MA1512 Cheatsheet - Summary Differential Equations for Engineering PDF

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Summary

1 First Order Differential Equations1 Separable Equations y'=M(x)N(y)Separate variables by moving dependent variables to RHS and independent variables to LHS∫1N(y)dy=∫M(x)dx1 Reduction to separable form y'=f(ax+by+c) 1. Let u=ax+by+c 2. du dx=d dx(ax+by+c)=a+bdy dx=a+bf(u)1 Integrating Factor dy dx+...

Description

1

2.1

First Order Differential Equations

1.1

Separable Equations '

Separate variables by moving dependent variables to RHS and independent variables to LHS

Case 2 Real Distinct Roots

( λ=a ± bi )

2.

1.3

2.2

u= ax + by + c dy du d = (ax + by + c )= a+b =a+bf (u) dx dx dx

Let

dy + P ( x ) y =Q ( x ) dx

solution of the nonhomogeneous equation,

2.2.1

2.

1.4

∫ P ( s ) ds

Ry=∫ RQ

1.

1.

y ' + p ( x ) y=q ( x ) y n

2.

where n is any real number. If otherwise it is nonlinear. 1.

Rewrite this as

y

1−n

−n

y

n=0∨1

'

y +y

1−n

, the equation is linear,

p ( x ) =q(x )

and set

=z ( 1−n ) y −n y ' =z '

2.

Then,

3.

The given equation becomes

3.

A.

B.

'

Second Order Differential Equations ' y +P ( x ) y +q ( x ) y=F (x)

yp

'

u=

3

y= A x +B x +Cx + D (k ≠ 0)

Try

3.1

kx

C.

''

λ1

or

λ2

e =cos ax +i sin ax

iax

, a particular solution is given

|

− y2 F ( x )

Solve for value

'

v=

y1 F ( x )

y 1 y 2' − y 1' y 2 y1 F ( x) v =∫ dx y 1 y'2− y 1' y 2

θ

when

'' f ( a) ( x−a )2+ … 2!

´θ=0

a to replace sin f ( θ) cos f (θ ) to get ´θ=B ( θ−a ) ´ . Therefore . ´ϕ=θ a ≠ 0 , let =θ−a ϕ ´ϕ=Bϕ

Use Taylor’s Theorem around

3.

If

Unstable Equilibrium

´ϕ= g ϕ L

( √gL )t + B e−(√

θ= A e A

(Equilibrium Point). Call this

a

2.

, guess

u to be a polynomial with highest power same as p (x ) If the value of k tallies with either simple root λ1 or λ2 , guess u to be a polynomial with highest power +1 of highest power of p (x ) If the value of k tallies with repeated root λ , guess u to be a polynomial with highest power +2 of highest power of p (x )

Trigonometric Case iax

is neither

'

y2 = y 1 y'2− y 1' y 2 y2 '

Pendulum 1.

k

3.

''

z + z + z =e iax z=u e y=ℜ z

Try

Oscillation

y=u e ' ' kx kx y =u e + ku e '' '' kx ' kx 2 kx y =u e + 2 k u e + k u e

If the value of

y1 y1 '

f ( x )=f ( a ) + f ' ( a)( x−a ) +

2

2.

y 1 y 2' − y 1' y 2 y 2 F (x ) u=−∫ dx y 1 y'2− y'1 y 2

OR

z + ( 1−n ) p( x ) z=( 1−n ) q ( x ) 2

Try

Exponential Case

Reduction to linear form

For a Bernoulli equation of the form

|

kx

3

1.

y p ( x ) =u ( x ) y 1 ( x ) +v ( x ) y2 (x )

, and a

(k =0)

iax

by,

y (x )= y h ( x ) + y p ( x )

Polynomial Case

R ( x ) =e

yh

'

y h (x ) =c 1 y 1( x )+c 2 y 2( x )

Given

Method of undetermined coefficients

F ( x ) = p ( x )e

Define a new function

'

Method of variation of parameters

W=

Find the general solution of the homogeneous equation,

x

1.

''

Try

3.

Nonhomogeneous Equations

General Solution:

Integrating Factor

λ2 x

λ1 x

Repeated Roots

'

1.

General Solution

y =c1 e + c 2 e y= ( c1 +c 2 x) e λx ax y=e ( c 1 cos bx +c 2 sin bx

2 Complex Roots

y =f (ax +by +c )

''

y + y + y =cos ax

For

z + z + z =e iax z=u e y=ℑ z

2.

2.2.2

Reduction to separable form

'

1.

For a homogeneous linear differential equation, the sum of 2 solutions is also a solution

1 dy= M ∫ ( x ) dx N ( y)

1.2

''

y + y + y =sin ax

For

, the linear differential equation is homogeneous,

otherwise it is nonhomogeneous

y =M ( x ) N ( y )

∫

Homogeneous Equations

F ( x ) =0

If

is the amplitude,

angular frequency

or

Stable Equilibrium (SHM)

´θ= −g θ=−ω2 θ L g )L t +π θ=C cos ( ωt )+D sin ( ωt) ¿ A cos(ωt−δ) 2π L is the period, ω is the =2 π g ω

√

3.2

Simple Harmonic Oscillator

3.4

m ´x +kx =0 m

k

is the mass, while

( )

B 2−4 sE=0

For SHM,

x ( t )=C cos ωt + D sin ωt= A cos(ωt−δ ) D A= √C 2+ D 2 , tan δ= C Angular frequency: ω ; Amplitude: A ; Phase Angle: δ 1 2π Frequency= Period= =2 π m T k ω k ω= m

√

√

Buoyancy

m ´x =−kx

m

( )

d 1 2 x´ =−kx dx 2

4

Modelling

4.1

Malthus Model of Population

Ne N ( t )= ^

kt

4.2

m x´ + ρAgx=0

Logistic Model

N ∞=

Forced Oscillation

B B , s= N∞ s

m ´x +kx =F 0 cos αt

Case

ω=

√

k m

^ N N∞

cos αt

^ N=N ∞

Particular Solution:

x ( t )= A (t ) sin

(

α +ω t 2

)

4.3

Particular Solution:

α=ω

(

)

(

N ∞ − Bt e ^ N

' L ( f (t ) )=sF ( s)−f ( 0 ) ct L ( e f (t ) )= F ( s−c ) n! L ( ect tn ) = (s−c )n +1 s−c ct L ( e cos wt )= 2 2 (s−c ) +w −as L ( δ ( t−a ) )=e

N −Bt 1+ ∞ −1 e ^ N N∞ 1− 1−

)

N ( t )=N ∞

dN = (B−sN ) N−E=−s N 2+BN −E dt

Resonance

Resonance occurs when

N (t )=

N∞

Laplace Transforms k at 1 L ( k )= L (e )= , s>a s−a s w wt L (sin wt )= 2 2 s s +w cos ¿= 2 2 s +w L¿ s a L ( cosh at ) = 2 2 , s >¿ a L (sinh at ) = 2 2 , s >¿ a s −a s −a n n! L ( t ) = n+1 s n−1 n−1 n−2 ' L ( f ( n) ) =s n L (f )−s f (0 ) −s f (0 )−…−f ( ) (0 ) '' ' 2 L ( f ( t )) =s F ( s) −sf (0 ) −f ( 0 )

Harvesting

Case

3.3.1

5

1 1 E= m x´ 2+ k x 2 2 2

^ N=Current Population , k=B− D

m x´ =mg− ρAgd − ρAgx ( t ) =−ρAgx (t )

The natural frequency that the system has is given by

> 0if β1...