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• Carl Lemuel Javellana

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DIFFERENTIAL EQUATION - an equation containing at least one derivative or which contain differentials. Examples: Order 1

Degree 1

Type ODE

3

2

ODE

3

1

ODE

∂2 y 2 ∂2 y ( ) + 4 =3 4. ∂ x 2 ∂x ∂ y

2

2

PDE

5. fyy +fxy=1

2

1

PDE

√

3

6

ODE

d3 y 2 dy ( ) = +4 3 7. d x dx

√

3

2

ODE

√

3

1

ODE

8.

dy d3 y = +4 3 dx dx

9.

dy =cos x−tan y dx

1

1

ODE

1.

dy 2 =2−6 x +x dx

' '' 2 '' 3 ' 5 2 2. ( y ) +( y ) +( y ) =x

3.

d 3 y dy 3 −( ) =x+ 3 d x 3 dx

d3 y 2 3 d3 y ( 6. d x 3 ) = d x 3 +1

CLASSIFICATION BY ORDER AND DEGREE 1. Order of Differential Equation - the order of the highest-ordered derivative in the equation. 2. Degree of Differential Equation - the power of exponent of the highest ordered derivative in the equation

TYPE OF DIFFERENTIAL EQUATION 1. Ordinary Differential Equation

1

- involves one or more dependent variable with one or single independent variable. 2. Partial Differential Equation - involves two or more independent variable and therefore contains partial derivatives.

CLASSIFICATION AS LINEAR OR NON - LINEAR 1. Linear - conditions are: a. The dependent variable y (or x) and all its derivatives must be of degree 1. b. There's no product of y (or x) and the derivative of y (or x). c. There's no transcendental functions involving y (or x). 2. Non-Linear - equation that is not linear. Examples: Linearity Linear

DV x,y

IV y,x

Order 1

Degree 1

Type ODE

dy =9−5 x + y 2 2. dx

NL

x,y

y,x

1

1

ODE

'' ' 2 2 3. y +( y ) =x +3 y

NL

x,y

y,x

2

1

ODE

d3 y dy 4. d x 3 −x dx = y

Linear in y

x,y

y,x

3

1

ODE

∂2 y 5 ∂y ( 5. ∂ x 2 ) =10 ∂ x

NL

x,y

y,x

2

5

PDE

d 2 y 3 d2 y 6. d x 2 = d x2 +1

NL

x,y

y,x

2

6

ODE

4 7. 2 y( W )xx −Wy=x y

NL

W

x,y

2

1

PDE

1

NL

x,y

y,x

2

1

ODE

Linear in y

x,y

y,x

1

1

ODE

1.

xdy + ydx=0

√

'' 3 8. ( x+ 2 y ) 3 =( y ) x 9. dy + ( xy−e ) dx=0

2

NL

10. dy + ( 2 xy−cos y ) dx=0

x,y

y,x

1

1

ODE

DETERMINATION OF DIFFERENTIAL EQUATION - a solution of DE defines a function such that when substituted into the DE reduces the equation into an identity. Two Kinds of Constant a. Absolute Constant- has fixed value b. Arbitrary Constant - with unknown value

I.

Conditions are: a. the no. of arbitrary constants are equal to the order of the DE b. consistent with the DE c. free from derivatives or differentials

II.

Solutions can be distinguished as explicit or implicit. Examples: Explicit y=a x 3 +b x 2+ cx +d y=C e 4 x +C2 e−4 x y=cos ⁡( 4 x +2)

III.

Implicit 2

2

x + y =r

2

a y 2 +by +ct=0 y 4 ( 1+9 e−6 x ) =C6

Types of Solution of an ODE A. General Solution (Complete/Primitive) - contains a no. of arbitrary constants equal to the order of DE. - constitute a family of solution (family of curves) Example: 2

3

x =c y

3

is the GS of 2 ydx−3 xdy=0

B. Particular Solution - obtained from the GS by specifying the values of the arbitrary constants, they free from arbitrary constants. Example: if C = 9. then from the GS:

x 2=C y 3

C. Singular Solution - can't be obtained from the GS whatever values are assigned to the arbitrary constant. Example: GS:

1 y= x 2 +C ; for any value of C, the trivial solution y=0 can't be obtained, 8

thus y=0 is a singular solution. D. Trivial Solution - solution of an ODE that is identically zero, that is, y = 0

IV.

To prove that the algebraic equation is a solution of DE. a. AE → (differentiation) →DE b. BE → (integration) → AE

V.

Algebraic Equation with Arbitrary Constants a. differentiate the AE n times. b. the corresponding DE is of order.

Examples:

4

x 2=c y3

1.) Verify if

Proof: if

2

is a solution of 2 ydx−3 ydy=0 .

x2 c= , then y3

3

x =c y

2 x =c ( 3 y 2 y ' ) 2x=

x2 ( 2 ' ) 3y y 3 y

2x=

x ( ') 3y y

2 y=

x2 ( ' ) 3y x

2

2 y=3 x

( dydx )

2 ydx−3 xdy=0

1.) Prove that

y=( x+ c) e−x

is a solution of

−x

y ' + y=e

.

y=( x +c ) e−x

Proof:

y ' =( x+ c ) (−e−x ) + e−x y ' =−( x+ c ) (e−x )+e−x y ' =− y+e−x y ' + y=e−x 2.) Show that AE

y= A sin 3 x+ 3 cos 3 x

y= A sin 3 x+ 3 cos 3 x

Proof:

5

is a solution of the DE

y ' ' +9 y=0 .

