AMATH 351 Differential Equations and Applications

Contents
  • Differential equations (DE) - relationship between an unknown function yy and its derivative
    • F(y(x),y(x),y(x),)=0F(y(x), y'(x), y''(x), …) = 0
  • Order - order of the highest order derivative in the DE
  • Ordinary differential equation (ODE) - contains derivative with respect to only 1 variable
  • Partial differential equation (PDE) - contains derivative with respect to multiple variables
  • Linear - unknowns of DE do not appear as argument of nonlinear functions or multiply with each other or themselves
  • Valid solution - a solution of an ODE that does not…
    • have a complex number
    • have limxa=\lim\limits_{x\to a} = \infty (limx=\lim\limits_{x\to \infty} = \infty is ok)
    • have division by zero
    • have undefined operation (e.g. ln(2)\ln(-2))
  • Interval of validity - the largest interval where the solution is valid.
  • Distinct solution - y1(x)y2(x)y_{1}(x) \not = y_{2}(x) for some xx
  • 1st order ODE
  • Consider the IVP y=f(t,y)y' = f(t, y), y(0)=0y(0) = 0.
    • first order ODE only
  • If ff and fy\frac{\partial f}{\partial y} are continuous for ta,yb\lvert t \rvert \le a, \lvert y \rvert \le b,
    • then there exists a tha\lvert t \rvert \le h \le a such that there exists a unique solution y(t)=ϕ(t)y(t) = \phi(t) of the IVP.

Given some initial values y(0)=a0,y(0)=a1,,y(n)(0)=any(0) = a_{0}, y'(0) = a_{1}, …, y^{(n)}(0) = a_{n}. Solve the ODE F(y(x),y(x),y(x),)=0F(y(x), y'(x), y''(x), …) = 0

  1. Find the general solution to the ODE (or verify a given solution).
  2. Plug in any initial values to determine the values of unknown constants in the ODE general solution.
  3. Check if there is any additional prompt to do with the solution of the ODE.
  • 1st order
  • Nonlinear, linear
  • Separable
  1. Write the DE in the form of separable DE dydx=g(x)h(y)\dfrac{dy}{dx} = g(x) h(y).
  2. Check if h(y)=0h(y) = 0 generates trivial solutions.
  3. Separate the xx and yy terms: dyh(y)=g(x)dx\dfrac{dy}{h(y)} = g(x) dx. (Don’t consider int. const.)
  4. Integrate both sides and add constant of integration: dyh(y)=g(x)dx\displaystyle\int\dfrac{dy}{h(y)} = \int g(x) dx.
    1. Solve for yy for explicit solution.
  5. IVP
  • 1st order
  • Linear
  1. Write the DE in the form of y(x)+p(x)y(x)=q(x)y'(x) + p(x)y(x) = q(x).
  2. Find the integrating factor μ(x)=exp(p(x)dx)\mu(x) = \exp(\int p(x)dx).
  3. The solution is in the form y(x)=μ(x)q(x)dx+Cμ(x)y(x) = \dfrac{\displaystyle\int \mu(x)q(x)dx + C}{\mu(x)}.
  4. IVP
  • 1st order
  • Nonlinear, linear
  • Exact
  1. Write the DE in the form of N(x,y)y+M(x,y)=0N(x, y)y' + M(x, y) = 0
    1. Let y=f(x,y)y = f(x, y).
    2. Let M(x,y)=fxM(x, y) = \dfrac{\partial f}{\partial x}, and N(x,y)=fyN(x, y) = \dfrac{\partial f}{\partial y}
  2. Check if the DE is an exact DE by verifying My=Nx\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}
  3. Find the solution f(x,y)f(x, y) using one of the following methods:
    • Method A1

      1. Find f(x,y)=f=M x=fmain(x,y)+h(y)\displaystyle f(x, y) = \int \partial f = \int M \ \partial x = f_{\text{main}}(x, y) + h(y).
      2. Solve for h(y)h'(y) in N=fy=fmainy+h(y)N = \dfrac{\partial f}{\partial y} = \dfrac{\partial f_{\text{main}}}{\partial y} + h'(y).
      3. Solve for h(y)h(y) and plug in f(x,y)f(x, y).
    • Method A2

      1. Find f(x,y)=f=N y=fmain(x,y)+g(x)\displaystyle f(x, y) = \int \partial f = \int N \ \partial y = f_{\text{main}}(x, y) + g(x).
      2. Solve for g(x)g'(x) in M=fx=fmainx+g(x)M = \dfrac{\partial f}{\partial x} = \dfrac{\partial f_{\text{main}}}{\partial x} + g'(x).
      3. Solve for g(x)g(x) and plug in f(x,y)f(x, y).
    • Method B

      • Do not consider int. const. g(x)g(x) and h(y)h(y) in the following steps.
