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MATH 4B Differential Equation

Differential Equation Cheatsheet

Tue, Mar 18, 2025 4 min read

Ordinary Differential Equation

y′=F(x,y)y'=F(x,y), F(x,y)=0F(x, y)=0 are equilibrium solutions.

Separable ODE

dydx=F(x,y)=P(x)Q(y)\frac{dy}{dx} = F(x, y) = \frac{P(x)}{Q(y)}

∫Q(y)dy=∫P(x)dx+C\int Q(y)dy = \int P(x)dx + C

First Order Linear ODE

y′+p(x)y=q(x)y' + p(x)y = q(x)

The Method of Integrating Factors

μ(x)=e∫p(x)dx\mu(x) = e^{\int p(x)dx}

y(x)=1μ(x)[∫μ(x)q(x)dx+C]y(x) = \frac{1}{\mu(x)} [\int \mu(x)q(x)dx + C]

Modeling

y′=ay+by' = ay + b y=(y0+ba)eax−bay = (y_0 + \frac{b}{a})e^{ax} - \frac{b}{a}

y′=kyy' = ky y=y0ekty = y_0e^{kt}

Logistic Equation

y′=ry(1−yL)y' = ry(1-\frac{y}{L})

y=LAA+e−rt,A=y0L−y0y = \frac{LA}{A+e^{-rt}}, A = \frac{y_0}{L-y_0}

y=0y = 0 is unstable y=Ly = L is asymptotically stable

Autonomous ODE

y′=F(y)y'=F(y) y(t)=a,F(a)=0y(t) = a, F(a)=0 are equilibrium solutions.

Phase Diagram: The sign of F(y)F(y) indicates increasing or decreasing.

Classification of Equilibria:

  1. F′(a)<0F'(a) < 0 asymptotically stable
  2. F′(a)>0F'(a) > 0 unstable
  3. F′(a)=0F'(a) = 0 semi-stable or no conclusion

Existence and Uniqueness Theorem for Linear ODE

y′+p(t)y=q(t),y(t0)=y0y' + p(t)y = q(t), y(t_0) = y_0

p(t)p(t) and q(t)q(t) are continuous functions and t0t_0 is on that interval.

y′=F(t,y),y(t0)=y0y' = F(t, y), y(t_0) = y_0

F,dFdyF, \frac{dF}{dy} are continuous.

Euler's Method y=y0+F(x0,y0)(x−x0)y = y_0 + F(x_0, y_0)(x - x_0)

Second Order Linear ODE

y′′+p(t)y′+q(t)y=g(t)y'' + p(t)y' + q(t)y = g(t) p,q,gp, q, g are continuous functions.

Constant Coefficients

Homogeneous

ay′′+by′+cy=0ay'' + by' + cy = 0

Superposition Principle: y=c1y1+c2y2y = c_1y_1 + c_2y_2, y=yh+ypy = y_h + y_p Wronskian: W(y1,y2)=∣y1(t0)y2(t0)y1′(t0)y2′(t0)∣W(y_1, y_2) = \begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{vmatrix} Characteristic Equation: ar2+br+c=0ar^2 + br + c = 0

  1. y=[er1t, er2t]y = [e^{r_1t},\ e^{r_2t}]
  2. y=[ert, tert]y = [e^{rt},\ te^{rt}]
  3. r=λ±iμr = \lambda \pm i\mu, then y=[eλtcos⁡μt, eλtsin⁡μt]y = [e^{\lambda t}\cos\mu t,\ e^{\lambda t}\sin\mu t]

Euler's Formula: eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta, eiπ+1=0e^{i\pi} + 1 = 0

Amplitude and Phase Angle:

c1cos⁡(ωt)+c2sin⁡(ωt)=Acos⁡(ωt−θ)A=c12+c22, cos⁡θ=c1c12+c22c_1\cos(\omega t) + c_2\sin(\omega t) = A\cos(\omega t - \theta) \\ A = \sqrt{c_1^2 + c_2^2},\ \cos\theta = \frac{c_1}{\sqrt{c_1^2 + c_2^2}}

