M4T|-| 2I5

use method of variation of parameters when...

its not easy to guess a particular soln for the given g(t)

component form of vector solns for systems of diffl eqns

just break up the vectors vector form is the vectors not broken up

x°'=A(x°) for complex eigenvalues

For a real polynomial with a complex root, conjugate also a root. That is, if r₁=λ+iu where λ,u real, then r₂=λ-iu is also a soln.

Let γ₁° be the complex eigenvector corresponding to r₁.Then, (A-r₁I)γ₁°=0°. Taking the conjugate of the eqn to find r₂ we get: (A-r₁_I)(λ¹°_)=0°. Then, x(¹)°= γ₁°exp(r₁t) and x(²)°= γ₁°_exp(r₁_t) are 2 complex solns.

However, we can just find 2 complex solns by expanding x(¹)° or x(²)°(get same result): γ(¹)°=a°+ib° where a°,b° real valued vectors. Then, x(¹)°=(a°+ib°)(exp((λ+iu)t) =exp(λt)(a°cos(µt)-b°sin(µt))+iexp(λt)(a°sin(µt)-b°cos(µt)) The first vector is u°,2nd is v°, which are both lin. indep solns by 7.4.5

for the fnc g(t) = d(t) ={1/2T for -T≤t≤T,0 otherwise}

I(t) = area of rectangle =1 for any T as long as h=1/2T. the total impulse remains constant even as the domain gets smaller and the fnc gets taller, ie. as T→0⁺

phase portrait for systems with 2 real distinct eigenvales with same sign

If eigenvalues>0 → inf along 2 lines if <0→origin along 2 lines -origin = a node

Thrm 7.4.5: consider the system x°'=P(t)(x°) where Pij(t)'s are real and cts.

If x°=u°(t)+iv°(t) is a soln, then u° and v° are both solns by linear combinations.

method of reduction of order can also be used for eqns of the form: y''+p(t)y'+q(t)y=0

In this case the eqn in r reduces to: y₁r''+(2y₁'+py₁)r'=0. Then, we can make r'=W turning iit into 1st order diffl eqn in W. Then, find W, solve for r with int factor/separation

L is a linear operator for functions whose laplace transforms all exist

L{5exp(-2t)-3sin(4t)}= 5L{exp(-2t)}-3L{sin(4t)} =5[1/(s+2)]-3[4/(s²+4²)] s>0

Laplace transform of Uc(t)

L{Uc}=(0∞)∫exp(-st)Uc =(0c)∫exp(-st)dt =[exp(-sc)]/s s>0

L{f} where f=1 0≤t<1 f=k t=1 f=0 t<0

L{f} = (01)∫exp(-st) = (1-exp(-s))/s for s>0

Rmrk: if for an nxn system we have n real and diff eugenvalues, we get n lin indep eigenvectors and thus W≠0

Not always the case as there may be complex/repeated roots.

units for w in u=Rcos(wt-δ) is s⁻¹

SI: when mass is given in lb*s²/ft this is alr slugs so dont need divide by 32

Locally linear systems to approx. non linear sys: Suppose x°=Ax°+g(x) and x°=0° is an isolated critical pt. That is, there is a circle around x°=0° such that 0° is the only critical pt inside. In addition, assume det A≠0 so 0° is the only crit pt. of Ax° and g has cts partial derivatives and satisfy |g(x°)|/|x°|→0 as x°→0. g(x°) is a small change in the matrix so that λ's change, but not too much.

Such a system is called a locally linear system in the neighborhood of x°=0° these are conditions ensuring that we can do 1st order taylor approx in order to locally linearize the system around the critical point, 0°.

Thrm 7.4.2: If the vectors x(¹)...x(ⁿ) are lin. indep solns of the system x°'=P(t)(x°), for point in the interval a<t<b, then each soln x° of the system can be written as a lin. comb of x(¹)...x(ⁿ) in exactly one way.

This soln x°, the linear combination, is called the general soln whereas the set of solns {x(¹)...x(ⁿ)} is called the fundamental set of solns.

laplace transform: F(s)=L{f(t)}=(0∞)∫exp(-st)f(t)dt whenever the improper integral is defined

Thrm 6.1.2 states when its defined: suppose that 1) f is piecewise cts on interval 0≤t≤A for any positive A 2) there exists real constants C,a,M with K,M>0 s.t. |f(t)|≤Kexp(at) when t≥M (ie. bounded by exponential) Then, the laplace L{t}=F(s) defined exists for s>a. Such funcions f, are 'piecewise cts fns of exp order as t→∞' ex. exp(t²) not of exp order

Thrm 7.4.3 Abels Thrm: If x(¹)...x(ⁿ) are solns of the system x°'=P(t)(x°) on the interval a<t<b then in this interval W[x(¹)...x(ⁿ)] either always =0 or never

Thrm 7.4.4: Let e(¹)=[100..0]⁺,..e(ⁿ)=[00..01] and let x(¹)...x(ⁿ) be solns of the system x°'=P(t)(x°) satisfying IC x(^i)(t₀)=e(^i) for i=1,2...n where t₀ ∈ (a,b), then x(¹)...x(ⁿ) form the set of fund. solns

When we have a general system of n first order linear diffl eqns x₁'=P₁₁(t)x₁+...+P₁n(t)xn+g₁(t) ... xn'=Pn₁(t)x₁+...+Pnn(t)xn+gn(t)

We can write this system as x°'=P(t)x°+g(t) where x°=[x₁...xn]⁺, g°=[g₁....gn]⁺, and P(t) is the nxn matrix of n eqns and x°'=[x'...xn]⁺

Forward, explicit Eulers method formula positive error means expected > approx value

Y(n+₁)=Yn+hf(tn,yn) for small step size h f(tn,yn) = the differential eqn at the point yn= the fnc values

backwards euler method

Yn+₁=Yn+hf(tn+₁,yn+₁)

1) real unequal evals of same sign x°=C₁T(¹)exp(r₁t)+C₂T(²)exp(r₂t) where r₁,₂ both > or < 0 crit pt=node or nodal sink/source

a) if both <0, x°→0 as t→∞. If soln starts on T¹or T², it remains on the lines. The solns are going to converge to 0 but also end towards the line through T², the larger eigenvector. b) if 0<r₁<r₂, the trajectory will have the same pattern but away from the origin

a)r₁>r₂>0 b)r₁<r₂<0 c)r₂<0<r₁ d)r₁=r₂>0 e)r₁=r₂<0 ---r₁,r₂=λ±iµ--- f)λ>0 g)λ<0 h)λ=0

a) node, unstable b) node, asymptotically stable c) saddle point, unstable d)proper/improper node, unstable e) proper.improper node, asymptotically stable f)spiral point, unstable g) spiral point, asymptotically stable h) center, stable