      1. Find f1(x,y)f=M xfmixed-terms(x,y)+fy-terms(y)\displaystyle f_{1}(x, y) \equiv \int \partial f = \int M \ \partial x \equiv f_{\text{mixed-terms}}(x, y) + f_{\text{y-terms}}(y)
      2. Find f2(x,y)f=N yfmixed-terms(x,y)+fx-terms(x)\displaystyle f_{2}(x, y) \equiv \int \partial f = \int N \ \partial y \equiv f_{\text{mixed-terms}}(x, y) + f_{\text{x-terms}}(x)
      3. Match up the terms to get the solution
        f(x,y)=fmixed-terms(x,y)+fx-terms(x)+fy-terms(y)=Cf(x, y) = f_{\text{mixed-terms}}(x, y) + f_{\text{x-terms}}(x) + f_{\text{y-terms}}(y) = C
  4. IVP
  • 1st order
  • Nonlinear, linear
  1. Substitute function yy with uu to find an easy-to-solve ODE by…
    1. Given DE y=f(x,y)y' = f(x, y),
      • write the DE in terms of yy'
      • find the substitution u=u(x,y(x))u = u(x, y(x))
      • find the inverse substitution y=y(x,u(x))y = y(x, u(x))
    2. By chain rule, find u=ux+uyyu' = \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} y'.
    3. Substitution (replace yy' with original ODE):
      u=ux+uyyyy=f(x,y)u' = \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} y' \xleftarrow{y'} y' = f(x, y)
      u=ux+uyf(x,y)F(x,y,u)u' = \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} f(x, y) \equiv F(x, y, u)
    4. Substitution (replace yy with inverse substitution):
      u=F(x,y,u)yy(x,u(x))u' = F(x, y, u) \xleftarrow{y} y(x, u(x))
      u=F(x,y(x,u),u)G(x,u)u' = F(x, y(x, u), u) \equiv G(x, u)
  2. Solve the ODE u=G(x,u)u' = G(x, u) for u(x)u(x).
  3. Substitution:
    y(x,u(x))u(x)u(x)y(x, u(x)) \xleftarrow{u(x)} u(x)
    y(x)y(x)
  4. IVP
Description Equations
DE of radioactive decay dNdt=kN(t)\dfrac{dN}{dt} = -k N(t)
Number of nuclei over time (solution) N(t)=N(0)ekt=N0ektN(t) = N(0)e^{-kt} = N_{0}e^{-kt}
Half life N(τ)=N02N(\tau) = \dfrac{N_{0}}{2}
Decay constant k=ln2τk = \dfrac{\ln2}{\tau}
Radioactive dating t=ln(N1(t)N2(t)N2(0)N1(0))ln2τ1τ2τ1τ2t = \dfrac{\ln \left( \dfrac{N_{1}(t)}{N_{2}(t)} \dfrac{N_{2}(0)}{N_{1}(0)} \right)}{\ln 2} \dfrac{\tau_{1}\tau_{2}}{\tau_{1} - \tau_{2}}
Description Equations
Logistic equation
(r,k>0r, k > 0)
P(t)=r(1Pk)PP'(t) = r \left( 1-\dfrac{P}{k} \right)P
Bernoulli equation y+p(t)y=q(t)yny' + p(t)y = q(t)y^{n}
General solutions of Bernoulli equation
(n=1,2,3n = 1,2,3)
n=0:y(t)=ep(t)dtq(t)dt+Cep(t)dtn=1:y(t)=Ceq(t)p(t) dtn=2:y(t)=ep(t)dtep(t)dtq(t)dt+Cn = 0: y(t) = \dfrac{\displaystyle\int e^{\int p(t) dt} q(t) dt + C}{e^{\int p(t) dt}} \newline n = 1: y(t) = Ce^{\int q(t)-p(t) \ dt} \newline n = 2: y(t) = \dfrac{-e^{-\int p(t) dt}}{\displaystyle\int e^{-\int p(t) dt} q(t) dt + C}
General solutions of Bernoulli equation n:y(t)=(1nf(t)q(t)f(t)dt)1/(1n)\forall n: y(t) = \left( \dfrac{1-n}{f(t)} \displaystyle\int q(t)f(t) dt \right)^{1/(1-n)}
where f(t)=e(1n)p(t)dtf(t) = e^{\int (1-n) p(t) dt}
Riccati equation y=q0(t)+q1(t)y+q2(t)y2y' = q_{0}(t) + q_{1}(t)y + q_{2}(t)y^{2}
Solving Riccati equation known particular solution y1y_{1} y=y1+1v(t)y = y_{1} + \dfrac{1}{v(t)}, where vv satisfies
v(t)=(q1+2q2y1)vq2v'(t) = -(q_{1} + 2q_{2}y_{1})v - q_{2}
General solutions of Riccati equation
known particular solution y1y_{1}
y=y1+μ(t)μ(t)q2dt+Cy = y_{1} + \dfrac{\mu(t)}{-\displaystyle\int \mu(t)q_{2} dt + C}