'

y =3 A cos 3 x−9 sin 3 x y ' ' =−9 A sin 3 x−27 cos 3 x y ' ' =−9 ( A sin 3 x+ 3 cos 3 x ) y ' ' =−9 y ''

y +9 y =0 y=2( e2 x −cos x−2 sin x) is a solution of

2.) Verify if

y ' ' + y=10 e 2 x .

y=2 ( e 2 x −cos x−2 sin x )

Proof:

y ' =2 ( 2 e2 x +sin x−2 cos x ) y ' ' =2 ( 4 e 2 x +cos x +2 sin x ) ''

2x

y + y=10 e

but:

x 8 e +2 cos x+ 4 sin ¿ ¿ x 2x 2 e −2 cos x−4 sin¿ ¿ ¿ 2x

10 e2 x =10 e2 x 3.) Show that

x=2 t+c

x=2 t+C , let t =

Proof:

x=2

y −3 C

y=ct+3 is a solution of 2( y ' )2−xy =3− y .

y −3 C

+C

Cx=2 y −6+C 6

and

2

'

C=2 y

2 y' x=2 y−6+2 ( y '2 )

2( y ' )2−xy =3− y

4.) Prove that the DE:

x y ' =x 2 + y

x y ' =x 2 + y , but

Proof:

x

y'=

dy dx

dy =x 2+ y dx

xdy − ydy =x 2 dx 2 x

∫

xdy− ydy =∫ dx 2 x

y =x +c x y=x 2 +c

7

is a solution of the AE:

y=x 2 +cx .

ORIGIN OF DIFFERENTIAL EQUATION I.

ELIMINATION OF ARBITRARY CONSTANTS Given: GS of DE Solution: Differentiate the given GS - As many as there are arbitrary constants until a DE free from arbitrary constant is obtained 3 Techniques or Method: A. Algebraic Elimination - may use of elimination by addition or subtraction of elimination by substitution. Example: 1.)

y=C 1+ C2 +C 3 e−3 x

Solution:

y ' =−3C 3 e−3 x - Equation 1 y ' ' =9 C3 e−3 x - Equation 2 '' '

−3 x

y =−27 C3 e

-Equation 3

Equate Equation 1 and 2: −3 x

3(¿¿ ' ' =9 C 3 e

y '' −3 x )→ 3 y =27 C 3 e ¿

y ' ' ' =−27 C3 e−3 x → y ' '' =−27 C 3 e−3 x ''

' ''

3 y + y =0 2.)

y=C 1 cos 3 x +C 2 sin 3 x

Solution:

'

y =−3C 1 sin 3 x +3 C2 cos 3 x ''

y =−9 C 1 cos 3 x−9C 2 sin3 x

y ' ' =−9 ( C 1 cos 3 x +C 2 sin 3 x ) y ' ' =−9 y y ' ' +9 y =0

B. Eliminant Method - may use of discriminant of matrices. Recall: System of Equation Example: 1.)

y=C 1 e2 x +C 2 e 3 x

Solution:

y ' =2 C1 e2 x +3 C 2 e3 x y ' ' =4 C 1 e 2 x +9 C 2 e 3 x

[ ] 1 1 2 3 4 9

y y ' =0 '' y

( 3 y '' + 4 y ' +18 y ) −( 12 y+ 9 y ' +2 y ' ' ) =0 y ' ' −5 y ' +6 y=0 2.)

y= A x 4 + B x3

' 3 2 Solution: ( y =4 A x +3 B x ) x ''

2

( y =12 A x + 6 Bx) x

2

[

]

1 1 y 4 3 x y ' =0 2 '' 12 6 x y

( 3 x 2 y ' ' +12 x y ' +24 y ) −( 36 y +6 x y ' +4 x2 y '' ) =0 2

''

'

- x y +6 x y −12 y =0 C. Isolation of Equation Example: 3 2 1.) C y =3 x + y

Solution: C=

3 x2 + y y3

0=

y 3 ( 6 x+ y ' ) −( 3 x 2+ y )( 3 y 2 y ' ) 6 y

y 2 (6 xy−2 y y' −9 x2 y ' )=0 6 xy−

dy ( 2 y +9 x 2 )=0 dx

6 xydx−( 2 y +9 x 2 ) dy=0 2.)

x 2+( y−k)2 =r 2

Solution: 2 x +2 ( y−k ) y '=0 '

'

x+ y y −k y =0 x + y y' k= y'

y ' ( 1+ y y ' ' + y ' y ' )−( x + y y ' ) y ' ' 0= ' 2 (y ) 3

y ' + ( y ' ) −x y ' ' =0 II.

