differential equations - ode of first order

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

−

= +

= − += + +