 1
Example 1 (cont.)
1 du  1 
7) Multiply by

  2  u  1
x dx  x 

1


8) Product Rule in reverse   u    1
x

1
9) Integrate  u    x  C
x
x–1
2
u


x
 Cx
10) Multiply by x 
11) Substitute back for y (Recall u = y–1)
1
y–1 = –x2 + Cx  y  2
 x  Cx
Example 2
dy
2
2
Find the general solution to 2 xy  y  x
dx
Example 3 – Initial Value Problem
y
Solve the differential equation y   y  0 with the initial condition
x
y(1) = 0.
Still Confused?
Watch this video:
https://www.youtube.com/watch?v=7MmhoqvM9_Q
It helped me!