Thrm 9.33: Let r₁,₂ be the eigenvalues of the linear system x°'=Ax° corresponding to the locally linear system x°'=Ax°+g(x°). Then the type & stability of the critical pts of the linear sys and locally linear sys is given as

a)r₁>r₂>0 (Lin: node/unstable) (LocLin: same) b)r₁<r₂<0 (L: node/asymp.stable) (LL:same) c)r₂<0<r₁ (?) d)r₁=r₂>0 (L: PN or IN, unstable)(LL: N or SpP, unstable) e)r₁=r₂<0 (L:PN or IN, asymp.stable)(LL:N or SpP, asymp.stable) ---r₁,r₂=λ±iµ--- f)λ>0 (L:SpP, unstable)(LL:same) g)λ<0 (L: SpP, asymp.stable)(LL: same) h)λ=0 (L:center, stable) (LL:center, indeterminate)

importance of critical points:

critical points correspond to eq solns where x(t) and y(t) are constants, ie. the system doesnt change for such solns. for linear sys: crit pt at origin determines behaviour of trajectories. not the case for non-linear, however can still determine behaviour around crit pts

4)complex evals with non zero real part roots = λ±iµ with λ≠0, µ>0 spiral points: λ<0 - spiral sink λ>0 - spiral source

direction of spiral : calculate derivative at given point. ex. given x'=ax+by y'=cx+dy and [x,y]=[0,1] , we see x'=b, y'=d since both b,d>0 the direction is CW

the trajectories of 2x2 autonomous systems of the form dx/dt=F(x,y) and dy/dt=G(x,y) can be determined by solving

dy/dx=(dy/dt)/(dx/dt) = G(x,y)/F(x,y) and getting a soln of the form H(x,y)=C

Find the trajectories to the system dx/dt=4-2y dy/dt=12-3x²

dy/dx=(dy/dt)/(dx/dt)=(12-3x²)/(4-2y) (12-3x²)dx=(4-2y)dy 4y-y²=12x-x³+C =>H(x,y) = 4y-y²-12x+x³=C

L⁻¹{exp(-cs)F(s)}=Uc*L{f(t-c)}

ex. L⁻¹{exp(-2t)/(s²+1)} = sin(t-2)

when you see Uc*f(t) the goal is to make f(t) into f(t-c) in order to pull out an exp(-ct). L{Uc*f(t-c)}=exp(-cs)L{f}

ex. exp(t) = exp(π)exp(t-π) -sin(t) = sin(t-π) in order to get (t-π) term

case where expected particular soln is already soln to homo eqn:

ex. y''-2y'+y=2exp(t) particular soln guess: not exp(t) & texp(t) since these are homo solns ∴guess y(t) = At²exp(t)

another way to check that the limit |g(x°)|/|x°|→0 as x°→0 is by using an inequality, that is, if a limit "bigger" than |g(x°)|/|x°|→0 as x°→0, then so does |g(x°)|/|x°|

ex. |(sinx-x)|/√(x²+y²) ≤ |(sinx-x)|/|x| and |(sinx-x)|/|x| →0 as x°→0 dont need to change to polar cause alr in terms of one var

ex. apply thrm 2.4.2 to y'=y^(1/3) y(0)=0

f continuous but d/dy DNE when y=0 so for given IC, thrm doesnt apply --> at least 2 solns, with y=0 being one of them

"jumping" method of expressing piecewise cts fts ex. f(t)= 2 0 ≤t<4 5 4≤t<7 -1 7≤t<9 1 t≥9

f starts at 2, jumps to 5 at t=4, -1 at t7, 1 at t9 => f=2+3u₄-6₇+2u₉ start at 2, up 3 at 4, down 6 at 7, u2a9

Ch 7.5 : Homo linear sys with const coefficients. systems of the form x°'=A(x°) where A is a constant nxn matrix. x° is an eq soln if Ax°=0. 0° is the only eq soln if det A≠0

for n=2 matrices, we can make the phase plane (and portrait) by plotting Ax in plane for every x. The typical trajectories on the plane makes it a phase portrait.

x'=F(x,y) y'=G(x,y) Thrm 9.3.2: The above system is locally linear in the neighborhood of a critical pt (x₀, y₀) whenever the fncs F&G have cts partial derivatives up to order 2

for system (**) du°/dt =(u°)df(x°)/dx°+η(x°) where u°=[u₁, u₂]⁺=[x-x₀,y-y₀]⁺, and df(x°)/dx°=jacobian for (x₀,y₀) the crit pt we say that du°/dt =(u°)df(x°)/dx° approximates the corresponding non linear system. assume det J≠0 so that this pt is an isolated cirtical pt of the system (**)

9.1 phase plane - linear systems

for x°'=Ax°,the points x° for which Ax°=x°'=0 are called equillibrium (constant) solns or critical points assume det A ≠0 so its not singular and thus x°=0 is the only equillibrium point.

if a first order diffl eqn is not linear, then use separation

form: M(x,y) + N(x,y)dy/dx = 0 seperable means M(x,y) = M(x) and N(x,y) = N(y)

undamped free vibrations - no Fd or Fe mu''+ku=0 having char eqn. mr²+k=0 r=±i√(k/m)

gen soln: u=Acos(w₀t)+Bsin(w₀t) with w₀=√(k/m) with trig subs, u=Rcos(w₀t-δ) where A=Rcosδ; B=Rsinδ; R=√(A²+B²); tanδ=B/A

locally linear systems: thrm 9.3.1

given that det A≠0, the critical pt x°=0° of system x°'=Ax° is: 1) asymptotically stable if evals r₁,₂ are both real and negative or complex w neg real parts 2) stable, but not asympt stable if roots are pure imaginary 3) unstable if r₁,₂ are real and at least one is pos or complex w positive real parts

critical points of x' °= f(x°) are stable when

given ε>0 there is δ>0 such that for every soln x°=x°(t) satisfying |x°(0)-x°⁰|<δ both exists and satisfies |x°(t)-x°⁰| <ε for all t ≥0. ie. solns that start "sufficiently close" to x°⁰ sty "close" to x°⁰ crit point not stable = unstable crit point asymptotically stable if it is stable and there exists a δ₀ (δ₀>0) such that if a soln x°=x°(t) satisfies |x°(0)-x°⁰|<δ₀, then lim(t→∞) x°(t) = x°(0) ie. solns not only stay "close," but converge to x°⁰ as t→∞

error analysis

global truncation error: En=∅(tn)-yn after n steps local truncation error: error in one step en=∅(tn)-yn assuming yn_₁=∅(tn_₁) en+₁=∅(tn+₁)-yn_₁=[½∅''(tn~)h²] where tn<tn~<tn+₁ pick tn~=tn+₁ for upper bound on error size of error: |en+₁|