where μ(t)=eq1+2q2y1 dt\mu(t) = e^{\int q_{1}+2q_{2}y_{1} \ dt}
  1. Plot P(t)P'(t) vs P(t)P(t).
  2. Find fixed point PP^{*} such that P=0P' = 0
    • x-intercept of P(t)P'(t) vs P(t)P(t) plot
    • a solution with initial value P(0)=PP(0) = P^{*} is constant over time
  3. Draw flow arrows near fixed points
    • P>0P' > 0 flows to the right
    • P<0P' < 0 flows to the left
  4. Identify stability of fixed points by flow arrows
    • stable - solution approaches the fixed point
    • unstable - solution diverges from the fixed point
    • semi-stable - solution approaches the fixed point from one side and diverges from another
  5. Sketch solution P(t)P(t) vs. tt so that
    • PP increases in region flow to the right
    • PP decreases in region flow to the left
    • solution approaches to stable fixed point
    • solution diverges from unstable fixed point
  • Second order linear differential equation - r(x)y+p(x)y+q(x)y=g(x)r(x)y'' + p(x)y' + q(x)y = g(x)
  • Homogeneous - g(x)=0g(x) = 0
  • Non-homogeneous - g(x)0g(x) \not= 0
  • Linearly independent - c1f(x)+c2g(x)=0c_{1}f(x) + c_{2}g(x) = 0 can only be satisfied by choosing c1=c2=0c_{1} = c_{2} = 0 for functions f,gf, g
  • Wronskian - W(f,g)(x)=fgfgW(f, g)(x) = fg' - f' g
  • 2nd order homogeneous ODE

  • Consider 2nd order homogeneous ODE r(x)y+p(x)y+q(x)y=0r(x)y'' + p(x)y' + q(x)y = 0.

  • If Wronskian W(y1,y2)0W(y_{1}, y_{2}) \not = 0 (y1y_{1} and y2y_{2} are linearly independent solution of the ODE),

    • then the general solution of the ODE is y(x)=c1y1(x)+c2y2(x)y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x).
  • Two functions ff and gg are linearly dependent
    • if their Wronskian W(f,g)(x)=fgfg=0W(f, g)(x) = fg' - f' g = 0.
  • Corollary
    • two linearly independent functions ff and gg has W(f,g)(x)0W(f, g)(x) \not= 0.
  • If y1y_{1} and y2y_{2} be any two solutions of y+p(x)y+q(x)y=0y'' + p(x)y' + q(x)y = 0,
    • then W(y1,y2)=xep(d)dxW(y_{1}, y_{2}) = xe^{-\int p(d) dx}.
  • Corollary: reduction of order formula
    • Known y1y_{1}, then y2=y1Wy12dxy_{2} = y_{1}\displaystyle\int\dfrac{W}{y_{1}^{2}}dx
  • eiβx=cos(βx)+isin(βx)e^{i\beta x} = \cos(\beta x) + i\sin(\beta x)
  • eiβx=cos(βx)isin(βx)e^{-i\beta x} = \cos(\beta x) - i\sin(\beta x)
  • 2nd order
  • Linear
  • Constant coefficient
  • Homogeneous
  1. Write the ODE in the form of ay+by+cy=0ay'' + by' + cy = 0.
  2. Write the characteristic equation aλ2+bλ+c=0a\lambda^{2} + b\lambda + c = 0.
  3. Solve the characteristic equation λ1,λ2=b±b24ac2a\lambda_{1}, \lambda_{2} = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}.
    • If λ1λ2\lambda_{1} \not = \lambda_{2} and λ1,λ2R\lambda_{1}, \lambda_{2} \in \Reals,
      • then the general solution is y(x)=c1eλ1x+c2eλ2xy(x) = c_{1}e^{\lambda_{1}x} + c_{2}e^{\lambda_{2}x}.
    • If λ1=λ2\lambda_{1} = \lambda_{2}, and λ1,λ2R\lambda_{1}, \lambda_{2} \in \Reals,
    • If λ1λ2\lambda_{1} \not = \lambda_{2}, and λ1,λ2C\lambda_{1}, \lambda_{2} \in \Complex, where λ1,λ2=α±iβ\lambda_{1}, \lambda_{2} = \alpha \pm i\beta,
      • then the general solution is y(x)=c1eαxcos(βx)+c2eαxsin(βx)y(x) = c_{1}e^{\alpha x} \cos(\beta x) + c_{2}e^{\alpha x} \sin(\beta x),
      • where α=b2a\alpha = -\dfrac{b}{2a}, β=4acb22a\beta = \dfrac{\sqrt{4ac-b^{2}}}{2a}.
