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Lesson 100 — Integration Workshop Trimester 10 (ODEs)

Integrating workshop on ODEs: separable, first-order linear, populations, second-order with constant coefficients, vibrations, RLC, numerical Euler, and Newton cooling.

Used in: AP Calculus BC (USA) · Leistungskurs Mathematik Klasse 12 (Germany) · H2 Mathematics (Singapore) · Spécialité Maths Terminale (France)

y′+p(x) y=q(x)⏟1st-order linear∣ay′′+by′+cy=q(x)⏟2nd-order const. coeff.\underbrace{y' + p(x)\,y = q(x)}_{\text{1st-order linear}} \quad \Big| \quad \underbrace{ay'' + by' + cy = q(x)}_{\text{2nd-order const. coeff.}}

Two structural models cover nearly every application from the trimester. The one on the left solves with integrating factor μ=e∫p dx\mu = e^{\int p\,dx}. The one on the right solves via the characteristic equation aλ2+bλ+c=0a\lambda^2 + b\lambda + c = 0. Non-analytic cases receive Euler or RK4.

Choose your door

Rigorous notation, full derivation, hypotheses

Map of techniques and ODE decision criteria

"The existence and uniqueness theorem says: there is exactly one solution curve through each point (x₀, y₀) where f and ∂f/∂y are continuous." — Lebl, Notes on Diffy Qs §1.2

"The idea behind the integrating factor is to write the left-hand side as an exact derivative." — OpenStax Calculus Vol. 2 §4.5

Techniques table for the trimester:

ODECanonical formTechniqueGeneral solution
Separabley′=g(x)h(y)y' = g(x)h(y)Separate and integrateImplicit
Linear 1sty′+py=qy'+py=qμ=e∫p\mu=e^{\int p}y=1μ∫μq dx+Cy = \frac{1}{\mu}\int\mu q\,dx + C
MalthusP˙=rP\dot P = rPSeparableP=P0ertP = P_0 e^{rt}
LogisticP˙=rP(1−P/K)\dot P = rP(1-P/K)Separable + partial fractionsP=K/(1+Ae−rt)P = K/(1+Ae^{-rt})
CoolingT˙=−k(T−Ta)\dot T = -k(T-T_a)SeparableT=Ta+(T0−Ta)e−ktT = T_a + (T_0-T_a)e^{-kt}
2nd homog.ay′′+by′+cy=0ay''+by'+cy=0Char. aλ2+bλ+c=0a\lambda^2+b\lambda+c=03 cases
2nd forceday′′+by′+cy=qay''+by'+cy=qyh+ypy_h + y_pUndetermined coeff.
Numerical EulerAnyyn+1=yn+hfy_{n+1}=y_n+hfSequence

SVG — ODE decision diagram

ODE arrivesOrder?1st order2nd orderSeparable?→ IntegrateLinear?→ Integr. factorConst. coeff?→ Char.Numericalnot analytic

Worked examples

Exercise list

45 exercises · 11 with worked solution (25%)

Application 19Understanding 7Modeling 12Challenge 5Proof 2
  1. Ex. 100.1Application

    General solution of y′=3xyy' = 3xy:

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    Solve onlineActive Calculus · §7.4 · 1 (y'=3xy)
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    Separable: dy/y=3x dxdy/y=3x\,dx. Integrate: ln⁡∣y∣=3x2/2+C1\ln|y|=3x^2/2+C_1. y=Ce3x2/2y=Ce^{3x^2/2}.
    Show step-by-step (with the why)
    1. Step 1. Separable: dy/y=3x dxdy/y = 3x\,dx.
    2. Step 2. Integrate both sides: ln⁡∣y∣=3x2/2+C1\ln|y| = 3x^2/2 + C_1.
    3. Step 3. Exponentiate: ∣y∣=eC1e3x2/2|y| = e^{C_1}e^{3x^2/2}. Absorb constant: y=Ce3x2/2y = Ce^{3x^2/2}.
    4. Trick: Separable = isolate dydy from dxdx, integrate each side.
  2. Ex. 100.2Application

    General solution of y′=−xy2y' = -xy^2:

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    Solve onlineActive Calculus · §7.4 · 2 (y'=xy²)
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    Separable: dy/y2=−x dxdy/y^2=-x\,dx. Integrate: −1/y=−x2/2+C1-1/y=-x^2/2+C_1, so 1/y=x2/2−C11/y=x^2/2-C_1. Rename −C1=C-C_1=C: y=1/(x2/2+C)y=1/(x^2/2+C).
  3. Ex. 100.3Application