"rectangular pulse" example

h(t)=Uπ(t)=U₂π(t) t≥0 0 for 0≤t<π 1 for π≤t≤2π 0 for t≥2π

note: in systems with complex eigenvalues, x(¹)°=u°+iv° is a soln and u°,v° are 2 lin. indep solns by thrm 7.4.5, but..

in the general soln, dont write the 'i', just a lin comb of u°,v°

If ∅ is defined on [a,b] and at least twice cts'ly diffl, then the local truncation error for ∅ for the Euler method is "order h²" on [a,b]. ie. |en+₁|=O(h²)

meaning, there is a constant K independent of h & tn such that |en+₁| ≤ Kh² for all sufficiently small h. global truncation error: "order h" / "first order" ie. |En|=O(h) or |En| ≤Ch for all sufficiently small h where C is independent of h.

non homo 2nd order diffl eqn

method of undertermined coefficients gen soln: homo soln + particular soln

ex. [x,y]'⁺=[1,0;0,½][x,y]⁺+[-x²-xy; -¾xy-¼y²]⁺ ie. x°'=Ax°+g(x°) is a locally linear sys around x°=0. the critical pts are (0,0), (0,2), (1,0), and (½,½) thus 0° is an isolated crit pt.

moreover, satisfying the 2nd condition, the limit |g(x°)|/|x°|→0 as x°→0 where g°=[g₁,g₂]⁺ Check that |g₁|/|x| and |g₂|/|x| both →0 as r→0 by making x=rcosθ, y=rsinθ to reduce to one variable. If this is the case, then |g(x°)|/|x°|→0 as x°→0 and its a locally linear system

+C

must specify if arbitrary constant, pos, neg given IC and domain

ordinary diffl eqn defn - all y',y'', depend on t

n times diffl fnc y depending on one variable, t, F(t,y,y',y'',y^(n))=0 where F is is a multivariable function

diffl eqns w/ discts forcing fncs

non-homo discts term - forcing fnc

when A has a repeated eigenvalue with multiplicity n, but with n lin indep. eigenvectors, the t multiple is not needed in the full soln and you dont need to do the process to find η.

note: real symmetric matrices ex. [ 211,121,112] always have full set of eigenvectors

phase portrait for systems with one eigenvalue/vector

one line trajectory positive → inf negative → origin origin=improper node

turning diffl eqns into system

order of diffl eqn is how many vars u need ex. y''+y'+2y=-4sin(12t) y(0)=2, y'(0)=2 x₂=y' x₁=y x₂'+x₂+2x₁=-4sin(12t) {x₂'=-x₂+2x₁-4sin(12t); x₂=x₁' {y(0)=2; y(0)=5

how to get upper bound for local truncation error

pick tn~=tn+₁ for upper bound on error size of error: |en+₁| recall |a+b| ≤ |a| + |b|

2nd order homo linear diffl eqn

plug in exp(rt) to get characteristic eqn ar²+br+c=0

critical points of x' °= f(x°)

points x° where x' °= f(x°) = 0 =constant/equillibrium solns of the system

for y'' + 6y' + 10y = t² + 2 what would u guess to be particular soln

polynomial of order 2 → At² + Bt + C

Find all pairs (t₀,y₀) such that the IVP is guaranteed to have a unique soln determine open interval where soln gauranteed to exist

put into standard form and use thrms ln |t| cannot equal 0 check initial conditions to see which soln is the correct one

3) equal evals, r₁=r₂=r a) lin indep evecs: x°=C₁T(¹)exp(rt)+C₂T(²)exp(rt) origin=proper node or star point

ratio x₂/x₁ indep of t ∴ no directions, constant with time -trajectories lie on straight line thru origin

2nd order homo linear diffl eqn : 2 real roots

solns (2): y=Cexp(rt)

2) Real eval of opp sign x°=C₁T¹exp(r₁t)+C₂T²exp(r₂t) with r₁>0,r₂<0 origin=saddle pt

solns on T¹ line away from origin solns on T² line approach origin other solns: x°→T¹ as t→∞ and T² as t→-∞

NULLCLINES (where dx/dt=0=dy/dt) can also be used to understand the behaviour of systems. They help us understand how dx/dt, dy/dt change -if pic is clear with just nullclines, then dont need to find eigenvalues

steps to understand a linear system:x°'=Ax° 1) Evals and Evecs to find general soln 2) Nullclines where dx/dt=dy/dt=0 and plot 3) Plot info from eigenvectors and nullclines tgt to get phase portrait

g(t) = t/2 for 0≤t<6 3 for t≥6 how to turn into step fnc

t/2 -(t/2)u₆ +3u₆ need to minus the fnc if u dont want it in the next interval

in a linear system with det A≠0 so 0° is the only critical pt to which all trajectories converge,

the crit pt is "globally asymptotically stable"

ex. oscillating pendulum

the eqn for angular momentum is d²θ/dt²+ (c/mL)dθ/dt + (g/L)sinθ=0 where c|dθ/dt| is the amplitude of the damping force. This can be rewritten as: d²θ/dt² + (γ) dθ/dt + (w²)sinθ=0 which can be made into a sys of eqns with x=θ, y=dθ/dt. crit pts are when y=0 and x=±nπ where n is an integer. Thus, the eqns correspond to the situations where θ=0 or θ=π

wronskian for vector solns x°(¹)

the matrix cols are the vectors. no derivatives involved

for a 2d sys with at least one asymp stable crit pt, we can find the set of all points P such that if a trajectory passing through P , it approaches the critical point as t→∞. such a set is called the "basin of attraction" of that critical point.

the trajectory that bounds a basin of attraction is called a seporatrice because it seperates the trajectories

solns to Ax°=x°' can be viewed as a parametric representation of a curve on the x₁x₂ plane

these soln curves can be seen as trajectories that show the path of a moving praticle on the plane with velocity dx°/dt. The plane with the trajectories is called the phase portrait.

