Free Online Flashcard Deck

7 Differential Equations and Dynamic Models Free Online FlashCards

Study 7 Differential Equations and Dynamic Models with 12 free online flashcards. Review key terms, definitions, and concepts with this interactive flashcard deck.

12 cards
01
Front

What does a differential equation describe?

Back

A differential equation relates an unknown function to one or more derivatives, describing how a quantity changes rather than specifying the quantity directly.

02
Front

What defines an initial-value problem?

Back

An initial-value problem consists of a differential equation together with an initial condition, such as y(0)=y0y(0)=y_0.

03
Front

What information does a slope field show?

Back

A slope field places a short segment with slope f(t,y)f(t,y) at each point (t,y)(t,y), showing the local directions solution curves follow.

04
Front

How are equilibria found in y′=g(y)y'=g(y)?

Back

For y′=g(y)y'=g(y), an equilibrium solution is a constant value satisfying g(y)=0g(y)=0. A system starting there remains there.

05
Front

Where does y′=y(4−y)y'=y(4-y) increase?

Back

For y′=y(4−y)y'=y(4-y), solutions increase when 0<y<40<y<4 and decrease when y<0y<0 or y>4y>4.

06
Front

When is a differential equation separable?

Back

A separable equation can be written as dydt=f(t)g(y)\frac{dy}{dt}=f(t)g(y), allowing the variables to be isolated: 1g(y)dy=f(t)dt\frac{1}{g(y)}dy=f(t)dt.

07
Front

Why check constant solutions before separating?

Back

Find constant solutions before dividing by g(y)g(y), because dividing by it can remove solutions where g(y)=0g(y)=0.

08
Front

Solve y′=2tyy'=2ty with y(0)=3y(0)=3.

Back

For dydt=2ty\frac{dy}{dt}=2ty with y(0)=3y(0)=3, the solution is y(t)=3et2y(t)=3e^{t^2}.

09
Front

What is the solution of P′=rPP'=rP?

Back

The exponential model dPdt=rP\frac{dP}{dt}=rP has solution P(t)=P0ertP(t)=P_0e^{rt}, where rr is the constant proportional growth rate.

10
Front

What are the doubling-time and half-life formulas?

Back

For exponential growth, the doubling time is ln⁡2r\frac{\ln 2}{r} when r>0r>0. For decay with constant k>0k>0, the half-life is ln⁡2k\frac{\ln 2}{k}.

11
Front

What does the carrying capacity KK represent?

Back

The logistic model is dPdt=rP(1−PK)\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right), where KK is the carrying capacity limiting long-term growth.

12
Front

At what population is logistic growth fastest?

Back

For the logistic model with 0<P<K0<P<K, growth is fastest at P=K2P=\frac{K}{2}, the inflection point of the S-shaped solution.