Inhomogeneous

Undetermined Coefficients

g(t)g(t) is

  1. polynomial, yp=A0+A1t+⋯+Antny_p = A_0+A_1t+\dots+A_nt^n
  2. exponential, yp=Aeαty_p = Ae^{\alpha t}
  3. trigonometric, yp=Asin⁡βt+Bcos⁡βty_p = A\sin\beta t + B\cos\beta t

When exception happens, multiply by tt

Variation of Parameters

Assumption: v1′y1+v2′y2=0v_1'y_1 + v_2'y_2 = 0

yp=v1y1+v2y2y_p = v_1y_1 + v_2y_2 W=y1y2′−y2y1′v1=−∫y2gWdtv2=∫y1gWdtW = y_1y_2' - y_2y_1' \\ v_1 = -\int\frac{y_2g}{W}dt \\ v_2 = \int\frac{y_1g}{W}dt \\

Modelling

Spring Mass System: mx′′+rx′+kx=F(t)mx'' + rx' + kx = F(t) RLC Circuit: LQ′′+RQ′+1CQ=E(t)LQ'' + RQ' + \frac{1}{C}Q = E(t)

High Order Linear ODE with constant coefficients are the same with second order.

First Order Systems of ODEs

x′=Ax+gx' = Ax + g

  1. Matrix AA's eigenvalues and eigenvectors are r1,v1r_1, v_1 and r2,v2r_2, v_2 x=[er1tv1⃗, er2tv2⃗]x = [e^{r_1t}\vec{v_1},\ e^{r_2t}\vec{v_2}]

  2. r,v,ur, v, u (A−rI)u=v(A-rI)u = v x=[ertv,ert(tv+u)]x = [e^{rt}v, e^{rt}(tv + u)]

  3. r1,2=λ±iμr_{1,2} = \lambda\pm i\mu and v⃗1,2=a⃗±ib⃗\vec{v}_{1,2} = \vec{a}\pm i\vec{b}, then x1=eλt[cos⁡(μt)a⃗−sin⁡(μt)b⃗]x2=eλt[sin⁡(μt)a⃗+cos⁡(μt)b⃗]x_1 = e^{\lambda t}[\cos(\mu t)\vec{a} - \sin(\mu t)\vec{b}] \\ x_2 = e^{\lambda t}[\sin(\mu t)\vec{a} + \cos(\mu t)\vec{b}]

Phase Portrait: test (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix}, x2′>0x_2' > 0: counterclockwise; x2′<0x_2' < 0: clockwise test the first column or AA, A2,1>0A_{2,1} > 0: counterclockwise; A2,1<0A_{2,1} < 0: clockwise

Stability of equilibria

  • Real eigenvalues
    • r1,r2<0r_1, r_2 < 0, asymptotically stable node
    • r1,r2>0r_1, r_2 > 0, unstable node
    • r1r2<0r_1r_2 < 0, unstable saddle
  • Complex eigenvalues
    • λ=0\lambda = 0, stable center
    • λ<0\lambda < 0, spiral sink
    • λ>0\lambda > 0, spiral source

High order linear DEs can be transformed to first order linear system.

Nonhomogenous

If gg is a constant vector, a particular equilibrium solution is xp=−A−1gx_p = -A^{-1}g

Undetermined Coefficients

Variation of Parameters

The fundamental matrix of homogeneous is Φ(t)\Phi(t) μ=∫Φ−1gdt\mu = \int \Phi^{-1}g dt xp=Φμx_p = \Phi \mu

Autonomous

x′⃗=Ax⃗+g⃗=0\vec{x'} = A\vec{x} + \vec{g} = 0

The stability can be determined by AA

Nonlinear Autonomous System of DEs

x′=f(x,y)y′=g(x,y)x' = f(x, y) \\ y' = g(x, y) \\ A=[fx(x0,y0)fy(x0,y0)gx(x0,y0)gy(x0,y0)]A = \begin{bmatrix} f_x(x_0, y_0) & f_y(x_0, y_0) \\ g_x(x_0, y_0) & g_y(x_0, y_0) \\ \end{bmatrix}

Boundary Value Problem

y(a)=y0,y(b)=y1y(a) = y_0, y(b) = y_1

nonhomogeneous has unique solution ↔ homogeneous has only trivial solution nonhomogeneous has no/infinite solution ↔ homogeneous has nontrivial solution

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