5) purely imaginary roots crit pt= center

trajectories are circles with center the origin, traveresed cw if µ>0 and ccw if µ< 0 all solns are periodic with period 2π/µ

ex. of finding trajectories of autonomous systems by finding dy/dx

v° ' =[0,1;10]v° where v°=[x,y] dx/dt=y, dy/dt=x dy/dx=(dy/dt)/(dx/dt)=x/y => ydy=xdx → ydy-xdx=0 =>2ydy-2xdx=0→y²-x²=C = hyperbolic curves for trajectories

eulers method: given y'=f(t,y), y(t₀)=y₀ and that f & δf/dy are cts in some given rectangle containing (t₀,y₀), we know that a unique soln esists in some interval around t₀.

when finding this soln is hard for a non linear system, use eulers method, an approximation

competing species models: ex. dx/dt =x(∈₁-σ₁x-α₁y) dy/dt=y(∈₂-σ₂y-α₂x)

where x,y denote the population of two species, and ∈₁,₂ denote the growth rates. ∈₁/σ₁, ∈₂/σ₂ denote saturation levels and α₁,₂ denote how the populations interact

autonomous systems dx/dt=F(x,y) dy/dt=G(x,y) where F &G are cts with cts partial derivatives in some domain D where the initial value (x₀,y₀) is given. Then, by Thrm .11, there is a unique soln x=x(t), y=y(t) satisfying the IC x(t₀)=x₀, y(t₀)=y₀ in an interval I of t that contains t₀. Then we can write the system as:

x' °=dx°/dt = f(x°), x°(t₀)=x°⁰ where x°=[x,y]⁺ f(x°)= [F(x,y), G(x,y)]⁺ x°⁰=[x₀,y₀]+ Here the fnc f does not depend on t explicitly, thus its "autonomous" system . autonomous systems have an associated direction field independent of time

If A has 2k complex eigenvalues r₁, r₁_...rk, rk_ with eigenvectors λ(¹),λ(¹)_...λ(^k),λ(^k)_ and n-2k real and distinct eigenvalues r(n-₂κ-₁)...rn with eigenvectors λ(ⁿ⁻²k⁺¹)...λ(ⁿ) then the general soln is:

x°=C₁u(¹)°+C₂v(¹)°+...C₂κ-₁u°(^k)(t)+C₂κu°(^k)(t) + C₂κ+₁λ(ⁿ⁻²k⁺¹)exp((rn-₂κ+1)t)+...Cnλ(ⁿ)exp((rn)t) where u° and v° are found by seperating one complex soln as above. ie. the general soln is the linear combination of complex solns and real solns

for solns of systems of diffl eqns, "converges to 0" means that the soln with the positive eigenvector coefficient =0

x₁(t) is the 1st component form soln, so x₁(0) means plug 0 into 1st soln to find C

partial diffl eqn defn

y is a multivariable function and partial derivatives exist in the eqn

guess for particular soln for non homo eqn where g(t) = -8exp(t)cos(2t)

y(t) = Aexp(t)cos(2t) + Bexp(t)sin(2t)

solution to logistic eqn 1 found by partial fractions

y(t) = Y₀k/(Y₀+(k-y₀)exp(-rt)

autonomous diffl eqns: exp growth

y(t) = pop of given species at time t. dy/dt=r*y rate of change prop. to current pop. r>0 solve => y(t)=Y₀e^(rt)

comparing quasi fq µ, Td and real fq w₀, T of undamped motion

µ/w₀ ≈ 1-(γ²/8km) for small γ²/4km Td/T=1+(γ²/8km) for small γ²/4km =>fq drop and T inc for damped vibrations ∴effect of damping not related to γ but its relation to 4km instead "for small γ²/4km" - small damping

what is the formula for ∅''(t)

ƒt(t,∅)+ƒy(t,∅)ƒ where ƒ is the diffl eqn and ƒt,ƒy are partial derivatives.

complete the square to find complex roots

'steady state' part = forced response so when asked to maximize amplitude of foced response given g(t), caculate R=sqrt(A²+B²) using A and B from particular soln guess for g(t)

To solve systems of the form x°'=A(x°) we are going to assume a soln of the form Texp(rt) where T and r are going to be determined.

(A-rI)T=0. r is eigenvalue of A and T is corresponding eigenvector.

for g(t)=t²cos(4t)+tsin(4t)+sin(5t) what would you guess as particular soln?

(t^s)[(At²+Bt+C)cos(4t)+(Dt²+Et+F)sin(4t)]+Gsin(5t)+Hcos(5t) the cos(4t) and sin(4t) can share one set of cos&sin in the particular guess even though one has t² and the other has t. just need one set of cos/sin for each argument. The 2nd order polynomial can be used in place of the first one for t. pick s=1,2... in case that part. soln = homo

Repeated eigenvalues - when theres multiplicity , one eigenvalue/vector/soln

-assume 2nd soln has form x(²)°=λ(¹)°texp(r₁t)+η°exp(r₁t) Plugging this into x°'=A(x°), we find: [λ(¹)°=(A-r₁I)η°] which we can use to solve for η°, the 'generalized eigenvector' corresponding to r₁

3b) one indep evec for eqal evals x°=C₁Texp(rt) +C₂(Ttexp(rt)+ηexp(rt)) where T is the evec, r is the eval and η is the generalized evec satisfying (A-rI)η=T -if r₁=r₂=r>0, crit pt=improper/degenerate node

-dom term as t→inf is C₂Ttexp(rt). Thus, if r<0, every traject approaches origin tangent to the line through the eigenvector, even when C₂=0 -as t→-inf, the slope of each traject approaches the slope of T -orientation of traject depends on T and η: ccw when η in Q4 btwn x axis and T, cw when η in Q4 NOT btwn x axis and T. -writing soln as y°=(C₁T+C₂η)+C₂Tt, we get lines in the direction of T passing through C₁T+C₂η

example about phase portrait behaviour x°'=[a 2;-2 0]x° contains the parameter a. Describe the values of a for which the qualitative behavior of the solns change.