      • by Euler’s formula
  4. IVP
  • 2nd order
  • Linear
  • Non-constant coefficient, constant coefficient
  • Homogeneous
  1. Given a solution y1y_{1}.
  2. Write the ODE in the form of y+p(x)y+q(x)y=0y'' + p(x) y' + q(x)y = 0.
  3. Calculate the Wronskian by Abel’s Theorem W=cep(x)dxW = ce^{-\int p(x)dx}.
  4. Find y2y_{2} by reduction of order formula y2=y1Wy12 dxy_{2} = y_{1} \displaystyle\int \dfrac{W}{y_{1}^{2}} \ dx.
  5. Pick a convenient coefficient cc (but c0c \not= 0).
  6. Write the general solution y(x)=c1y1(x)+c2y2(x)y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x).
  7. IVP
  • 2nd order
  • Linear
  • Non-constant coefficient (of special type)
  • Homogeneous
  • Euler equation
  1. Write the ODE in the form of x2y+αxy+βy=0x^{2}y'' + \alpha xy' + \beta y = 0.
  2. Write the indicial equation s2+(α1)s+β=0s^{2} + (\alpha - 1)s + \beta = 0.
  3. Solve the indicial equation s1,s2=1α±(α1)24β2s_{1}, s_{2} = \dfrac{1 - \alpha \pm \sqrt{(\alpha - 1)^{2} - 4\beta}}{2}.
    • If s1s2s_{1} \not = s_{2} and s1,s2Rs_{1}, s_{2} \in \Reals,
      • then the general solution is y(x)=c1xs1+c2xs2y(x) = c_{1}x^{s_{1}} + c_{2}x^{s_{2}}.
    • If s1=s2s_{1} = s_{2}, and s1,s2Rs_{1}, s_{2} \in \Reals,
    • If s1s2s_{1} \not = s_{2}, and s1,s2Cs_{1}, s_{2} \in \Complex, where s1,s2=η±iμs_{1}, s_{2} = \eta \pm i\mu,
      • then the general solution is
        y(x)=c1xηcos(μln(x))+c2xηsin(μln(x))y(x) = c_{1}x^{\eta} \cos(\mu \ln(x)) + c_{2}x^{\eta} \sin(\mu \ln(x)),
      • where η=1α2\eta = -\dfrac{1-\alpha}{2}, μ=4β(α1)22\mu = \dfrac{\sqrt{4\beta-(\alpha - 1)^{2}}}{2}.
      • by Euler’s formula
  4. IVP
  • 2nd order
  • Linear
  • Constant coefficient
  • Nonhomogeneous
  1. Write the ODE in the form of L[y]ay+by+cy=g(x)L[y] \equiv ay'' + by' + cy = g(x).
  2. Calculate the solution yHy_{H} to the homogeneous problem L[yH]=0L[y_{H}] = 0.
  3. Guess a particular solution yPy_{P} to the nonhomogeneous problem L[yH]=g(x)L[y_{H}] = g(x).
  4. Substitute yPy_{P} into the ODE and solve for any constants.
  5. Write the general solution y(x)=yH(x)+yP(x)y(x) = y_{H}(x) + y_{P}(x).
  6. IVP
  • Consider each term of nonhomogeneity g(x)g(x) separately.
  • Beware of the constants
    • Changes
      • αA\alpha \to A
      • βB\beta \to B
      • Pn(x)Sn(x)P_{n}(x) \to S_{n}(x) contains αiAi,n\alpha_{i} \to A_{i}, \forall n
      • Qm(x)Tm(x)Q_{m}(x) \to T_{m}(x) contains αiAi,n\alpha_{i} \to A_{i}, \forall n
    • No changes
      • k,ωk, \omega
  • If a term of guessed yPy_{P} conflicts with yHy_{H}, then multiply the term of guessed yPy_{P} (not the entire guess of yPy_{P}) by xx.
Term of Nonhomogeneity g(x)g(x) Term of guessed particular solution yPy_{P}
α\alpha AA
αekx\alpha e^{kx} AekxA e^{kx}
Pn(x)=i=0nαixiP_{n}(x) = \sum\limits_{i=0}^{n} \alpha_{i}x^{i} Sn(x)=i=0nAixiS_{n}(x) = \sum\limits_{i=0}^{n} A_{i}x^{i}
ekxPn(x)e^{k x}P_{n}(x) ekxSn(x)e^{k x}S_{n}(x)
αcos(ωx)+βsin(ωx)\alpha\cos(\omega x) + \beta\sin(\omega x) Acos(ωx)+Bsin(ωx)A\cos(\omega x) + B\sin(\omega x)
Pn(x)cos(ωx)+Qm(x)sin(ωx)P_{n}(x)\cos(\omega x) + Q_{m}(x)\sin(\omega x) Sn(x)cos(ωx)+Tm(x)sin(ωx)S_{n}(x)\cos(\omega x) + T_{m}(x)\sin(\omega x)
ekx[Pn(x)cos(ωx)+Qm(x)sin(ωx)]e^{k x}[P_{n}(x)\cos(\omega x) + Q_{m}(x)\sin(\omega x)] ekx[Sn(x)cos(ωx)+Tm(x)sin(ωx)]e^{k x}[S_{n}(x)\cos(\omega x) + T_{m}(x)\sin(\omega x)]
