differential equations - ode of first order
DESCRIPTION
(c) Engr. Reynaldo Ted Penas IIITRANSCRIPT
Differential Equations
Ordinary Differential Equations of Order OneOrder One
Ordinary Differential Equations of Order One
1. Variable-Separable Equations
2. HomogeneousEquations2. HomogeneousEquations
3. Exact Equations
4. Linear Differential Equations of the FirstOrder
General Form of Ordinary Differential Equations of the First Order
Consider the form
where bothM andN can be functions ofx, y, orbothx andy.
( ) ( ), , 0M x y dx N x y dy+ =
bothx andy.
Variable-Separable Equations
Given
If this equation can be expressed as
( ) ( ) 0A x dx B y dy+ =
( ) ( ), , 0M x y dx N x y dy+ =
then it is a variable-separable equation.
( ) ( ) 0A x dx B y dy+ =
Examples
Problems:
( )2
1. sin sin cos cos 0
2. cos sin
3. cos tan 0
x ydx x ydy
dr b dr r d
x ydx ydy
θ θ θ+ =
= +
+ =
Examples
Answers:
( )2 2 2
1. sin cos
2. 1 cos
3. tan
y C x
r C b
x y C
θ=
= −
+ =
Homogeneous Equations
Given
If each termof the equation has a total degree ofn (sum of exponents of all the variables in aterm), then the equation is a homogeneous
( ) ( ), , 0M x y dx N x y dy+ =
term), then the equation is a homogeneousdifferential equation of degreen.
Homogeneous Equations
To solve a homogeneous equation, one maychooseto substitutechooseto substitute
or
An advantagemay be gained if M has fewer
x vy dx vdy ydv= = +
y vx dy vdx xdv= = +An advantagemay be gained if M has fewer
terms thanN andx = vy is chosen. Same goesfor N has fewer terms andy = vx. The resultingequation becomes variable-separable.
Homogeneous Equations
Theorem 1. If M(x,y) and N(x,y) are bothhomogeneousand of the same degree, thehomogeneousand of the same degree, thefunction M(x,y)/N(x,y) is homogeneous ofdegree zero.
Theorem 2. Iff(x,y) is homogeneous of degreezeroin x andy, f(x,y) is aunctionof y/x alone.zeroin x andy, f(x,y) is aunctionof y/x alone.
( ) ( ) ( ) ( )0, , 1, 1,f x y f x vx x f v f v= = =
Examples
Problems:
( )2 21. 3 3 2 0x y dx xydy+ − =( )( )
( )
2 2
2 2
1. 3 3 2 0
2. 3 0
3. csc 0yx
x y dx xydy
xydx x y dy
x y dx xdy
+ − =
+ + =
− + = ( )3. csc 0xx y dx xdy − + =
Examples
Answers:
( )3 2 21. 9x C x y= +( )( )
( )
3 2 2
32 2 2 2
1. 9
2. 4
3. ln cos yxc x
x C x y
y x y C
= +
+ =
= ( )3. ln cosc x=
Exact Equations
Given
If the following partial differentials are equal,
( ) ( ), , 0M x y dx N x y dy+ =
M N∂ ∂=∂ ∂
then it is an exact differential equation.y kx k
y x ==
=∂ ∂
Exact Equations
To solve an exact differential equation, set
Then solve for F by integrating one of thefunctions with respect to its partial differentialindependentvariable (with the other variable
or F F
M Nx y
∂ ∂= =∂ ∂
independentvariable (with the other variabletreated as constant.
Exact Equations
If M was initially chosen, setT’(y) with functionterms of N with y variablesonly. If N wasterms of N with y variablesonly. If N wasinitially chosen, setT’(x) with function termsof M with x variables only. SolveT byintegrating the function obtained.
Thesolutionis thenThesolutionis then( ) ( )
( ) ( )
,
or
,
F x y T x C
F x y T y C
+ =
+ =
Exact Equations
Tips and tricks: the shortcuts
or
( ) ( ),y k
M x y dx N y dy C=
+ =∫ ∫
( ) ( ),x k
M x dx N x y dy C=
+ =∫ ∫
Examples
Problems:
( ) ( )( ) ( )( ) ( )
3 2 3 2
2 2
1. 2 cos cos 0
2. 0
3. 2 tan sec 0
x y xy dx x xy dy
w wz z dw z w z w dz
xy y dx x x y dy
+ + =
+ − + + − =
− + − =
Examples
Answers:
( )( )
2
22 2
2
1. sin
2. 4
3. tan
x xy C
w z wz C
x y x y C
+ =
+ = +
− =
Linear Differential Equation of the First Order
Given
( ) ( )+ =If this equation can be expressed as
( ) ( ), , 0M x y dx N x y dy+ =
( ) ( )
( ) ( ) or
dy yP x dx Q x dx+ =
+ =then it is a linear differential equation of the first
order.
( ) ( )dx xP y dy Q y dy+ =
Linear Differential Equation of the First Order
To solve the linear differential equation of thefirst order,determinetheintegratingfactorbyfirst order,determinetheintegratingfactorby
Then solve the equation
( ) ( ) or
P x dx P y dyv e v e∫ ∫= =
( )vy vQ x dx C= +∫
( ) or
vx vQ y dy C= +∫
Examples
Problems:
( )( )
2
1. ' csc cot
2. 2
3. 1 2 tan 0
y x y x
y y x dy dx
dx x y dy
= −
− =
− + =
Examples
Answers:
22
2
1. sin
2. 1
3. 2 cos sin cos
y
y x x C
x y Ce
x y y y y C
−
= +
= − += + +