    General solution of y′=ycos⁡xy' = y\cos x:

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    Solve onlineActive Calculus · §7.4 · 3 (y'=y cosx)
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    1. Step 1. Separable: dy/y=cos⁡x dxdy/y = \cos x\,dx.
    2. Step 2. Integrate: ln⁡∣y∣=sin⁡x+C1\ln|y| = \sin x + C_1.
    3. Step 3. Exponentiate: y=Cesin⁡xy = Ce^{\sin x}.
    4. Trick: The exponent is the antiderivative of cos⁡x\cos x, which is sin⁡x\sin x.
  4. Ex. 100.4Application

    General solution of y′=x/yy' = x/y (separable ODE):

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    Solve onlineActive Calculus · §7.4 · 4 (yy'=x)
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    Separable: y dy=x dxy\,dy=x\,dx. Integrate: y2/2=x2/2+C1y^2/2=x^2/2+C_1. y2=x2+Cy^2=x^2+C.
  5. Ex. 100.5Application

    General solution of y′=1+y2y' = 1 + y^2:

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    Solve onlineActive Calculus · §7.4 · 5 (y'=1+y²)
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    1. Step 1. Separable: dy/(1+y2)=dxdy/(1+y^2) = dx.
    2. Step 2. Integrate: arctan⁡y=x+C\arctan y = x + C.
    3. Step 3. Solve: y=tan⁡(x+C)y = \tan(x+C).
    4. Trick: ∫dy/(1+y2)=arctan⁡y\int dy/(1+y^2) = \arctan y — standard arctangent integral.
  6. Ex. 100.6Application

    Solution of y′=y2exy' = y^2 e^x, y(0)=1y(0) = 1:

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    Solve onlineActive Calculus · §7.4 · 6 (y'=y²e^x)
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    Separable: dy/y2=ex dxdy/y^2=e^x\,dx. −1/y=ex+C-1/y=e^x+C. IC y(0)=1y(0)=1: −1=1+C⇒C=−2-1=1+C\Rightarrow C=-2. y=1/(2−ex)y=1/(2-e^x).
    Show step-by-step (with the why)
    1. dy/y2=ex dxdy/y^2=e^x\,dx. Integrate: −1/y=ex+C-1/y=e^x+C.
    2. IC: −1=1+C⇒C=−2-1=1+C\Rightarrow C=-2.
    3. y=1/(2−ex)y=1/(2-e^x).
  7. Ex. 100.7ApplicationAnswer key

    Solution of yy′=xyy' = x, y(0)=1y(0) = 1:

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    Solve onlineActive Calculus · §7.4 · 7 (yy'=x y(0)=1)
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    1. Step 1. Rewrite: y dy=x dxy\,dy = x\,dx.
    2. Step 2. Integrate: y2/2=x2/2+Cy^2/2 = x^2/2 + C.
    3. Step 3. IC: y(0)=1⇒1/2=0+C⇒C=1/2y(0)=1 \Rightarrow 1/2 = 0 + C \Rightarrow C=1/2.
    4. Step 4. y2=x2+1y^2 = x^2 + 1, so y=x2+1y = \sqrt{x^2+1} (+ sign since y(0)=1>0y(0)=1>0).
    5. Trick: The IC selects the positive branch of the implicit hyperbola.
  8. Ex. 100.8ApplicationAnswer key

    Logistic equation P′=0.5P(1−P/3)P' = 0.5P(1-P/3), P(0)=1P(0)=1. Solution P(t)P(t):

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    Solve onlineActive Calculus · §7.6 · logística K=3 P0=1
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    Logistic with K=3K=3, r=0.5r=0.5, P0=1P_0=1. A=(3−1)/1=2A=(3-1)/1=2. P=3/(1+2e−0.5t)P=3/(1+2e^{-0.5t}).
  9. Ex. 100.9ApplicationAnswer key

    General solution of y′+3y=6y' + 3y = 6:

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    Solve onlineActive Calculus · §7.5 · y'+3y=6
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    Factor e3xe^{3x}. (e3xy)′=6e3x(e^{3x}y)'=6e^{3x}. y=2+Ce−3xy=2+Ce^{-3x}.
    Show step-by-step (with the why)
    1. Step 1. Standard form: y′+3y=6y' + 3y = 6. Integrating factor: μ=e3x\mu = e^{3x}.
    2. Step 2. Multiply: (e3xy)′=6e3x(e^{3x}y)' = 6e^{3x}.
    3. Step 3. Integrate: e3xy=2e3x+Ce^{3x}y = 2e^{3x} + C.
    4. Step 4. Divide by e3xe^{3x}: y=2+Ce−3xy = 2 + Ce^{-3x}.
    5. Trick: Particular solution = right side divided by coefficient of yy: 6/3=26/3=2.
  10. Ex. 100.10Application