FAMILY OF CURVES - an equation involving a parameter as well as one f the coordinate in a plane may represent a family of curves. Parameter - a constant, usually denoted by a letter, but unlike the arbitrary constant, it is not to be eliminated. Example: 1.) Find the DE of the given family of SL through (5,3). Solution:

y− y 1=m ( x−x 1 ) y−3=m ( x−5 ) m=

y−3 x−5

( x−5 ) y ' −( y−3 )=0 ( x−5 ) dy−( y −3 ) dx=0 2.) Find the DE of the family of SL with slope and y-intercept equal. Solution:

y=mx +b , but m=b y=mx +m

m=

y x+1

( x+1 ) y ' − y =0 ( x+1 ) dy − ydx=0

3.) Find the DE of the family of circles with center at (5,-1). 2 2 2 Solution: ( x−h ) + ( y−k ) =r

( x−5 )2 + ( y +1 )2=r 2 2 ( x−5 )+ 2 ( y +1 ) y '=0

( x−5 ) dx + ( y+ 1 ) dy =0 4.) Obtain the DE of the family of circles with center on x-axis. 2

2

Solution: ( x−h ) + ( y−k ) =r

2

( x−h )2 + ( y−0 )2=r 2 2 ( x−h ) +2 y y ' =0 ''

' 2

1+ y y +( y ) =0 5.) Find the DE of the family of cardioids r = a (1 - sin θ ) where a is the AC. Solution: r = a (1 - sin θ )

Let

x=r ; y=θ

x=a ( 1−sin y ) a=

x (1−sin y )

( 1−sin y )−x (−cos y ) y ' =0

( 1−sin y ) dx + x ( cos y ) dy=0 ( 1−sin θ ) dx +r cos θ dθ=0 6.) Find the DE of the family of parabolas having their vertices at the origin and there foci on the x-axis.

Solution:

2

y =4 ax 4 a=

y2 x

x ( 2 y y ' ) − y 2=0 1 [ y ( 2 x y ' − y )=0 ] y 2 xdy − ydx=0

7.) Obtain the DE of the family of parabolas with vertex on the x-axis, with x-axis parallel to the y-axis, and with distance from focus to vertex fixed as "a". 2 Solution: ( y−k ) =4 a ( x−h ) ; V(h,0)

( y−0)2=4 a ( x−h) 2 y y ' =4 a ( 1−0 ) y y' =2a ydy=2 adx

8.) Obtain the DE of the Family of parabolas with axis parallel to x-axis. 2

Solution: ( y−k ) =4 a ( x−h ) 2 ( y−k ) y ' =4 a ( 1−0 )

( y−k ) y ' =2 a 2

y y' ' −k y ' ' + ( y ' ) =0

' 2

''

y y +( y ) k= '' y 2

2

3 y ' ( y ' ' ) −( y ' ) y ' ' ' =0 '' 2

'

'' '

3( y ) − y y =0 9.) Find the DE of the family of SL with algebraic sum of intercepts fixed as k. Solution:

x y + =1 , where a+b=k a b x y + =1 k−b b 1 ( k−b ) ( x )+( 1b ) ( y )=1 1 y' + =0 k−b b b=

−k y ' 1− y '

x y + =1 ' ' ky −k y k+ ' ' 1− y 1− y

x ( 1− y ' ) y ( 1− y ' ) − =1 k−k y ' +ky ' k y' x y ' ( 1− y ' )− y ( 1− y' ) =k y ' I.

Variable Separable General Form: M(x,y)dx + N(x,y)dy = 0 Standard From: f(x)dx+g(y)dy = 0

Steps: -

Reduce the general formula of the equation to standard form. Integrate both sides of the standard equation, thus, the General Solution (GS) is: ∫ f(x)dx + ∫ g(y)dy = C (Where “C” is the constant of integration.)

Examples: 1.)

dy dx

= 80xy

Solution: dy = 80xydx 80 xydx −dy=0 y

∫ 80xdx - ∫

dy y

(General Form)

= ∫ 0 (Standard Form)

2

x 80 ( 2 ¿ - lny = C 40 x

2

- lny = C (General Solution)

Or 40 x

2

- lny + lnC= 0

Or y 2 40 x – ln c

=0

2 2.) (xy + x)dx + 2 x ydy = 0

Solution:

x ( y +1 ) dx+2 x 2 y dy=0 ( y +1 ) ( x 2 )

(General Form)

∫

dx 2 ydy +∫ = 0 (Standard Form) x y +1 ∫

[

ln x+∫ 2 1−

]

1 dy=C y +1

ln x+ 2 y −2 ln ( y +1 ) =C Or ln x+ 2 y −ln ( y +1 )2=C 3.)

6 x 5 ( 1+ y 2 ) dx=dy 6 x 5 ( 1+ y 2 ) dx−dy=0 2 1+ y

Solution:

(General Form)

dy

∫ 6 x 5 dx−∫ 1+ y 2 =∫ 0 6

(Standard Form)

x6 −tan −1 y+C=0 6

( )

x 6−tan−1 y +C=0 (General Solution) 4.)

2

3

tan ydy=sin xdx

Solution:

∫ sin3 xdx−∫ tan2 ydy =∫ 0

(Standard form)

x x sec ¿ y −1 ¿2¿ ¿ dy =∫ 0 ¿ ¿ sin ¿ dx−∫ ¿ sin2 ¿ ¿ ¿ ∫¿ x x sin ¿ dx−tan y + y=C ¿ 1−cos 2 ¿ ¿ ¿ ∫¿ xdx sin ¿−tan y + y=C ¿ ¿ x¿ cos 2 ¿ xdx−∫ ¿ sin ¿ ∫¿ Let u = cosx

du = -sinxdx 3

cos x −cos x+ −tan y + y =C 3 −3 cos x +cos 3 x −3 tan y +3 y=3 c

II.

Homogenous Differential Equations. Standard Form: M(x,y)dx + N(x,y)dy = 0, where M(x.y) and N(x,y) are homogenous of same degree. Note: A funtion f(x,y) is homogenous if (rx,ry) = r

n

· f(x,y), replacing rx and ry w/ x and y.