  • 2nd order
  • Linear
  • Non-constant coefficient, constant coefficient
  • Nonhomogeneous
  1. Write the ODE in the form of L[y]y+p(x)y+r(x)y=g(x)L[y] \equiv y'' + p(x)y' + r(x)y = g(x).
  2. Calculate the solution to the homogeneous problem L[yH]=0L[y_{H}] = 0
    • yH=c1y1+c2y2y_{H} = c_{1}y_{1} + c_{2}y_{2}
  3. Calculate the Wronskian W(y1,y2)W(y_{1}, y_{2})
  4. Calculate the particular solution yPy_{P} to the nonhomogeneous problem L[yH]=g(x)L[y_{H}] = g(x)
    • yP=y1g(x)y2W(y1,y2)dx+y2g(x)y1W(y1,y2)dxy_{P} = -y_{1} \displaystyle\int \dfrac{g(x)y_{2}}{W(y_{1}, y_{2})} dx + y_{2} \int \dfrac{g(x)y_{1}}{W(y_{1}, y_{2})} dx
  5. Write the general solution y(x)=yH(x)+yP(x)y(x) = y_{H}(x) + y_{P}(x).
  6. IVP
  • A mass on a spring moves vertically in a fluid bath on Earth.
  • Newton’s second law adds up all the forces
    • F=mx(t)\sum F = mx''(t)
    • Fdamper=γx(t)F_{\mathrm{damper}} = -\gamma x'(t)
    • Fspring=kx(t)F_{\mathrm{spring}} = -kx(t)
    • FexternalF_{\mathrm{external}}
  • ODE: mx(t)+γx(t)+kx(t)=Fext(t)mx''(t) + \gamma x'(t) + kx(t) = F_{\mathrm{ext}}(t)
  • No external force on the system: Fext(t)0F_{\mathrm{ext}}(t) \equiv 0
  • Homogeneous ODE: mx+γx+kx=0 (m>0,γ,k0)\boxed{mx'' + \gamma x' + kx = 0} \ (m > 0, \gamma, k \ge 0)
    • Overdamped system
      • γ24mk>0\gamma^{2} - 4mk > 0
      • λ1λ2R\lambda_{1} \not= \lambda_{2} \in \Reals
      • General solution: x(t)=c1eλ1t+c2eλ2tx(t) = c_{1}e^{\lambda_{1}t} + c_{2}e^{\lambda_{2}t}
    • Critically damped system
      • γ24mk=0\gamma^{2} - 4mk = 0
      • λ1=λ2R\lambda_{1} = \lambda_{2} \in \Reals
      • General solution: x(t)=c1eλ1t+c2teλ2tx(t) = c_{1}e^{\lambda_{1}t} + c_{2}te^{\lambda_{2}t}
    • Underdamped system
      • γ24mk<0\gamma^{2} - 4mk < 0
      • λ1λ2C\lambda_{1} \not= \lambda_{2} \in \Complex
      • General solution: x(t)=eγt/2m[c1cos(ωt)+c2sin(ωt)]x(t) = e^{-\gamma t/2m} [c_{1} \cos(\omega t) + c_{2}\sin(\omega t)]
      • Undamped spring
        • γ=0\gamma = 0
        • General solution: x(t)=c1cos(ωt)+c2sin(ωt)x(t) = c_{1} \cos(\omega t) + c_{2}\sin(\omega t)
        • Phase-amplitude form: x(t)=Acos(ωtφ)x(t) = A\cos(\omega t - \varphi)
          • amplitude A=c12+c22A = \sqrt{c_{1}^{2}+c_{2}^{2}}
          • phase φ=arctan(c2/c1)\varphi = \arctan(c_{2}/c_{1})
          • natural frequency ω=km\omega = \sqrt{\dfrac{k}{m}}
          • period T=2πωT = \dfrac{2\pi}{\omega}
        • Graph: oscillating wave with constant amplitude
      • Underdamped spring
        • γ>0\gamma > 0
        • General solution: x(t)=eγt/2m[c1cos(ωt)+c2sin(ωt)]x(t) = e^{-\gamma t/2m} [c_{1} \cos(\omega t) + c_{2}\sin(\omega t)]
        • Phase-amplitude form: x(t)=Aeγt/2mcos(ωtφ)x(t) = Ae^{-\gamma t/2m}\cos(\omega t - \varphi)
          • amplitude A=c12+c22A = \sqrt{c_{1}^{2}+c_{2}^{2}}
          • phase φ=arctan(c2/c1)\varphi = \arctan(c_{2}/c_{1})
          • natural frequency ω=km\omega = \sqrt{\dfrac{k}{m}}
          • period T=2πωT = \dfrac{2\pi}{\omega}