    General solution of y′′+y′−2y=0y'' + y' - 2y = 0:

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    Solve onlineActive Calculus · §7.1 · y''+ y'-2y=0
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    1. Step 1. Characteristic equation: λ2+λ−2=0\lambda^2 + \lambda - 2 = 0.
    2. Step 2. Factor: (λ+2)(λ−1)=0(\lambda+2)(\lambda-1) = 0. Roots: λ1=−2,λ2=1\lambda_1 = -2, \lambda_2 = 1.
    3. Step 3. Distinct real roots: y=C1e−2x+C2exy = C_1 e^{-2x} + C_2 e^x.
    4. Trick: Substitute y=eλxy = e^{\lambda x} into the ODE to get the characteristic equation directly.
  11. Ex. 100.11ApplicationAnswer key

    Equilibria and stability of y′=2y−y2y' = 2y - y^2:

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    Solve onlineActive Calculus · §7.1 · equilíbrios y'=2y-y²
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    y′=2y−y2=y(2−y)y'=2y-y^2=y(2-y). Equilibria: y=0y=0 (unstable, since f′(0)=2>0f'(0)=2>0) and y=2y=2 (stable, since f′(2)=−2<0f'(2)=-2<0).
    Show step-by-step (with the why)
    1. Step 1. Equilibria: solve 2y−y2=y(2−y)=02y - y^2 = y(2-y) = 0. Roots: y∗=0y^* = 0 and y∗=2y^* = 2.
    2. Step 2. f(y)=2y−y2f(y) = 2y - y^2. Derivative: f′(y)=2−2yf'(y) = 2-2y.
    3. Step 3. At y=0y=0: f′(0)=2>0f'(0)=2>0 — unstable. At y=2y=2: f′(2)=−2<0f'(2)=-2<0 — stable.
    4. Trick: Sign of f′(y∗)f'(y^*): positive = unstable (repeller); negative = stable (attractor).
  12. Ex. 100.12Application

    y′=y−ty' = y - t, y(0.2)=0.9y(0.2) = 0.9, h=0.2h = 0.2. Euler: y(0.4)≈y(0.4) \approx: (Resp: y1=0.9+0.2×0.7=1.04y_1 = 0.9 + 0.2 \times 0.7 = 1.04)

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    Solve onlineActive Calculus · §7.3 · Euler y'=y-t
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    Euler: f(0.2, 0.9)=y−t=0.9−0.2=0.7f(0.2,\,0.9)=y-t=0.9-0.2=0.7. y1=0.9+0.2×0.7=1.04y_1=0.9+0.2\times0.7=1.04.
  13. Ex. 100.13Understanding

    An ordinary differential equation (ODE) is:

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    Solve onlineActive Calculus · §7.1 · definição EDO
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    An ordinary differential equation (ODE) relates an unknown function y(t)y(t) with its derivatives with respect to the single independent variable tt.
  14. Ex. 100.14Understanding

    The difference between stable and unstable equilibrium is:

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    Solve onlineActive Calculus · §7.1 · estabilidade conceitual
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    Stable equilibrium (attractor): every solution nearby converges to it. Unstable (repeller): nearby solutions move away. Determined by the sign of f′(y∗)f'(y^*).
  15. Ex. 100.15UnderstandingAnswer key

    A 1st-order ODE is linear when:

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    Solve onlineActive Calculus · §7.1 · linearidade 1ª ordem
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    See the reference indicated in source for detailed solution.
  16. Ex. 100.16Understanding

    An ODE of the type y′=f(y/x)y' = f(y/x) is called homogeneous and is solved by:

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    Solve onlineActive Calculus · §7.4 · EDO homogênea
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    Homogeneous ODE: y′=f(y/x)y'=f(y/x). Substitution v=y/xv=y/x, y=vxy=vx, y′=v+xv′y'=v+xv'. Results in separable in vv.
  17. Ex. 100.17ModelingAnswer key

    Solution of the logistic equation P′=rP(1−P/K)P' = rP(1-P/K), P(0)=P0P(0) = P_0:

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    Solve onlineActive Calculus · §7.6 · solução logística geral
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    Logistic: P=K/(1+Ae−rt)P=K/(1+Ae^{-rt}) with A=(K−P0)/P0A=(K-P_0)/P_0. Rewriting: P=KP0/(P0+(K−P0)e−rt)P=KP_0/(P_0+(K-P_0)e^{-rt}).
  18. Ex. 100.18ModelingAnswer key

    Falling velocity with linear drag (mv˙=mg−kvm\dot v = mg - kv, v(0)=0v(0)=0):

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    Solve onlineActive Calculus · §7.5 · queda com arrasto
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    See the reference indicated in source for detailed solution.
  19. Ex. 100.19Modeling

    Coffee at 100°C in a room at 20°C. Temperature T(t)T(t):

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    Solve onlineActive Calculus · §7.5 · Newton cooling
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    Newton: T=Tamb+(T0−Tamb)e−kt=20+80e−ktT=T_{\text{amb}}+(T_0-T_{\text{amb}})e^{-kt}=20+80e^{-kt} with T0=100T_0=100, Tamb=20T_{\text{amb}}=20.
  20. Ex. 100.20Modeling

    ODE of the series RLC circuit:

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    Solve onlineActive Calculus · §7.3 · circuito RLC EDO
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    KVL in RLC circuit: sum of voltages = source. With I=q˙I=\dot q: Lq¨+Rq˙+q/C=EL\ddot q+R\dot q+q/C=E.
  21. Ex. 100.21Modeling

    The Malthus model P′=rPP' = rP is appropriate when:

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    Solve onlineActive Calculus · §7.1 · Malthus limitações
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    Malthus: P′=rPP'=rP, exponential growth without limit. Valid for P≪KP\ll K. For large PP: logistic is more realistic (limited resource).
  22. Ex. 100.22Modeling

    y′=xyy' = xy, y(0)=1y(0)=1, h=0.5h=0.5: Euler 1 step, y(0.5)≈y(0.5) \approx: (Resp: y1=1y_1 = 1, exact ≈1.133\approx 1.133)

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    Solve onlineActive Calculus · §7.3 · Euler y'=xy y(0)=1
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    Euler: f(0,1)=0⋅1=0f(0,1)=0\cdot1=0. y1=1+0.5×0=1.0y_1=1+0.5\times0=1.0. Exact: y=ex2/2y=e^{x^2/2}, y(0.5)=e0.125≈1.133y(0.5)=e^{0.125}\approx1.133. Truncation error large because h=0.5h=0.5 is large.
  23. Ex. 100.23Modeling

    Oscillator with m=1m=1 kg, b=4b=4, k=13k=13, x(0)=3x(0)=3, x˙(0)=0\dot x(0)=0. Solution x(t)x(t):

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    Solve onlineActive Calculus · §7.1 · oscilador sub-amortecido
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    See the reference indicated in source for detailed solution.
  24. Ex. 100.24UnderstandingAnswer key

    Comparison between the exponential and logistic models for population growth:

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    Solve onlineActive Calculus · §7.6 · comparação logística vs exponencial
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    See the reference indicated in source for detailed solution.
  25. Ex. 100.25Understanding

    A 1st-order ODE is separable when it can be written in the form:

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    Solve onlineActive Calculus · §7.4 · identificar separável
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    Separable ODE: y′/g(y)=f(x)y'/g(y)=f(x) — separate into different sides. Distractor py+qpy+q: linear, not separable in general. Distractor y2+xy^2+x: does not factor.
  26. Ex. 100.26Understanding

    The maximum growth rate of the logistic equation is:

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    Solve onlineActive Calculus · §7.6 · máximo taxa logística
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    P′=rP(1−P/K)P'=rP(1-P/K). Maximum at: P=K/2P=K/2. Maximum value: r(K/2)(1/2)=rK/4r(K/2)(1/2)=rK/4.
  27. Ex. 100.27Modeling

    Maximum sustainable yield (MSY) in the logistic equation P′=rP(1−P/K)P' = rP(1-P/K):

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    Solve onlineActive Calculus · §7.6 · colheita sustentável
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    Logistic equation with harvest hh: equilibria where rP(1−P/K)=hrP(1-P/K)=h. Maximum sustainable harvest: hmax⁡=rK/4h_{\max}=rK/4. Stock balance is integration of the solution.
  28. Ex. 100.28Application

    General solution of y′=tan⁡xy' = \tan x:

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    Solve onlineActive Calculus · §7.4 · y'=tanx
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  29. Ex. 100.29Application