Example: Test for Homogenity f (x , y) =

1.)

x 2+5 xy

f ( rx , ry ) = ( rx )2+ 5 ( rx )( ry ) f ( rx , ry )=r 2 x 2+5 r 2 xy 2 2 f ( rx , ry )=r (x +5 xy)

Homogeneous of 2nd degree; r 2.)

n

,where n is degree

f ( x , y ) =x3 −9 x 2 y f ( rx , ry )=r 3 x 3−9 r 3 x 2 y ¿ r 3 (x 3−9 x 2 y ) Homogeneous of 3rd degree

3.)

f ( x , y ) = y−xtany f ( rx , ry )=ry −( rx ) (rtany) ¿ ry−r 2 xt any ¿ r ( y−rxtany)

Not homogeneous Steps: 1.) Choose the simpler term for dy or dx

2.) Let y=vx or x=vy if the given ODE is homogenous of the same degree 3.) Take the differentials: dy = vdy + xdv or dx = vdy + ydv 4.) Substitute in the given ODE 5.) Reduce the equation to a form solvable by variable separable 6.) Express the general solution in terms of the original variable of the given ODE 7.) Solve for the particular solution of any given conditions Examples: 1.)

2 2 xydx−( x +3 y ) dy=0 (Homo of 2nd degree)

Solution:

Let x=vy ; dx =vdy + ydv

[ ( vy ) ( y ) ] ( vdy + ydv )−[ ( vy )2+ 3 y 2 ] dy =0 2

2

3

2

2

2

v y dy + v y dv−v y dy−3 y dy=0 v y3 dv−3 y 2 dy =0( VS) 3 y 2

vdv −

3 y dy =0 y3

∫ vdv−3∫ 2

v −3lny=C 2 v 2−6lny=C but v =

x y

dy =0 y

x y ¿ ¿ ¿ x 2−6 y 2 lny=C y 3 2.)

( x 2+ 6 y 2) dx −2 xydy =0 ; y ( x )=2 (Homo of 2nd degree) y=vx ; dy=vdx + xdv

Solution: Let

¿ [ x 2+3 ( xv )2 ] dx−[ 2 x ( vx ) ]( vdx + xdv) 2

2 2

2

2

3

¿ x dx +3 x v dx−[2 x v dx +2 x vdv ] ¿ x 2 dx +3 x 2 v2 dx−2 x 2 v 2 dx−2 x 3 vdv ¿ x 2 dx + x 2 v 2 dx−2 x 3 vdv

¿

x 2 ( 1+ v 2 ) dx−2 x 3 vdv ( 1+ v 2 ) (x 3)

∫

dx 2v −∫ dv=∫ 0 2 x (1+v )

lnx−2 ∫

vdv =C ; u=1+ v 2 ; du=2 vdv 2 1+ v

lnx−ln ( 1+ v 2 )=lnC x

e

ln 1+v

2

=e lnC

x y =C ; but v= 2 x 1+ v

x y 2 1+( ) x

=C

x3 =C x2 + y2 III.

Exact Differential Equation Standard Form: M(x,y)dx + N(x,y)dy = 0 General Solution: P(x,y) = 0 Steps: Check if the equation is exact: ∂M ∂N = ∂ y x=k ∂ x y=k

Integrate equation partially with respect to x (or y) holding y (or x) as constant: ∂F =M (x , y ) ∂ x y=k

or

∂F =N (x , y) ∂ y x=k

∫ dF=M ( x , y ) dx

or

∫ dF=N ( x , y ) dy

F=f ( x , y ) +T ( y )

or

F=f ( x , y ) +T (x)

Differentiate equation partially with respect to y (or x) holding x (or y) as constant: ∂F ∂ = [f ( x , y ) +T ( y ) ] ∂Y ∂ y

or

Integrate T’(y) or T’(x) to obtain T(y) or T(x): Substitute in to obtain the General Solution Examples: 1.)

( 4 x3 y 3−2 xy ) d x + ( 3 x 4 y 2−x 2 ) dy=0

∂F ∂ = [f ( x , y ) +T ( x ) ] ∂ X ∂x

∂M =4 x 3 y 3 −2 xy ∂ y x=k

Solution:

3

∂N =3 x 4 y 2−x 2 ∂ x y=k

2

¿ 4 x (3 y )−2 x

3

2

3

¿ 12 x y −2 x

∂F =M (x , y ) ∂ x y=k

∫ dF y=k =∫ (4 x3 y 3−2 xy)dx x4 x2 )−2 y ( )+T ( y) 4 2

F=x 4 y 3−x 2 y +T ( y) ∂F ∂ 4 3 2 = [ x y −x y +T ( y ) ] ∂ y x=k ∂ y

3y (¿¿ 2)−x 2+T ' ( y ) 3 x 4 y 2−x 2=x 4 ¿

∫T ' ( y)

∫0

=

T ( y) =C 4

3

2

F=x y −x y +C Method 2:

2

¿ 12 x y −2 x

Method 1:

F=4 y 3 (

4x 2 (¿¿ 3) y −2 x ¿ 3¿

Check for exactness:

∂F =N (x , y) ∂ y x=k

∫ dF x=k=∫ (3 x 4 y 2−x 2) dy F=3 x 4 (

3

y )−x 2 ( y)+T ( x) 3

F=x 4 y 3−x 2 y +T ( x) ∂F ∂ 4 3 2 = [x y −x y +T ( x ) ] ∂ x y=k ∂ x

4x 3 (¿¿ 3) y −(2 x) y +T ' ( x ) 4 x 3 y 3−2 xy=¿

∫T ' ( x)