-need to look at eigenvalues to see where behaviour changes. 1) find char eqn and roots of det(A-rI)=0 in terms of a 2) behaviour changes at a=-r,r when √(b²-4ac)=0 in quad eqn 3) a ∈[-r,r] →complex eigenvalues. in this region, a<0→negative Re(exp)=spiral in and a>0=spiral out. 4)∴0,-r,r are places where the phase portrait changes behaviour

turning step fnc back into piecewise

-see what gets kept/minused moving from one interval to the next. if it gets minused once, unless it's added again, its removed indefinately

phase portrait for systems with complex eigenvalues

-solns form a spiral -if real part of the exponent>0 → spirals to inf and origin is unstable -Re(exp)<0→spirals to origin, origin is asymptotically stable -Re(exp)=0→forms circles, origin stable but not asymptotically stable

phase portrait for systems with opposite sign eigenvalues (real)

-there is a line for each eigenvector, found by making C₁=0, then C₂=0. -eigenvalue >0 → diverges to inf along line (1st and 3rd quads) -eigenvalue<0→converges to 0 along line (2nd and 4th quads) -origin is saddle point

steps for solving x°'=A(x°) for complex eigenvalues

0) find char eqn to find eigenvalues by (A-rI)T=0 1) conjugate also a soln 2) find eigenvector for r₁, then take conjugate to get eigenvector for r₂ 3) form a solution with eigenvalue/vector x(¹)° 4)Break up x(¹)° by seperating Re and Im parts 5) Write linear combination of u°,v° 6) Check W 7) draw phase plane by drawing Ax for every x

ex. find soln to 2y''+y'+2y=g(t)

0)make g(t) in terms of step fncs 1)take laplace of eqn both sides 2) make L{f}=Y(s) 3) plug in IC to simplify 4) isolate Y(s) 5)break L{g} into exponentials*H(s) 6)L⁻¹{Y(s)} → expression in terms of h(t) still unknown 7)L⁻¹{H(s)} by rearranging so it looks like the chart ex. if theres a quadratic term in denom → sin/cos by completing square, partials etc 8) now having h(t), plug this into (6) to get ans

Thus, we have seen the trajectories behave in 3 ways:

1) all traj → 0 as t →∞ when evals ( real&neg) or (complex w neg real part) 2) bounded but does not →0 as t→∞ when evals are pure imaginary 3) some traj (not 0°) unbounded (→∞) as t→∞ when at least one λ is (real&pos) or (complex w pos real part)

steps for reduction of order to find 2nd complimentary soln in case of repeated roots

1) assume soln y₂=v(t)y₁ 2) find y₂',y₂'' 3) write the 2nd order diffl eqn in standard form (coefficient on y'' is 1) 4) plug in y₂,y₂',y₂'' to the eqn and simplify to get rid of the v, leaving v'',v',and y₁(','') 5)v'=W, now first order, solve w int. factor 6)∫W to find v and plug v into (1) 7)wronskian for y₁,y₂

given IVP, solving damped free vibs probs

1) char eqn to find gen soln 2) IC 3) rewrite gen soln in trig form 4) find phase angle δ w trig idents for A,B 5) find µ,Td 6)time when fnc passes thru eq first is when cos portion of fnc =0, when arguement = π/2 if in first quad

"Discuss the qualitative behaviour of the (nonlinear) system" dx/dt=F(x,y) dy/dt=G(x,y)

1) find crit pts where F(x,y)=0=G(x,y) 2) Calculate Jacobian 3) Plug in critical points and find Evals, Evecs around each critical point 4) classify the critical points (unstable/stable, node, etc) -NOTE: when crit pt not 0, [u,v]⁺=[x-x₀,y-y₀]⁺ where x₀,y₀ is the non zero point 5) draw the trajectories , seperatrices

steps in doing repeated eigenvalue x°'=A(x°) problems

1) find eigenvalue by (A-r₁I)T=0 2) find associated eigenvector 3) form a solution with eigenvalue/vector pair 4)Find 2nd soln by finding η° T(¹)°=(A-r₁I)η° 4) plug η° into x(²)°=T(¹)°texp(r₁t)+η°exp(r₁t) 5) expand and delete parts of x(²)° that are k*first soln 6) write gen soln as lin comb of x(¹)°,x(²)°. 7) draw phase plane by drawing Ax for every x

steps in solving x°'=A(x°) for n distinct eigenvectors and eigenvalues

1) find eigenvalues by (A-rI)T=0 2) find associated eigenvectors 3) form a solution with eigenvalue/vector pairs and write the gen soln as lin. comb. 4) Check W 5) draw phase plane by drawing Ax for every x

modelling first order diffl eqns

1. construct model 2. analyze model by understanding diffl eqns 3. compare with experimental data

solving spring probs

1. determind k, m, (γ) 2. write IVP 3. write gen soln, find coefficients 4. find natural fq w₀, period, amplitude, phase angle δ 5. graph function

some particular solns for ay''+by'+cy=g(t)

1. g(t) = a₀tⁿ+a₁tⁿ⁻¹+...+a₉ (polynomial) →y(t) = (t^s)(A₀tⁿ+A₁tⁿ⁻¹+...A₉) 2. g(t) = exp(at)(a₀tⁿ+a₁tⁿ⁻¹+...+a₉) →y(t) = (t^s)(A₀tⁿ+A₁tⁿ⁻¹+...A₉)exp(at) 3.g(t) = exp(at)(a₀tⁿ+a₁tⁿ⁻¹+...+a₉)sin(Bt) or g(t) = exp(at)(a₀tⁿ+a₁tⁿ⁻¹+...+a₉)cos(Bt) →y(t) = (t^s)[(A₀tⁿ+A₁tⁿ⁻¹+...A₉)exp(at)cos(Bt)+(B₀tⁿ+B₁tⁿ⁻¹+...B₉)exp(at)sin(Bt)] where s=0,1,2

How to find particular soln in non homo 2nd order diffl eqn ex. g(t) = 2sint

1. guess Y(t) = Asint + Bcost to cancel out cos. if we chose Y(t) = Asint alone, y' = cost which wont cancel out.Y 2. Find Y', Y'', plug in and expand out. Collect sin and cos terms and match coefficients to RHS (2 for sin), (0 for cos) 3. Now having the particular soln, find complimentary soln and add to get general soln.

relationship btwn laplace and step fncs - Thrm 6.3.1: If the laplace transform of f(t), F(s)=L{f} exists for s>a≥0 & if c is a positive constant, then L{Uc*f(t-c)}=exp(-cs)*L{f} s>a

Conversely, if f(t) is the inverse laplace transform of F(s), f(t) = L⁻¹{F(s)} then Uc*f(t-c)=L⁻¹{exp(-cs)*F(s)}

another property of the laplace Thrm 6.3.2: if F(s) = L{f} exists for s>a≥0 & if c is a constant, then L{exp(ct)f(t)}=F(s-c) s>a+c

Conversely, if f(t)=L⁻¹{F(s)} then exp(ct)f(t)=L⁻¹{F(s-c)}

for g(t)=texp(-t), using method of undetermined coefficients, what would you guess to be the form of the soln?