        • Graph: oscillating wave with exponentially decreasing amplitude
  • Has external force on the system: Fext(t)0F_{\mathrm{ext}}(t) \not= 0
  • Investigate a special case of oscillating external force Fext(t)F0cos(Ωt)F_{\mathrm{ext}}(t) \equiv F_{0}\cos(\Omega t)
  • Non-homogeneous ODE: mx+γx+kx=F0cos(Ωt) (m,γ,k0)\boxed{mx'' + \gamma x' + kx = F_{0}\cos(\Omega t)} \ (m, \gamma, k \ge 0)
    • No damping, no resonance
      • γ=0,Ωω0=km\gamma = 0, \Omega \not= \omega_{0} = \sqrt{\dfrac{k}{m}}
      • General solution: x(t)=(c1+F0m(ω02Ω02))cos(ω0t)+c2sin(ω0t)x(t) = \left( c_{1} + \dfrac{F_{0}}{m(\omega_{0}^{2} - \Omega_{0}^{2})} \right) \cos(\omega_{0}t) + c_{2}\sin(\omega_{0} t)
      • Graph: modulated wave + beats pattern
    • No damping, with resonance
      • γ=0,Ω=ω0=km\gamma = 0, \Omega = \omega_{0} = \sqrt{\dfrac{k}{m}}
      • General solution: x(t)=c1cos(ω0t)+(c2+F02ω0mt)sin(ω0t)x(t) = c_{1}\cos(\omega_{0}t) + \left( c_{2} + \dfrac{F_{0}}{2\omega_{0}m}t \right) \sin(\omega_{0}t)
      • Graph: oscillating wave with linearly increasing amplitude
  • Linear combination - x=c1x1+c2x2++cnxn\mathbf{x} = c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + … + c_n\mathbf{x}_n
  • Linearly dependent - vectors satisfy the equation c1x1+c2x2++cnxn=0c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + … + c_n\mathbf{x}_n = \mathbf{0} such that the constants are not all zero
  • Linearly independent - vectors that are not linearly dependent
  • Wronskian - W[x1,x2,,xn]=det([x1,x2,,xn])=det(X)W[\mathbf{x}_1, \mathbf{x}_2, …, \mathbf{x}_n] = \det([\mathbf{x}_1, \mathbf{x}_2, …, \mathbf{x}_n]) = \det(X)
  • Checking linear independence
    • If W[x1,x2,,xn]0W[\mathbf{x}_1, \mathbf{x}_2, …, \mathbf{x}_n] \not= 0
      • then they are linearly independent
  • Inverse of a square matrix - A1A^{-1} thats satisfies AA1=A1A=InAA^{-1} = A^{-1}A = I_n
  • Finding matrix inverse
    • A=[abcd],A1=1adbc[dbca]A = \begin{bmatrix} a & b \cr c & d \end{bmatrix}, A^{-1} = \dfrac{1}{ad-bc} \begin{bmatrix} d & -b \cr -c & a \end{bmatrix}
  • Solving systems of equations using inverse
    • Ax=bA\mathbf{x} = \mathbf{b}
    • x=A1b\mathbf{x} = A^{-1}\mathbf{b}
  • Finding matrix determinant
    • A=[abcd],det(A)=adbcA = \begin{bmatrix} a & b \cr c & d \end{bmatrix}, \det(A) = ad-bc
    • A=[abcdefghi],det(A)=aefhibdfgi+cdeghA = \begin{bmatrix} a & b & c \cr d & e & f \cr g & h & i \end{bmatrix}, \det(A) = a\begin{vmatrix} e & f \cr h & i \end{vmatrix} - b\begin{vmatrix} d & f \cr g & i \end{vmatrix} + c\begin{vmatrix} d & e \cr g & h \end{vmatrix}
  • Singular matrix - matrix with a determinant of 0
  • Equivalent statements
    • det(A)=0\det(A)=0
    • AA is singular
    • A1A^{-1} does not exist
    • Ax=bA\mathbf{x} = \mathbf{b} has either no solution or infinitely many solutions
    • columns of AA are linearly dependent
    • rows of AA are linearly dependent
  • Eigenvector - vector v\mathbf{v} such that Av=λvA\mathbf{v} = \lambda\mathbf{v} for square matrix AA