    Solution of y′=x(1−y)y' = x(1-y), y(0)=0y(0) = 0:

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    Solve onlineActive Calculus · §7.4 · y'=(1-y)x
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    See the reference indicated in source for detailed solution.
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    1. Step 1. Separable: dy/(1−y)=x dxdy/(1-y) = x\,dx.
    2. Step 2. Integrate: −ln⁡∣1−y∣=x2/2+C-\ln|1-y| = x^2/2 + C.
    3. Step 3. ∣1−y∣=e−Ce−x2/2|1-y| = e^{-C}e^{-x^2/2}. Absorb: 1−y=Ae−x2/21-y = Ae^{-x^2/2}.
    4. Step 4. IC: y(0)=0⇒1−0=A⇒A=1y(0)=0 \Rightarrow 1-0=A \Rightarrow A=1.
    5. Step 5. y=1−e−x2/2y = 1 - e^{-x^2/2}.
    6. Trick: Substitute u=1−yu = 1-y: u′=−y′=−xuu' = -y' = -xu — more direct equation.
  30. Ex. 100.30ApplicationAnswer key

    Solution of y′=ysin⁡xy' = y\sin x, y(0)=1y(0) = 1:

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    Solve onlineActive Calculus · §7.4 · y'=y sinx y(0)=1
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    1. Step 1. Separable: dy/y=sin⁡x dxdy/y = \sin x\,dx.
    2. Step 2. Integrate: ln⁡∣y∣=−cos⁡x+C\ln|y| = -\cos x + C.
    3. Step 3. y=Ae−cos⁡xy = Ae^{-\cos x}. IC: y(0)=1⇒A=ecos⁡0=e1=ey(0)=1 \Rightarrow A=e^{\cos 0}=e^1=e.
    4. Step 4. y=e⋅e−cos⁡x=e1−cos⁡xy = e \cdot e^{-\cos x} = e^{1-\cos x}.
    5. Trick: The exponent is −cos⁡x+1=1−cos⁡x-\cos x + 1 = 1 - \cos x — compact and elegant.
  31. Ex. 100.31Modeling

    Octopus: K=1000K=1000, r=0.3r=0.3/year. Logistic ODE and maximum growth rate:

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    Solve onlineActive Calculus · §7.6 · polvo logístico
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    1. Step 1. Logistic ODE: P˙=rP(1−P/K)=0.3P(1−P/1000)\dot P = rP(1-P/K) = 0.3P(1-P/1000).
    2. Step 2. Growth rate is a function of PP: g(P)=0.3P(1−P/1000)g(P) = 0.3P(1-P/1000).
    3. Step 3. Maximum when g′(P)=0g'(P)=0: 0.3(1−2P/1000)=0⇒P=K/2=5000.3(1-2P/1000)=0 \Rightarrow P=K/2=500.
    4. Step 4. Maximum rate: g(500)=0.3×500×0.5=75g(500) = 0.3 \times 500 \times 0.5 = 75 octopuses/year.
    5. Trick: In logistics, maximum rate always occurs at P=K/2P = K/2.
  32. Ex. 100.32Modeling

    Octopus: K=1000K=1000, r=0.3r=0.3, P(0)=100P(0)=100. P(10)≈P(10) \approx: (Resp: ≈691\approx 691 octopuses)

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    Solve onlineActive Calculus · §7.6 · polvo P(10)
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    Logistic: A=(K−P0)/P0=(1000−100)/100=9A=(K-P_0)/P_0=(1000-100)/100=9. P(10)=1000/(1+9e−0.3×10)=1000/(1+9e−3)≈691P(10)=1000/(1+9e^{-0.3\times10})=1000/(1+9e^{-3})\approx691 octopuses.
  33. Ex. 100.33Challenge

    For linear ODE y′+p(t)y=0y' + p(t)y = 0, the Wronskian of two solutions satisfies:

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    Solve onlineActive Calculus · §7.1 · wronskiano 1ª ordem
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  34. Ex. 100.34Challenge

    The solution of y′−y=1/xy' - y = 1/x involves:

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    Solve onlineActive Calculus · §7.5 · y'-y=1/x integral especial
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  35. Ex. 100.35Application

    General solution of y′=(y2−1)/yy' = (y^2-1)/y:

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    Solve onlineActive Calculus · §7.4 · y'=(y²-1)/y
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    Separate: y/(y2−1) dy=dxy/(y^2-1)\,dy=dx. Integrate: 12ln⁡∣y2−1∣=x+C1\tfrac12\ln|y^2-1|=x+C_1, so ln⁡∣y2−1∣=2x+C\ln|y^2-1|=2x+C, i.e., y2−1=Ce2xy^2-1=Ce^{2x}.
  36. Ex. 100.36ApplicationAnswer key

    General solution of xy′=1/yxy' = 1/y (or equivalently yy′=1/xyy' = 1/x):

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    Solve onlineActive Calculus · §7.4 · yy'=1/x
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  37. Ex. 100.37Challenge

    For xy′=y+xxy' = y + x (homogeneous ODE), the solution is:

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    Solve onlineActive Calculus · §7.4 · EDO homogênea difícil
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    Substitute v=y/xv=y/x, y=vxy=vx, y′=v+xv′y'=v+xv'. Then xv′+v=v+1xv'+v=v+1, so xv′=1xv'=1, v=ln⁡x+Cv=\ln x+C. Back: y=x(ln⁡x+C)=Cx+xln⁡xy=x(\ln x+C)=Cx+x\ln x.
  38. Ex. 100.38Modeling

    Oscillator with m=1m=1 kg and k=10k=10 N/m. Critical damping bcb_c:

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    Solve onlineActive Calculus · §7.1 · amortecimento crítico massa 1 mola 10
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    See the reference indicated in source for detailed solution.
  39. Ex. 100.39Challenge

    Bacteria: per capita 3 when P=6P=6 and per capita 2 when P=9P=9. Logistic ODE and carrying capacity:

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    Solve onlineActive Calculus · §7.6 · 6 (bactérias per capita)
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    Per capita rate is linear: P′/P=a−bPP'/P=a-bP. System: a−6b=3a-6b=3 and a−9b=2a-9b=2. Subtract: 3b=1⇒b=1/33b=1\Rightarrow b=1/3, a=5a=5. Thus P′=P(5−P/3)P'=P(5-P/3); carrying capacity K=a/b=15K=a/b=15.
  40. Ex. 100.40Proof

    The mathematical justification for the separation of variables method is:

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    Solve onlineActive Calculus · §7.4 · justificativa da separação
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    See the reference indicated in source for detailed solution.
  41. Ex. 100.41Application

    Doubling time of P′=rPP' = rP (r>0r > 0):

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    Solve onlineActive Calculus · §7.1 · tempo de dobramento
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    See the reference indicated in source for detailed solution.
  42. Ex. 100.42ApplicationAnswer key

    Empty tank of volume VV, brine cinc_{\text{in}} enters at rate QVQ_V. Amount of salt Q(t)Q(t):

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    Solve onlineActive Calculus · §7.5 · tanque de sal vazio
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    ODE: Q˙+(QV/V)Q=QVcin\dot Q+(Q_V/V)Q=Q_Vc_{\text{in}}. With Q(0)=0Q(0)=0: Q=cinV(1−e−QVt/V)Q=c_{\text{in}}V(1-e^{-Q_Vt/V}). Saturates at cinVc_{\text{in}}V.
  43. Ex. 100.43Modeling

    Present value of USD 50 000 to be received in 20 years, continuous rate r=5%r=5\%: (Resp: =50000e−1≈= 50000e^{-1} \approx USD 18 394)

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    Solve onlineActive Calculus · §7.5 · valor presente
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    Continuous present value: PV=50000 e−0.05×20=50000 e−1≈18394PV=50000\,e^{-0.05\times20}=50000\,e^{-1}\approx18394 USD.
  44. Ex. 100.44Challenge

    Species with minimum survival threshold TT (Allee effect): ODE and behavior:

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    Solve onlineActive Calculus · §7.6 · efeito Allee
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    Allee effect (survival threshold TT): for P<TP<T population collapses; for P>TP>T grows to KK. Model: P′=rP(1−P/K)(1−T/P)P'=rP(1-P/K)(1-T/P).
  45. Ex. 100.45Proof

    Prove that P=KP=K is globally stable equilibrium of the logistic equation for P0>0P_0 > 0, r>0r > 0:

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    Active Calculus · §7.6 · prova estabilidade K
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    For r>0r>0, P0>0P_0>0: P′=rP(1−P/K)>0P'=rP(1-P/K)>0 for P<KP<K and P′<0P'<0 for P>KP>K. By first-order stability theorem, P=KP=K is globally stable for P0>0P_0>0.

Sources

Updated on 2026-05-06 · Author(s): Clube da Matemática

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