(t^s)(At+B)exp(-t) because texp(-t) is (polynomial of order 1)*(exp) pick s=0..1..2 in the case that the particular guess is alr one of the homo solns

method of variation of parameters overview eqn: y''+p(t)y'+q(t)y=g(t) where g(t) hard to guess particular soln

0. put diffl eqn into standard form 1. knowing complimentary soln, y=c₁y₁+c₂y₂, we assume a soln to actual eqn of the form y=k(t)y₁(t)+u(t)y₂(t) where k,u are fncs 2. take y' and require [k'y₁+u'y₂=0] => new y' 3. take y'' and plug into eqn, simplify to get 4. k'y₁+u'y₂=0 from (2) & k'y₁'+u'y₂'=g(t) from (3) 5. Check W for y₁,y₂≠0=>can solve for k',u' uniquely 6. integrate k',u' 7.plug (6) into eqn from (1) to get partic. soln to eqn

ex. find the inverse laplace of F(s)=[1-exp(-2s)]/s² and graph

1) break up by linearity 2) use table to inverse laplace 3) write the piecewise fnc 4) graph it

ex. find inverse laplace transform of G(s)= 1/(s²-4s+5)

1) comoplete square in demoninator: 1/[(s-2)²+1] 2)notice this is F(s-2) for F(s)=1/(s²+1) 3) use table → L⁻¹{F}=sint 4) use thrm → L⁻¹{G}=exp(2t)sint

steps to solving a diffl eqn w impulse force ex. 2y''+y'+2y=δ(t-5)

1) take laplace of both sides 2)isolate Y(s) =L(y) 3) inverse laplace to find y(t)

how laplace is used to solve IVPS , works for non homo fncs as well. when we use laplace to solve diffl eqn, take laplace of the EQN and solve for laplace of the FNC

1) take laplace of the diffl eqn by linearity 2)apply the corollary to higher order derivatives 3)call L{y}=Y(s), collect Y(s), y'(0),y(0) 4) plug in IC 5) rearrange Y(s), (ex. partial fracs, taking out coeffs) to make it look like basic laplace transform 6)inverse laplace →L{f} using table 7) ans y(t)=f

To use the method of variation of parameters:

1.find complimentary soln 2. find W for y1 y2 3. use eqns for k' u' 4. integrate to get k u 5. gen soln= comp+partic

L{exp(at)}

1/(s-a) s>a

a diffl eqn is linear if

:(y) = F(t,y,..) is linear in y, meaning if y₁ and y₂ are solutions then linear combinations of y₁ and y₂ are also solutions. just need linearity wrt y, ex. e^t, sin t can be in it and it can still be linear

proof of why seperation works

?????

systems of 1st order linear eqns of the form : x₁'=F₁(t,x₁,x₂,...xn) x₂'=F₂(t,x₁,x₂...xn) ... xn'=Fn(t,x₁,x₂...xn) with given IC x₁(t₀)=x₁⁰,x₂(t₀)=x₂⁰...xn(t₀)=xn⁰

A soln to this system on I=a<t<b consists of n fncs x₁=∅₁(t), x₂=∅₂(t)... xn=∅n(t) These solns satisfy the IC and form an IVP

general linear ordinary diffl eqn of order n is of the form

A(t)yⁿ+B(t)yⁿ⁻¹+C(t)yⁿ⁻²...+D(t)y = g(t) linear combination of derivatives of y

Thrm calculating Wronskoin w/o knowing solutions

Abel's Thrm: If y₁ and y₂ are solns to L(y) where p,q are cts on interval I, then W[y₁,y₂](t) = Cexp(-∫p(t)dt) where C depends only on y₁ & y₂, not on t. W(t) = 0 (C=0) or ≠0 for all t ∈ I

Thrm 7.1.1: Let each of the n *fncs* F₁...Fn in the first order matrix and the *n² partial derivatives* δF₁/δx₁...δF₁/δxn, δFn/δx₁, be cts in a region T of t,x₁...xn space defined by a<t<b, a₁<t<b₁...an<t<bn and let (t₀,x₁⁰...xn⁰) be in R. Then, there is an interval (t-t₀)<h in which there exists a *unique soln* x₁=∅(t)...xn=∅n(t) satisfying the IVP.

If every fnc Fi in the system is linear, then the system is called a linear system. Otherwise, it is non linear.

Principle of supersposition for first order linear eqn systems

If the vector fncs x(¹)° and x(²)° are solns to the eqn x°'=P(t)(x°), then linear compinations are also solns for any constants

transforms: given a piecewise cts fnc f, an integral transform is a relation of the form: F(s) = (ab)∫k(s,t)f(t)dt where a,b ∈ R or a=-inf, b=inf

K(s,t)=kernel of the transformation and F is the transform of f

L{sin(at)} = a/(s²+a²) s>0

L{cos(at)}=s/(s²+a²)

Assume f and f' to be cts & of exp order, then L{f'} = (0∞)∫exp(-st)f'dt by IBP...

L{f'}=-f(0)+sL{f} => when we apply laplace to a diffl eqn, it turns into algebraic eqn in s

newtons law of motion applied to spring system mu''=Fnet=mg+Fs+Fd+Fe =mg-k|L+u|-γu'+Fe where Fe is external force

Since mg=kL we get: mu''+γu'+ku=Fe where m,γ,k>0

Corollary 6.2.2: suppose f, f', .. f(ⁿ⁻¹) are cts and f(ⁿ) is piecewise cts on any interval 0≤t≤A. Suppose further that there exists constants k,a,M st |f(t)|≤Kexp(at), |f'|≤Kexp(at)...|f(ⁿ⁻¹)|≤Kexp(at) for t≥M

Then, L{f(ⁿ)} exists for s>a & is given by L{f(ⁿ)}=(sⁿ)L{f}-(sⁿ⁻¹)f(0)-(sⁿ⁻²)f'(0)..-sf(ⁿ⁻²)(0)-f(ⁿ⁻¹)(0)

the diroc delta fnc δ is a 'generalized fnc' defined as the limit of d(t) as T→0⁺. for T=0, δ(t)=inf NOTE: t≠T. T is tau, the boundary, and t is the variable. δ≠d. δ(t)=0 for t≠0 and I(t)=1. this is the unit impulse function.

