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De-Lecture 9 by Engineer Marcos

1) Separable Method: Separate the variables and integrate both sides to get the general solution y=cx. 2) Homogeneous Method: Make the equation homogeneous by letting z=y/x and solve the resulting linear equation to get the general solution y=cx. 3) Exact Method: Check if the equation is exact and if so, integrate to find the general solution y=cx.

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186 views5 pages

De-Lecture 9 by Engineer Marcos

1) Separable Method: Separate the variables and integrate both sides to get the general solution y=cx. 2) Homogeneous Method: Make the equation homogeneous by letting z=y/x and solve the resulting linear equation to get the general solution y=cx. 3) Exact Method: Check if the equation is exact and if so, integrate to find the general solution y=cx.

Uploaded by

melenyo tuquero
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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BERNOULLI’S EQUATION

Given y ' + Px ( y )=Q( x ) y n →Equation (a)

If n = 1, The variables are separable

If n 1, Equation (a) may be written in the form

−n
y
+ Py 1−n=Q
dx

Let z = y1−n

dz=(1−n) y−n dy

−n dz
¿ y dy=
( 1−n )

dz
Therefore , + P ( z )=Q
( 1−n ) dx

multiplying the Equation by (1 – n), the Equation becomes,

dz
+ ( 1−n ) P ( z ) dx=( 1−n ) Q
dx

or,dz + ( 1−n ) P ( z ) dx=( 1−n ) Qdx ; which is a Linear Equation in Standard Form.

SAMPLE PROBLEMS:

[Example 1]

Find the PS (when x=2, y=1)


(𝐲𝟒- 𝟐𝐱𝐲)𝐝𝐱 + 𝟑𝐱𝟐𝐝𝐲 = 𝟎

SOLUTION:
1
(𝐲𝟒- 𝟐𝐱𝐲)𝐝𝐱 + 𝟑𝐱𝟐𝐝𝐲 = 𝟎] dx
3 x2
4
dy ( y −2 xy ) −4
+ =0 ¿ y
dx 3x
2

−4 dy 1 2 y−3
y + 2− =0
dx 3 x 3x

−4 dy 2 y −3 −1
y − = 2
dx 3 x 3x

Let z = y−3

−4
dz=−3 y dy

dz
y− 4 dy=
−3

dz 2z 1
− = 2 ¿−3
−3 dx 3 x 3 x
dz 2 z 1
+ = → linear differential equation
dx x 2

∫ 2 dx
x 2
z=e =x

2 x2 =∫
( ) 1 2
x
2
x dx+C

−3
but z= y
2
x
= x+C → General solution
y3

When x=2; y=1

22
=2+C
13

4 = 2+C

C=2

Substitute C in the general solution


2
x
= x+2
y3
2 3
x = y ( x +2 ) D.E

[Example 2]

y ( 6 y 2−x−1 ) dx +2 xdy=0

SOLUTION:

1
y(6𝐲𝟐 − 𝐱 − 𝟏)𝐝𝐱 + 𝟐𝐱𝐝𝐲 = 𝟎]
2 xdx

dy 6 y 2 y ( x +1 )
+ − =0 ¿ y−3
dx 2 x 2x

−3 dy −2 ( x+ 1 ) −6
y −y =
dx 2x 2x

let z= y−2

−3
dz=−2 y dy

dz −3
= y dy
−z

dz z ( x +1 ) 6
− = ¿−2
−2 dx 2x 2x
dz z ( x+1) 6
+ =
dx x x

using this formula ; y e∫ =∫ Q e∫


Pdx Pdx
dx +C

∫ (x+x1)dx ∫ dx +∫ dxx
ℷ=e =e

¿ e x+ ¿ ( x )=e x Inx

xz e x =∫ ( 6x ) x e dx +C x

xz e =6 ∫ e dx +C
x x

−2
but z= y

[ xe x
y2
=6 e
x
+C
y2
ex ]
2 2 −x
x=6 y +C y e

x= y ( C e +6 )
2 −x
D.E

[Example 3]

dy
= y ( y +3 x )
3 2 2
2x
dx

SOLUTION:

[ 2x
3 dy
dx
= y ( y +3 x )
2 2 1
2x
3 ]
dy
= y ¿¿
dx

[ ]
2 2
dy y 3x −3
= 3+ 3 y
dx 2 x 2 x
2 −2
−3 dy 3 x y 1
y − 3
= 3
dx 2x 2x
−2
let z= y
−3
dz=−2 y dy

dz
= y−3 dy
−2
[ dz 3z 1
− = 3 −2
−2 dz 2 x 2 x ]
dz 3 z 1
+ = →linear differential equation
dx x x 3
dx
3∫
ℷ=e x
=x 3

x z=∫
3
( x1 ) x dx+C
3
3

3
x z=−x+C
3
x
=−x+C
y2

x 3=−xy 2 +C y 2
3 2
x = y ( C−x ) D.E.

[Example 4]

Find the PS (when x=1, y=1)


(𝟐𝐲𝟑- 𝐱𝟑)𝐝𝐱 + 𝟑𝐱𝐲𝟐𝐝𝐲 = 𝟎

SOLUTION:
¿

( dy ( 2 y −x )
)
3 3
2
+ =0 y
dx 3 xy 2

3 2
2 dy 2 y x '
y + = → bernoull i sequation
dx 3 x 3
3
let z= y
2
dz=3 y dy
dz 2
= y dy
3

[ dz 2 z x 2
+ = 3
3 dx 3 x 3 ]
dz 2 z 2
+ =x → linear differential equation
dx x

dz
2∫
x 2
ℷ=e =x

x z=∫ ( x ¿¿ 2) ( x ) dx +C ¿
2 2

x z=∫ x dx +C
2 4

5
2 x
x z= +C
5
[ 2 x5
x y = +C 5
5
3
]
2 3 5
5 x y =x +5 C → general solution

substitute the value of x∧ y

2
5 (1) ¿
5=1+ 5C
4=5 C
4
C=
5

Substitute the C

5 x 2 y 3=x 5 +5 ( 45 )
2 3 5
5 x y =x +4 D.E.

Assign:

Solve problem no. 27 in three methods pp. 77

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