    • 0\mathbf{0} is not an eigenvector by convention
  • Eigenvalue - constant λ\lambda corresponding to the eigenvector v\mathbf{v}
  • Finding eigenvalues and eigenvectors
    1. Solve for λ\lambda in det(AλI)=0\det(A-\lambda I) = 0
    2. Substitute λ\lambda into Av=λvA\mathbf{v} = \lambda\mathbf{v} to find relationship between components of eigenvectors
  1. Define nn auxiliary variables y1y_1, …, yny_n for nnth order ODE
    1. Let y1y_1 be the original function in the ODE
    2. Let y2=y1y_2 = y_1'
    3. Let yn=yn1y_n = y_{n-1}'
  2. Rearrange the ODE to isolate the highest order derivative, and write it in terms of the auxiliary variables.
  3. Write a system of ODEs with derivatives of auxiliary variables on the left hand side and their expression on the right hand side in terms of the auxiliary variables
    1. y1=y2y_1' = y_2 (by definition)
    2. yn1=yny_{n-1}' = y_{n} (by definition)
    3. yn=y_{n}' = highest order derivative in step 2
  • {y1=a11y1+a12y2++a1nyn+b1y2=a21y1+a22y2++a2nyn+b2yn=an1y1+an2y2++annyn+bny=Ay+b\begin{cases} y_1' = a_{11}y_1 + a_{12}y_2 + … + a_{1n}y_{n} + b_1 \cr y_2' = a_{21}y_1 + a_{22}y_2 + … + a_{2n}y_{n} + b_2 \cr \vdots \cr y_n' = a_{n1}y_1 + a_{n2}y_2 + … + a_{nn}y_{n} + b_n \end{cases} \Rightarrow \boxed{\mathbf{y}' = A\mathbf{y} + \mathbf{b}}
    • where y=[y1y2yn],y=[y1y2yn],b=[b1b2bn],A=[a11a12a1na21a22a2nan1an2ann]\mathbf{y} = \begin{bmatrix} y_1 \cr y_2 \cr \vdots \cr y_n \end{bmatrix}, \mathbf{y}' = \begin{bmatrix} y_1' \cr y_2' \cr \vdots \cr y_n' \end{bmatrix}, \mathbf{b} = \begin{bmatrix} b_1 \cr b_2 \cr \vdots \cr b_n \end{bmatrix}, A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \cr a_{21} & a_{22} & \cdots & a_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}
  • Homogeneous - b=0\mathbf{b} = \mathbf{0}
  • Nonhomogeneous - b0\mathbf{b} \not= \mathbf{0}
  • Superposition principle
    • If the vectors x1,x2,,xn\mathbf{x}_1, \mathbf{x}_2, …, \mathbf{x}_n are linearly independent solutions of the homogeneous system x=Px\mathbf{x}' = P\mathbf{x}
      • then the general solution x\mathbf{x} is the linear combination of them
        x=c1x1+c2x2++cnxn\mathbf{x} = c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + … + c_n\mathbf{x}_n
  • ODE system
  • 1st order
  • Linear
  • Constant coefficient
  • Homogeneous
  1. Write the system of ODEs in the form of x=Px\mathbf{x}' = P\mathbf{x}
  2. For Pv=λvP\mathbf{v} = \lambda\mathbf{v}, find the eigenvalues of λ\lambda by solving det(PλIn)=0\det(P-\lambda I_n) = 0
  3. For Pv=λvP\mathbf{v} = \lambda\mathbf{v}, find the eigenvectors of by substitution of λ\lambda
  4. Write the general solution
    • n\le n distinct λR\lambda\in\R; nn distinct real v\mathbf{v}
      • General solution: x=c1v1eλ1t++cnvneλnt\mathbf{x} = c_1\mathbf{v}_1 e^{\lambda_1 t} + … + c_n\mathbf{v}_n e^{\lambda_n t}
    • n\le n distinct λC\lambda\in\Complex; nn distinct complex v\mathbf{v}
      • Known: x1=c1v1eλ1t=Re(x1)+iIm(x1)\mathbf{x}_1 = c_1\mathbf{v}_1 e^{\lambda_1 t} = \mathrm{Re}(\mathbf{x}_1) + i\mathrm{Im}(\mathbf{x}_1)
      • General solution: x=c1Re(x1)+c2Im(x1)\mathbf{x} = c_1 \mathrm{Re}(\mathbf{x}_1) + c_2 \mathrm{Im}(\mathbf{x}_1)
    • n\le n distinct λ\lambda; <n<n distinct v\mathbf{v}
      • General solution: x=c1veλt+c2(vteλt+ηeλt)\mathbf{x} = c_1 \mathbf{v} e^{\lambda t} + c_2(\mathbf{v}te^{\lambda t} + \vec{\eta}e^{\lambda t})
        • Find η\vec{\eta} (relationship between its components) by substituting into the ODE