Then, any unit impulse function at t=t₀ is δ(t-t₀). The laplace transform of δ(t) is the limit of the laplace transform of d(t) as T→0⁺. L{δ(t-t₀)}=exp(-st₀) L{f(t)}=1

unit step fnc (heaviside fnc)

Uc(t) = 0 t<c Uc(t) = 1 t≥c opposite: y=1-Uc(t) looking at laplace transforms: domain t≥0

for y''-y'=t³ what would you guess to be the particular soln

can solve by making W=y' and making it first order or guess particular soln to be y(t) = At⁴ + Bt³ + Ct² + Dt + E => t(At³+Bt²+Ct+D). Need order 4 bc y dependence starts with y', not y.

Using the MVT for integrals for δ(t), we can show [-inf,inf]∫δ(t-t₀)f(t)dt=f(t₀) for f cts.

can think of diroc delta as 'derivative' of step fncs

Nth order diffl eqn → N diff lin. indep solns

check that they're lin. indep by checking that det(W)≠0

ex. for characteristic eqn (r²+2r+2=0)

completing the square: (r+1)²=-1 r= -1 ± i λ = -1 µ = 1

γ=2√(km) - critically damped γ>2√(km) - over damped

critically damped or overdamped - one eq position and convergence, no oscillation

max/min of logistic graph f(y) vs. y = inflection points

differentiate f(y) to check where the inflection points are

autonomous diffl eqn

dy/dt = f(y) ie. eqn doesnt depend on independent variable, t.

autonomous diffl eqns: logistic eqn

dy/dt = r(1-y/k)y eq solns when f(y)=0; y=0,k downward parabola with roots 0,k phase lines indicate direction of pop. change

population growth with threshold

dy/dt=-r(1-y/T)(1-y/k)y r>0,T<K,y>0 f(y) = cubic eqn with roots 0,T,K

We can rewrite a 2nd order const. coeff eqn as a sys of diffl eqns

ex. ay''+by'+cy=g(t) x₁=u, x₂=u' => ax₂'+bx₂+cx₁=g(t) and the fncs x₁,x₂ satisfy the system x₁'=x₂ isolating x₂' = -(b/a)x₂-(c/a)x₁+g(t)/a

in spring problems, for δ, need to consider what quadrant the soln is in

ex. if tan=pos/neg, δ is in quad II so need to add pi to angle in quad I also remember to change all lengths to ft in SI

ex. f(t)= sin(t) 0≤t<π/4 sin(t)+cos(t-π/4) π/4≤t sketch f and find L{f}

f starts at sin(t) then jumps to cos(t-π/4) at π/4 => f(t)=sin(t)+U(π/₄)cos(t-π/4) =graph of sin up to pi/4 then graph of cos L{f}=L{sint}+L{U(π/₄)cos(t-π/4)}... use table

find L{1}

f(t)=1 for t≥0 L{1}=(0∞)∫exp(-st)dt lim(A→∞)(0A)∫exp(-st)dt =1/s for s>0 1<Cexp(at) for a,t≥0,C>1 =>bounded =>L{1}=1/s

impulse fncs

for eqns of the form ay''+by'+cy=g(t), impulse fncs are where g(t) is localized in t. Namely, g(t)=0 for t⊄(t₀-T,t₀+T). In this interval, g(t) can be very large. Then the integral I(T) from [t₀-T,t₀+T] = [-inf,inf] for g(t) is called the *total impulse* of the force g(t)

thrm 3.6.1 ensures k'y₁+u'y₂=0 so that y=ky₁+uy₂ is the general soln

for non homo 2nd order lin diffl eqn y''+p(t)y'+q(t)y=g(t), if p,g,q are cts in open interal I and if fncs y₁,y₂ form fundamental set of solns to homo eqn, then a particular soln is: y(t)=-y₁k+y₂u where k,u (t₀,t)∫ and t₀ is any chosen point in I Then, the general soln is Y=c₁y₁+c₂y₂+y(t)

in spring systems we have either spring at rest (length k) or spring w/ mass m attached at rest (k+L). if move spring by 'u' amnt, length is (k+L+u)

forces on spring: 1)Fs=-k|L+u| hookes law 2)Fg=mg =>mg-kL=0 3)viscous damping force, a resistive force α velocity and opp diretion Fd=-γu'(t) γ>0 u=pos, u'=vel, u''=acc

non homogenous eqns and method of undetermined coefficients

form: L(y) = y'' + p(t)y' + q(t)y = g(t) where p,q,g are given cts fncs on I/ y'' + p(t)y' + q(t)y = 0 is the homogenous part of L(y). Let Y₁, Y₂ be two solns to the eqn L(y) (non homo). by L(Y₁)-L(Y₂), we see that (Y₁-Y₂) satisfies the homogenous part of the eqn. General eqn to L(y): y=C₁Y₁ + C₂Y₂ + y(t) Y₁, Y₂ = fundamental set of solns to homo eqn = complimentary soln y(t) is a particular fnc satisfying the equation.

for a given fnc f defined for t≥0 we can consider the related fnc g defined by

g(t) =0 t<c f(t-c) t≥c just the function but shifted by c g(t)=Uc*f(t-c)

for linear first order diffl eqns, use *integrating factor* method dy/dt + p(t)y = g(t) with p and g given

goal: change LHS to full derivative by reverse product rule, then integrate 1. standard form 2. int factor = exp(∫p(t)) constant unimportant 3. mult diffl eqn by int. factor 4. now, LHS is full derivative (uy)' 5. integrate RHS and divide by int factor to find y

order of diffl eqn

highest order of derivatives appreaing in the eqn

How to find particular soln in non homo 2nd order diffl eqn ex. g(t) = 3exp(2t)

method of undetermined coefficients: guess form of soln but leave coeff unspecified. Doesnt work → try another guess. ex. y''-3y'-4y=3exp(2t) guess: y(t) = Aexp(2t). Find y', y'', plug in and solve for A → result: particular soln for L(y)

ex. mass 4N stretches spring 2cm. stretched another 6cm in pos. direction and released. medium exerts resistance 6N when u(t)=3. formulate IVP => find m,γ,k assuming Fe=0

