Bernoulli’s Differential Equation

Standard Form
dy n
+ y P ( x )= y Q ( x ) (Eq. 1)
dx
or
dx n
+ x P ( y )=x Q ( y ) (Eq. 2)
dy

Solution for Bernoulli DE

For Eq. 1: For Eq. 2:

z ɸ = ∫ ɸQ r ( x ) dx +C z ɸ = ∫ ɸQ r ( y ) dy +C

Where: Where:
ɸ = e∫ P ( x )dx
r
ɸ = e∫ P ( y ) dy
r

z = y1-n z = x1-n
Pr (x) = (1- n) P(x) Pr (y) = (1- n) P(y)
Qr (x) = (1 –n) Q (x) Qr (y) = (1 –n) Q (y)

y dx 2 2
Example 1: dy + =3 x y dx
x

[ dy +
y dx
x
=3 x 2 y 2 dx
1
dx]; we multiply
1
dx
so we can transform it into standard form.

dy y dx 2 2 dx
+ =3 x y ; simplify
dx x dx dx
dy y 2 2
+ =3 x y
dx x
dy
dx
+y
1
x ()
= y 2 ( 3 x 2 ) ; it is now in a standard form.

Identify the values of P(x), Q(x), and n

dy n
+ y P ( x )= y Q ( x )
dx
1
P(x) = ;Q(x) = 3 x 2; and n = 2
x
Solve for Pr(x), Qr(x), and z
Pr(x) = (1- n)P(x) Qr(x) = (1-n)Q(x) z = y1-n

Pr(x) = (1- 2) ( 1x ) Qr(x) = (1-2)( 3 x 2 ) z = y1-2

I can do all things through Christ who strengthens me. –Philippians 4:13
Pr(x)= (-1) ( 1x ) Qr(x) = (-1)( 3 x 2 ) z = y-1

Pr(x)= ( −1x ) Qr(x) = (−3 x 2 ) z=

1
y

Solve for integrating factor ɸ.

ɸ = e∫ P ( x )dx ; substitute the value of Pr ( x )
r

−1
ɸ = e∫ x
dx

ɸ = e−ln x
−1
ɸ = e ln x
−1 1
ɸ=x =
x
Solve for the solution using Bernoulli Equation.

z ɸ = ∫ ɸQ r ( x ) dx +C ; substitute the solved values.

( 1y )( 1x )=∫ ( 1x ) (−3 x ) dx+ C; simplify and integrate.

2

1
=∫ (−3 x ) dx +C
xy
1 −3 2
= x +C
xy 2

2
Example 2: dx – 2xy dy = 6x3y2e−2 y dy

[ dx – 2 xy dy=6 x 3 y 2 e−2 y dy ] dy1

2

dx dy 3 2 −2 y dy 2

−2 xy =6 x y e
dy dy dy
dx 3 2 −2 y 2

−2 xy =6 x y e
dy
dx
+ x (−2 y )=x ( 6 y e )
2
3 2 −2 y
dy
2
P(y) = -2y Q(y) = 6 y 2 e−2 y n=3
Solve for Pr(x), Qr(x), and z
Pr(y) = (1- n)P(y) Qr(y) = (1-n)Q(y) z = x1-n

Qr(y) = (1-3)( 6 y2 e−2 y )

2
Pr(y) = (1- 3)(−2 y ) z = x1-3

Qr(y) = (-2)( 6 y2 e−2 y )

2
Pr(y)= (-2)(−2 y ) z = x-2

1
Qr(y) = (−12 y 2 e−2 y )
2
Pr(y)= ( 4 y ) z= 2
x

I can do all things through Christ who strengthens me. –Philippians 4:13
Solve for ɸ:
ɸ = e∫ P ( y ) dy ;
r

ɸ = e∫ 4 ydy
2
ɸ = e2 y

Solve for the solution:

z ɸ = ∫ ɸQ r ( y ) dy +C

( x1 ) ( e
2
2y
2

)= ∫ ( e 2 y )(−12 y 2 e−2 y ) dy+C

2 2

( )
2

e2 y
2
= ∫ ( −12 y 2 ) dy +C
x

( )
2

e2 y
= −4 y 3+ C
x2
2

e 2 y =−4 x 2 y 3 +C x 2

I can do all things through Christ who strengthens me. –Philippians 4:13