=

∫0

T (x ) =C F=x 4 y 3−x 2 y +C Method 3: ❑

( 4 x3 y 3−2 xy ) dx+ ¿ ∫ (3 x 4 y 2−x 2 )dy x=0

❑

∫¿

y=k

4 y3

4

x4 x2 −2 y +C=0 4 2

( ) ( ) 3

2

x y −x y +C=0 Method 4:

❑

( 4 x3 y 3−2 xy ) dx+ ¿ ∫ (3 x 4 y 2−x 2 )dy x=k

❑

∫¿

y=0

3 x4

y3 −x 2 ( y ) +C=0 3

( )

x 4 y 3−x 2 y +C=0 2.)

( 2 x3 −xy 2−2 y+ 3 ) dx−( x 2 y +2 x ) dy=0 ∂M 3 2 =2 x −xy −2 y +3 ∂ y x=k

Solution:

∂N 2 =−x y −2 x ∂ x y=k

¿−x ( 2 y )−2

¿− y (2 x )−2

¿−2 xy−2

¿−2 xy−2

Method 1:

∫ dF y=k=∫ (2 x3 −xy 2−2 y+ 3) dx F=

x4 x2 y2 − −2 xy +3 x +T ( y ) 2 2

∂F ∂ x4 x2 y2 = [ − −2 xy +3 x+ T ( y )] ∂ y x=k ∂ y 2 2 2

2 −x y−2 x = −x y−2 x ( 1 ) +T ' ( y)

∫T ' ( y) T ( y) =C

=

∫0

Check for exactness:

F=x 4−( xy)2 −4 xy +6 x +C

Method 2:

∫ dF x=k=∫ (−x 2 y −2 x )dy F=

−x 2 y 2 −2 xy+ T (x ) 2

∂F ∂ −x 2 y 2 = [ −2 xy +T (x )] ∂ x y=k ∂ x 2 3

2

2

2 x −xy −2 y +3=−xy −2 y+T '( x )

∫ 2 x3 +3

∫T ' ( x)

=

T (x ) =

x +3 x+ C 2

4

4

2

F=x −( xy) −4 xy +6 x +C Method 3: 2 x 3−xy 2−2 y +3 ❑

(¿)dx− ∫ ( x 2 y+ 2 x ) dy x=0 ❑

∫¿

y=k

x4 x2 y2 − −2 xy +3 x+C=0 2 2 x 4−( xy )2−4 xy +6 x +C=0 Method 4:

3

2

2 x −xy −2 y +3 ❑

(¿)dx− ∫ ( x 2 y+ 2 x ) dy x=k ❑

∫¿

y=0

x4 x2 y2 − −2 xy +3 x+C=0 2 2 x 4−( xy )2−4 xy +6 x +C=0

3.)

( 2 xy−tan ( y ) ) dx + ( x 2−x sec2 y ) dy=0 ∂M =2 xy−tan ( y ) ∂ y x=k

Solution:

∂N 2 2 =x −x sec y ∂ x y=k

¿ 2 x−sec 2 y

¿ 2 x−sec 2 y

Method 1:

∫ dF y=k=∫ (2 xy−tan ( y ) )dx F=x 2 y−xtan ( y )+ T ( y ) ∂F ∂ 2 = [ x y −xtan( y )+T ( y)] ∂ y x=k ∂ y x 2−x sec 2 y =

x 2 ( 1 )−x sec 2 y +T '( y)

∫ T ' ( y ) =∫ 0 T ( y )=C F=x 2 y−xtan ( y )+ C Method 2:

Check for exactness

∫ dF x=k=∫ ( x 2−x sec2 y)dy F=x 2 y−xtan ( y )+ T (x ) ∂F ∂ = [x 2 y−xtan( y )+T ( x )] ∂ x y=k ∂ x 2 xy −tan ( y )=2 xy−tan ( y ) +T ' ( x )

∫ T ' ( x )=∫ 0 T ( x )=C F=x 2 y−xtan ( y )+ C Method 3: 2 xy−tan ( y ) ❑

(¿)dx− ∫ ( x 2−x sec 2 y ) dy x=0

❑

∫¿

y=k

x 2 y−xtan ( y )+ C=0 Method 4: 2 xy−tan ( y ) ❑

(¿)dx− ∫ ( x 2−x sec 2 y ) dy x=k

❑

∫¿

y=0

x 2 y−xtan ( y )+ C=0

IV.

Integrable Combination 2 1.) xdy + ydx=3 x dx ∫ d ( xy )= ∫ 3 x 2 dx

Solution:

3

x xy=3 3 xy=3

xy−x 3 +c=0 2.) y(2xy + 1) dx – xdy 2 xy 3 dx+ ydx−xdy=0 2 Solution: y 2 xdx +

ydx−xdy =0 2 y

∫ 2 xdx+ ∫ d

( xy )=0

2

2

( )

x x + =c 2 y

2 x x + =c y

x 2 y + x +cy=0

3.)

x 2 y 2 ( ydx+ xdy ) + 2

dx =0 x

Solution : ( x 2 y 2 ) d ( xy ) +

dx =0 x

( xy )2 d ( xy ) +

dx =0 x

let u=xy du=xdy+ ydx=d ( xy )

∫ u2 du+ ∫

dx =∫0 x

u3 +lnx+C=0 3 u3 +3 lnx+C=0 (xy)3 +3 lnx +C=0 3

3

x y +3 lnx+c=0 V.