mg=4 → m=2/5kg γ: Fd=-γu' γ>0 → 6=γ3 →γ=2 k:kL=mg=4 → k=mg/L=4/0.02m=200 =>(2/5)u''+2u'+200u=0 → u''+5u'+600u=0 with IC u(0)=0.06m u'(0)=0 no init. velocity

forced periodic vibrations - Fe, periodic external force => 2nd order linear non homo diffl eqn w non linearity g(t)=Fcos(wt)

mu''+γu'+ku=F₀cos(wt) where F₀ & w >0 describe amplitude and fq of force by undertermined coeffs, gen soln: u=c₁u₁+c₂u₂+Acos(wt)+Bsin(wt) note: Uc, comp soln →0 as t→∞ Uc = transient soln U (partic soln) = steady-state soln or 'forced response' of the system

ex. modelling salt in a bucket

rate in: (rate of flow)(concentration) rate out: (rate flow)(concentration) note: sometimes the volume is not constant, in which case conentration out =Q/volume and volume not constant

damped free vibrations - damping effect incl but no Fe =>mu''+γu'+ku=0 linear 2nd order const coeff homo eqn

roots = [-γ±√(γ²-4km)]/2m =(γ/2m)[-1±√(1-(4km/γ²))] 1)when √>0 → u=Aexp(r₁t)+Bexp(r₂t) 2)√=0→u=(A+Bt)exp(-γt/2m) 3)complex roots √<0 →u=exp(-γ/2m)(Acos(µt)+Bcos(µt)) where µ=(1/2m)√(4km-γ²) >0

2nd order homo linear diffl eqn : 2 complex roots eulers formula

roots: r= λ ± iµ solns: e=exp(λ ± iµ) general soln: y(t) = C₁exp(λt)cos(µt)+Cexp(λt)sin(µt)

guess for particular soln for non homo eqn where g(t) = g₁(t)+g₂(t)+g₃(t)`

suppose Y₁ and Y₂ are particular solns to the eqns ay₁'' + by₁' + cy₁ = g₁(t) respectively and ay₂''+...=g₂(t) respectively. Then, Y₁+Y₂ is a particular soln to ay''+...=g₁(t)+g₂(t).

Thrm 6.2.1: Suppose that f is cts and f' is piecewise cts on interval 0≤t≤A and there exists constants k,a,M st |f(t)|≤ Kexp(at) for t≥M

then, L{f'} exists for all s>a and L{f'}=sL{f}-f(0) Apply the same calc repeatedly for higher derivatives

existence and uniqueness for 1st order linear eqns

thrm 2.4.1: if fncs p,g, are cts on open interval I: α< t <β containing the point t=t₀ then there exists a unique soln ∅(t) to the eqn y' + p(t)y = g(t) for each t ∈ I satisfying the initial condition y(t₀)=y₀, y₀ ∈ R to solve such 1st order eqns, we rly on the integratability of µ, g, p. Conditions in this thrm gaurantee existence of these expressions

existence and uniqueness thrm for 1st order *non linear* diffl eqn

thrm 2.4.2: let fncs f and δf/δy be cts in some rectangle α<t<β & λ<y<δ containing point (t₀,y₀). Then, in some interval t₀-h < t < t₀+h contained in α<t<β, there is a unique soln y=∅(t) of the IVP y'=f(t,y) y(t₀)=y₀

damped free vibs case 3 complex roots

trig subs => u=Rexp(-γt/2m)cos(µt-1) soln tends to 0, but oscillates fnc*cos →cos enveloped btwn ±fnc motion not periodic but µ=quasi fq and Td=2π/µ quasi period

what does the general eqn of undamped free vibs tell us

u=Rcos(w₀t-δ) motion is periodic (simple harmonic) period T=2π/w₀=2π√(m/k) w₀=natural fq of vibration rad/time max displacement R - amplitude of motion δ- phase angle

laplace allows us to solve piecewise cts diffl eqns bc the thrm only requires that fncs are piecewise cts and of exp order

use step fncs to write piecewise cts fncs in a way that we can find their laplace transforms

"write as prod of 2 trig functions" if make all one term, easier to see the behaviour/periodicity of the function

use trig identities cos(A+B)=cosAcosB-sinAsinB cos(A-B)=cosAcosB+sinAsinB u=(128/45)sin(t/2)sin(15t/2) graph looks like oscillations with outer envelope being (128/45)sin(t/2), the one with bigger fq and amplitude

If P and g are cts on an interval I=a<t<b if all the componements of the matrix P and vector g are cts in I. The system is homo if g(t)=0 and non homo otherwise.

x(ⁿ)(t)=[x₁₁(t) x₂₁(t)...xn₁(t)]⁺ x(^k)(t)=[x₁k(t) x₂k(t) ...xnk(t)]⁺ are solns of the system x°'=P(t)(x°) Then, xij(t) = xi(^j)(t) is the ith component of the j'th solution.

2nd order homo linear diffl eqn : repeated root

y₁=exp(-bt/2a) y₂=r(t)y₁ = texp(-bt/2a)

proof that y₂=r(t)y₁ in case of repeated roots

y₂' = r'y₁+ry₁' y₂'' = r''y₁+2r'y₁' + ry₁'' plugging this into og eqn and collecting r terms results in r(og eqn) =0 and ar''y₁+2ar'y₁'+br'y₁=0 so the order of r is reduced plugging in y₁,y₁',y₁'' → r''=0 so r is linear r(t) = C₁+C₂t

Thrm 7.1.2 If the fncs P₁₁,P₁₂,...Pnn & g₁...gn are cts on I: a<t<b, ,then there exists a unique soln x₁=∅₁(t)...xn=∅n(t) of the first order diffl eqn system that satisfies the IC x₁(t₀)=x₁°...xn(t₀)=xn° where t₀∈I and x₁°...xn° are any prescribed numbers.

The soln exists throughout the interval I.

Consider a matrix X(t) with cols x(¹)°...x(ⁿ)°, the solns of the system x°'=P(t)(x°)

The wronskian of the n solns W[x(¹)°...x(ⁿ)°] =det X(t). The solns are linearly independent at a point t₀ iff W(t₀)≠0

consider IVP given, which IC pairs would make it have no soln, inf soln, one soln

put eqn into standard form no soln - where y(t) DNE and Yo =! 0 inf soln - where y(t) DNE and Yo = 0 one soln - where t =! x such that the functions p and g or f and df/dy are cts by the thrms

damped free vibs case 1,2 case of real and repeated roots

since k,m,γ>0; γ²-4km<γ² in case 1. case 1,2, the solns converge to 0 no oscillation