  • Linear
  • Non-constant coefficient (of special type)
  • Homogeneous
  • Euler system
  1. Write the system of ODEs in the form of x=Px\mathbf{x}' = P\mathbf{x}
  2. For Pv=λvP\mathbf{v} = \lambda\mathbf{v}, find the eigenvalues of λ\lambda by solving det(PλIn)=0\det(P-\lambda I_n) = 0
  3. For Pv=λvP\mathbf{v} = \lambda\mathbf{v}, find the eigenvectors of by substitution of λ\lambda
  4. Write the general solution
    • n\le n distinct λR\lambda\in\R; nn distinct real v\mathbf{v}
      • General solution: x=c1v1tλ1++cnvntλn\mathbf{x} = c_1\mathbf{v}_1 t^{\lambda_1} + … + c_n\mathbf{v}_n t^{\lambda_n}
    • Other conditions are not discussed
  • Laplace transform - L[f(t)]=F(s)=0estf(t) dt\mathcal{L}[f(t)] = F(s) = \displaystyle\int_{0}^{\infin} e^{-st}f(t) \ dt
  • Heaviside function (unit step function)
    • uc(t)=u(tc)={0t<c1tcu_c(t) = u(t-c) = \begin{cases} 0 & t < c \cr 1 & t \ge c \end{cases}
  • Laplace transform is linear
    • L[c1f(t)+c2g(t)]=c1L[f(t)]+c2L[g(t)]\mathcal{L}[c_{1}f(t)+c_{2}g(t)] = c_{1}\mathcal{L}[f(t)] + c_{2}\mathcal{L}[g(t)]
  • Laplace transforms of derivatives incorporate initial conditions
    • L[f(t)]=sF(s)f(0)\mathcal{L}[f'(t)] = sF(s) - f(0)
    • L[f(t)]=s2F(s)sf(0)f(0)\mathcal{L}[f''(t)] = s^2F(s)-sf(0)-f'(0)
    • L[f(n)(t)]=snF(s)sn1f(0)sn2f(1)(0)f(n1)(0)\mathcal{L}[f^{(n)}(t)] = s^nF(s)-s^{n-1}f(0) - s^{n-2}f^{(1)}(0) - … - f^{(n-1)}(0)
  • Heaviside function has a simple Laplace transforms
    • L[uc(t)]=escs\mathcal{L}[u_c(t)] = \dfrac{e^{-sc}}{s}
  • Time domain translation
    • L[f(tc)uc(t)]=escL[f(t)]\mathcal{L}[f(t-c)u_{c}(t)] = e^{-sc}\mathcal{L}[f(t)]
    • L1[escL[f(t)]]=f(tc)uc(t)\mathcal{L}^{-1}[e^{-sc}\mathcal{L}[f(t)]] = f(t-c)u_c(t)
  • Laplace domain translation
    • L[ectf(t)]=F(sc)\mathcal{L}[e^{ct}f(t)] = F(s-c)
    • L1[F(sc)]=ectf(t)\mathcal{L}^{-1}[F(s-c)] = e^{ct}f(t)
  1. Time domain: difficult ODE
    • Laplace transform (tst \to s)
  2. Laplace domain: easy algebra problem
    • Solve the algebra problem
  3. Laplace domain: solution to algebra problem
    • Inverse Laplace transform (sts \to t)
  4. Time domain: solution of difficult ODE
    • Problem solved
Inverse L.T.
f(t)f(t)
Laplace Transform
F(s)F(s)
Inverse L.T.
f(t)f(t)
Laplace Transform
F(s)F(s)
11 1s\dfrac{1}{s} eate^{at} 1sa\dfrac{1}{s-a}
tnt^{n} n!sn+1\dfrac{n!}{s^{n+1}} t\sqrt{t} π2s3/2\dfrac{\sqrt{\pi}}{2s^{3/2}}
sin(at)\sin(at) as2+a2\dfrac{a}{s^{2}+a^{2}} tsin(at)t\sin(at) 2as(s2+a2)2\dfrac{2as}{(s^{2}+a^{2})^{2}}
cos(at)\cos(at) ss2+a2\dfrac{s}{s^{2}+a^{2}} tcos(at)t\cos(at) s2a2(s2+a2)2\dfrac{s^{2}-a^{2}}{(s^{2}+a^{2})^{2}}
sin(at)atcos(at)\sin(at)-at\cos(at) 2a3(s2+a2)2\dfrac{2a^{3}}{(s^{2}+a^{2})^{2}} cos(at)atsin(at)\cos(at)-at\sin(at) s(s2a2)(s2+a2)2\dfrac{s(s^{2}-a^{2})}{(s^{2}+a^{2})^{2}}
sin(at)+atcos(at)\sin(at)+at\cos(at) 2as2(s2+a2)2\dfrac{2as^{2}}{(s^{2}+a^{2})^{2}} cos(at)+atsin(at)\cos(at)+at\sin(at) s(s2+3a2)(s2+a2)2\dfrac{s(s^{2}+3a^{2})}{(s^{2}+a^{2})^{2}}
sinh(at)\sinh(at) as2a2\dfrac{a}{s^{2}-a^{2}} sin(at+b)\sin(at+b) ssin(b)+acos(b)s2+a2\dfrac{s\sin(b)+a\cos(b)}{s^2+a^2}
cosh(at)\cosh(at) ss2a2\dfrac{s}{s^{2}-a^{2}} cos(at+b)\cos(at+b) scos(b)asin(b)s2+a2\dfrac{s\cos(b)-a\sin(b)}{s^2+a^2}
eatsin(bt)e^{at}\sin(bt) b(sa)2+b2\dfrac{b}{(s-a)^2+b^2} eatsinh(bt)e^{at}\sinh(bt) b(sa)2b2\dfrac{b}{(s-a)^2-b^2}
eatcos(bt)e^{at}\cos(bt) sa(sa)2+b2\dfrac{s-a}{(s-a)^2+b^2} eatcosh(bt)e^{at}\cosh(bt) sa(sa)2b2\dfrac{s-a}{(s-a)^2-b^2}
uc(t)u_{c}(t) escs\dfrac{e^{-sc}}{s}