Homogenous Leading to Exact

Standard Form:

M ( x , y ) dx + N ( x , y ) dy=0

(Homogeneous)

Steps: a. Check if the given ODE is a Homogeneous ODE. 1 μ= Mx+ Ny ≠ 0 ; μ = Integrating factor b. Let Mx+ Ny ; where μ to the given ODE d. Check for exactness; ( μ )(ODE) = EXACT

c. Multiply

e. Proceed as in solving for exact ODE f. Solve for the PS if there is a given condition Examples: 1.)

xydx−( x 2 +2 y 2) dy =0 (Homo of 2nd degree)

Solution:

(

μ=

1 ( xy ) x + (−x 2−2 y ) y

μ=

1 2 3 x y− x y−2 y

μ=

−1 2 y3

2

xydx−( x 2 +2 y 2) dy =0 ¿(

−1 ) 3 2y

−x x 2 +2 y dx + dy=0 2 3 2y 2y

(

)

∂ M −x x = = ∂ y 2 y 2 y3

;

∂ N x 2+ 2 y x = = 3 3 ∂x 2y y

∂M x ∂N x 2+2 y ∨ y=k ∫ – 2 dx+ ∨x=0 ∫ dy ∂x ∂y 2y 2 y3 −1 x 2 x2 1 + ∫ dy +∫ dy 2 3 y 2y 2 2y

( )

−x 2 + ln ( y )+C=0 2 4y VI.

Determination of Integrating Factor (For Non-Exact Ordinary Differential Equation) Standard Form: M(x,y)dx + N(x,y)dy = 0 Steps: a) Check for exactness of the given O.D.E.

(ODE IS EXACT)

b) If the ODE is not exact, then let

∅=

∂M ∂N − ∂ y ∂x

c) Multiply ∅ with 1 ∅=f (x) N

1 M

or

or

1 N

to obtain f(x) of g(y), that is:

1 ∅=g( y ) M

* if not

1 ∅ , then use N

1 ∅, vice M

versa d) Obtain the integrating factor (µ) ∫ µ= e

f ( x ) dx

∫ µ= e

−g ( y ) dy

or

e) Multiply µ with the given O.D.E. f) Check for exactness g) If exact, proceed as solving for exact O.D.E. µ (O.D.E.) = Exact D.E. Examples: 1.)

y ( 2 x + y 3 ) dx−x ( 2 x− y3 ) dy =0

Solution:

∂M 3 =2 x +4 y ∂ y x=k

* O.D.E. is not exact ∅=¿

2 x + 4 y 3−(−4 x+ y 3 )

∅=¿

6 x+ 3 y 3

∅=¿

3 (2 x + y3)

∂N 3 =−4 x+ y ∂ x y=k

Check for exactness

2 x+ y 3 ( y )(¿¿ 3)3 ( 2 x+ y ) 1 1 ∅= ¿ M

2 x− y 1 (−x)(¿¿ 3)3 ( 2 x + y 3 ) ∅ = N 1 ¿

= f (x,y)

g (y)

¿

3 y

[should only be f (x)] µ= e

dy ∫ −3 y

−3 ln ⁡(y) µ= e

µ=

y−3

1 [ y ( 2 x + y 3 ) dx−x ( 2 x− y 3 ) dy =0 ] 3 y

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

Check for exactness again

∂M −2 =2 x y + y ∂ y x=k −3 = −4 x y +1

∂N 2 −3 =−(2 x y −x) ∂ x y=k −3 = −4 x y +1

* O.D.E. is exact ❑

( 2 x y + y ) dx−¿ ∫ (2 x2 y−3−x)dy=0 −2

x=0 ❑

∫¿

y=k

2

3

2

x + xy + cy =0

2.)

( x 3 y 3 +1 ) dx + ( x 4 y 2 ) dy=0 ∂M 3 2 =3 x y ∂ y x=k

Solution:

∂N 3 2 =4 x y ∂ x y=k

O.D.E. is not exact 3

∅=¿

2

3

3 x y −4 x y

2

3 2 = −x y

1 ∅ = N

1 (−x 3 y 2 ) 2 x y 4

−1 x

=

dx ∫ −1 x

µ= e

−ln ⁡( x) µ= e

µ=

x−1

x−1 [( x 3 y 3+1 ) dx + ( x 4 y 2 ) dy=0]

( x 2 y 3 + x−1 ) dx+ ( x3 y 2 ) dy=0 Check for exactness again ∂M 2 3 −1 =x y + x ∂ y x=k 2 2 = 3x y

O.D.E. is exact

∂N 3 2 =x y ∂ x y=k 2 2 = 3x y

Check for exactness

x (¿ ¿ 3 y ) dy=0 2

2

3

❑

( x y + x ) dx−¿ ∫ ¿ −1

x=0

❑

∫¿

y=k

x 3 y 3 +3 l n ( x ) +C=0 VII.

Linear Differential Equation dx + y∗P ( x )=Q ( x ) (Linear in ‘y’) dy

Standard Form:

General Form: Where

yμ=∫ Q ( x )∗μ dx +C

μ=e P ( x ) dx

OR dx + x∗P ( y )=Q ( y ) (Linear in ‘x’) dy

Standard Form:

General Form: Where

xμ=∫ Q ( y )∗μ dy+C

μ=e P ( y )dy

RECALL: Conditions in Linearity: a. The variable y(or x) and all its derivatives must be of degree one. b. There is no product of y(or x) and its derivatives c. There is no transcendental functions involving y(or x) Examples: 1.

y

( dxdy )+3 x−xy +2=0

(Linear in ‘x’)

dx 3 x−xy 2 + + =0 dy y y

Solution:

dx 3 x− y −2 +x = dy y y

(

P ( y )=

)

(STANDARD FORM)

3− y −2 ; d ( y )= y y

∫ 3 −y y dy

μ=e

−∫ dy ∫ 3 dy y

=e

=e3 ln ( y )− y

3

e ln ( y ) e (−y )= y 3 e−y

x y 3 e− y =∫

( −2y ) ( y e

3 −y

) dy +C (GENERAL FORM)

x y3 2 −y =−2 ∫ y e dy +C ey x y3 −y 2 −y −y =−2 (−e y −2 y e −2 e ) +C y e x y 3=−2 y 2+ 4 y + 4+C e y VIII.

Bernoulli’s Equation Std. Form:

dy + yP ( x )= y n Q(x) dx

∫ Where: u=e

( 1−n) P ( x ) dx

General Formula: Or

n ≠ 0,1

;Q ( x )=k ; F ( x)

y 1−n u=(1−n)∫ Q ( x ) udx +C

Std. Form:

dx + xP ( y )=x n Q( y ) dy

∫ Where: u=e

( 1−n) P ( y ) dy

General Formula:

y

n ≠ 0,1

; Q ( y )=k ; F ( x)

1−n

u=(1−n)∫ Q ( x ) udx +C

Example: 1.)

y ( 6 y 2−x−1 ) dx +2 xdy =0

Solution:

1 ( y ( 6 y 2 −x−1 ) dx+ 2 xdy=0) 2 xdx dy 6 y 3−xy− y + =0 dx 2x dy 6 y 3 −xy − y + + =0 dx 2 x 2x dy −xy − y −3 y 3 + = dx 2x x dy −x−1 3 + y( )=− y 3 ( ) dx 2x x P ( x )=

u=e∫

−x−1 −3 ; Q ( x )= ; n=3 ; 1−n=1−3=−2 2x x

( 1−n ) P ( x ) dx

dx ∫ −2( −x−1 2x )

u=e

dx ∫ ( x +1 x )

u=e

∫ dx+∫ dxx

u=e

u=e

x+lnx

x

u=e e u=xe

lnx

x

General Formula: −2

x

y xe =−2∫ ( x

−3 x )( xe )dx +C x

x

y xe =6∫ (e )dx +C −2

xe x =6 e x +C 2 y xe x =6 y 2 e x + y 2 C x

2

x

2

xe −6 y e − y C=0

IX.

Coefficient in Linear in Two Variables Standard form: Where:

a1 x b1 y +c 1 ¿ dx+ ( a2 x +b 2 y +c 2 ) dy=0 ¿ +

a1 , b1 , c1 , a2 , b2 ,c 2 →constants

Method of solution a1 x

+ b1 y +c 1=0 → L1 =M (x , y)

a2 x+ b2 y + c2 =0 → L2=N ( x , y) a1 b 1 ≠ ∨L1 intercepts at L2 at P(h , k) a2 b 2

Case 1: If

Then, Let

x=u+h ; dx =du

y=v+ k ; dy=dv

Where h and k can be obtained from the values of x and y whenever the z lines are equated. Substitute in the given ODE to form a homogeneous DE introduce a new variable, say w…. let u = wv a1 b 1 = ∨L1 /¿ L2 a2 b 2

Case 2: If

Then, Let

u=a1 x +b 1 y∨u=a2 x+ b2 y

du=a1 d +b1 dy∨du=a 2 dx +b 2 dy (Whichever simpler). Substitute in the given ODE to form an equation solvable by variable separable. Examples: 1.) ( x+ y+1 ) dx + ( 2 x+2 y +1 ) dy=0 a1 1 b 1 1 a1 b1 1 1 = = ≈ = ≈ = (use case2) a2 2 b 2 2 a2 b2 2 2

Solution:

Let u=x + y → x =u− y

du=dx +dy →dx=du−dy ( substitute )

[

( ¿+

u− y [ 2 ( u− y )+ 2 y +1 ) ] dy =0

y+1 ] ( du−dy )+

( u+1 ) ( du−dy )+ ( 2u+1 ) dy=0 ( u+1 ) du−( u+ 1 ) dy + ( 2 u ) dy + ( dy )=0 ( u+1 ) du−udy −dy +2udy + dy=0 ( u+1 ) du+udy=0 → VS ( u+ 1 ) du+ udy =0 u

∫

(u+1) du+∫ dy=∫ 0 u

u+lnu+ y+ c=0 ; but u=x+ y x+ y+ ln ( x+ y ) + y +C=0 x+ 2 y + ln ( x + y ) +C=0 2.)

( 2 xy − y ) dx+ ( 4 x + y−6 ) dy =0 ( eq . A )

Solution:

a1 2 b 1 −1 2 −1 = = ≈ ≠ ( use case 1 ) a2 4 b 2 1 4 1 ¿ solve h∧k :equate L1∧L2 ¿ solve h∧k :equate L1∧L2 L1=2 x− y=0 L2=4 x + y−6=0 6 x=6 x =1=h 2 ( 1 )− y=0 2= y =k Let x=u+h=u+1(eq . B)→ dx=du(eq . D)

y=v+ k → v +2 ( eq .C ) →dy =dv (eq . E) substitute eq . B , C , D , E∈eq . A ) ¿

[ 2 ( u+ 1 )−( v +2 ) ] du+ [ 4 ( u+1 ) + ( v +2 ) −6 ] dv=0 ( 2u+ 2−v−2 ) du+ ( 4 u+4 +v +2−6 ) dv=0 ( 2u−v ) du+ ( 4 u+ v ) dv=0 ( homo of 1 st degree ) Let u=vw ; du=vdw+wdv

[ 2 ( vw ) −v ] ( vdw+ wdv )+ [ 4 ( vw ) + v ] dv=0 2 v 2 wdw−v 2 dw+ 2 v w2 dv−vwdv+ 4 vwdv +vdv =0 v 2 ( 2u−1 ) dw+ v ( 2 w 2+3 w+1 ) dv=0 ( VS ) v 2 ( 2 u−1 ) dw+ v ( 2 w2 +3 w+ 1 ) dv=0 v 2 ( 2 w2 +3 w+ 1 )

∫ ∫

( 2 w−1 ) dw dv +∫ =∫ 0 2 v 2 w −3 w+1 ( 2 w−1 ) dw 2

( 2 w +1 ) ( w+1)

+∫

dv =∫ 0(by ∂ fraction) v

( 2 w−1 ) dw A B = + ( 2 w +1 ) (w+1) 2 w+1 w +1 2 w−1= A ( w+1 )+ B ( 2 w+1 ) 2 w−1= Aw + A +2 Bw +B

for w :2= A+ 2 B k : (−1+ A+ B )−1 3=B an d A=−4 −4 dw

3 dw

∫ 2 w+1 +∫ w+1 +∫

dv = 0 v ∫

−2 ln ( 2 w+1 ) +3 ln ( w+1 ) +lnv=lnC 2

3

−ln (2 w +1 ) + ln ( w+1 ) +lnv+lnC 3

e

ln

v(w+1 ) 2 (2 w+1)

=e lnC 3

v (w+ 1) u =C ; but w= ; but u=x−1, v= y−2 2 v ( 2 w+ 1 )

[

x−1 ( y −2 ) +1 y −2 2

[( ) ] x−1 2 +1 y−2

3

]

=C

( x+ y −3 )3 =C ( 2 x + y−4 )2 ( x+ y−3 )3=C ( 2 x + y−4 )2

X.

Simple Substitution Standard Form: M(x,y)dx + N(x,y)dy = 0 Let u = f(x,y)

v = g(x,y) General Form: f(x) dx +g(y)dy = 0 Steps: a) b) c) d) e)

If a new variable is introduced, remove only one variable If two new variables are introduced, remove both the original variable Choose the simpler term for dx or dy Form an equation solvable by variable separable Express the general solution in terms of its original variables

Examples: 1.) 2dx + (2x + 3y)dy = 0 Solution: Let u =2x + 3y; du = 2dx + 3dy 2dx = du – 3dy 2dx + udy = 0 du – 3dy + udy = 0 du + (u-3) dy = 0 du

∫ u−3 ∫ dy=∫ 0 ln(u-3) + y + C = 0 ln (2 x +3 y – 3)+ y +C=0 2.) (2x + y + 6) dx + (2x + y) dy = 0 Solution: Let u = 2x + y; du = 2dx + dy dy = du -2dx (u + 6) dx + u(du - 2dx) = 0 (u + 6 – 2u) dx + udu = 0 (-u + 6)dx + udu = 0 udu

∫ dx ∫ u−6 =∫ 0

−x + y – 6 ln(2 x + y – 6)+C=0 3.)

x 2 y 3 ( xdy + ydx ) =( xy+ 1 ) dx

Solution: Let u = xy ; du = xdy + ydx u x

y=

x 2 y 3 du= (u+ 1 ) dx 3

u du ( − u+1 ) dx=0 x u3 du ∫ u+1 −∫ xdx=∫ 0 u 1 x2 (¿ ¿ 2−u+ 1− )du− =C u +1 2 ∫¿ 2( xy )3 −3 ( xy )2+6 ( xy ) −6 ln ( xy+ 1 )−3 x 2+C=0 4.)

2 3 x y ( xdy + ydx ) =( xy+ 1 ) dy

Solution: Let u = xy ; du = xdy + ydx x=

u y

;

dy=¿

du− ydx x

x 2 y 3 ( du )=( u+ 1 ) dy u (¿¿ 2 y) du−( u+1 ) dy =0 ( u+1 ) ( y) ¿

2

u du dy +¿ ∫ =∫ 0 ( u+ 1 ) y ∫¿

( u−1 ) du+∫

du dy + ¿∫ =∫ 0 u+1 y ∫¿

u2 −u+ ln ( u+1 ) +ln ( y )=C 2 (xy)2 −2 xy +2 ln ( xy +1 ) +2 ln ( y )+C=0

5.)

x xy ( xdy − ydx )=( +1) dy y

Solution: Let

u=

x y ; du =

x=uy

ydx −xdy 2 y

;

( u y 4 ) du+(u+1)dy udu dy +¿ ∫ 4 =∫ 0 ( u+ 1 ) y ∫¿ u−ln (u+ 1 )+

1 =C 3 3y

−3 xy 2 +3 y 3 ln

( xy +1)+1+C y =0 3