Calculus II - Ordinary Differential Equations

16 flashcards · Mathematics · by luckydog789 (harrison holbrook)

1. Order

   The order of a differential equation is the highest order derivative of yyy appearing in the equation (or the highest order of whatever is being differentiated)

2. Degree

   The degree of a differential equation is the exponent of the highest-order derivative in the differential equation.

3. Linear differential equation

   If P0P_0P0​ , P1P_1P1​ ...Pn...P_n...Pn​ and QQQ are functions of the variable xxx , then an equation in the form P0(x)dnydxn+P1(x)dn−1ydxn−1+...+Pn−1(x)dydx+Pn(x)y=Q(x)P_0(x)\frac{d^ny}{dx^n}+P_1(x)\frac{d^{n-1}y}{dx^{n-1}}+...+P_{n-1}(x)\frac{dy}{dx}+P_n(x)y=Q(x)P0​(x)dxndny​+P1​(x)dxn−1dn−1y​+...+Pn−1​(x)dxdy​+Pn​(x)y=Q(x) is a linear differential equation of order nnn , provided P0≠0P_0 \neq 0P0​=0 . The term linear is used to describe these differential equations because Each derivative t…

4. Solution to a differential equation

   A function y=f(x)y=f(x)y=f(x) is called a solution to a differential equation on an interval III if, when yyy and its derivatives are substituted into the equation, the equation is satisfied for all xxx in the interval III

5. General solution to a differential equation

   The general solution of a differential equations represents all the solutions of the differential equation and is in the form of an equation with arbitrary constants. The number of arbitrary constants in the general solution agrees with the order of the differential equation.

6. Particular solution to a differential equation

   Any solution to a differential equation obtained from the general solution by assigning values to the arbitrary constants

7. Verifying the solution of a differential equation

   Find the derivative of the solution and plug them into the differential equation. If the left and right side of the equation is equal, then the function is a solution to the differential equation

8. Initial condition of the general solution

   A restriction on the variables of the general solution that leads to a particular solution (think initial condition of integral to find constant of integration)

9. First-order differential equation

   A first-order differential equation is of the form dydx=f(x,y)\frac{dy}{dx}=f(x,y)dxdy​=f(x,y) where fff is continuous on its domain.

10. Separable first-order differential equation

    A first-order differential equation is said to be separable if it can be written in the form dydx=M(x)N(y)\frac{dy}{dx}=\frac{M(x)}{N(y)}dxdy​=N(y)M(x)​ , where MMM is a function of xxx alone, NNN a function of yyy alone, and both MMM and NNN are continuous on their domains. In terms of differentials, this can be written as N(y)dy=M(x)dxN(y)dy=M(x)dxN(y)dy=M(x)dx , which can be trivially solved by integrating both sides

11. Growth models and decay models

    If a quantity AAA varies with time ttt in such a way that the rate of change of AAA with respect to ttt is proportional to AAA itself, then the situation can be modeled by the differential equation dAdt=kA\frac{dA}{dt}=kAdtdA​=kA , where k≠0k \neq 0k=0 is a real number. If 0"> k>0k>0k>0 , AAA is increasing, and the equation is a growth model if k<0k<0k<0 , AAA is decreasing, and the equation is a decay model The solution to these growth models are of the form A=A0ektA=A_0e^{kt}A=A0​ekt , where…

12. Finding time it takes to reach some value from exponential model

    Find kkk by substituting the given doubling (or other) time into the equation A(t)=A0ekt=2A0A(t)=A_0e^{kt}=2A_0A(t)=A0​ekt=2A0​ ekt=2e^{kt}=2ekt=2 Set the exponential function to the given value times the initial quantity (if we want to find tripling time, for example) A(t)=3A0A(t)=3A_0A(t)=3A0​ A0ekt=3A0A_0e^{kt}=3A_0A0​ekt=3A0​ ekt=3e^{kt}=3ekt=3 Substitute k from above and algebraically solve for t

13. Newton's law of cooling

    If u=u(t)u=u(t)u=u(t) is the temperature of the object at time ttt and TTT is the constant ambient temperature, dudt=k[u(t)−T]\frac{du}{dt}=k[u(t)-T]dtdu​=k[u(t)−T] and u(t)=(u0−T)ekt+Tu(t)=(u_0-T)e^{kt}+Tu(t)=(u0​−T)ekt+T

14. Slope fields

    The graph of the general solution to a differential equation created by evaluating dydx\frac{dy}{dx}dxdy​ at points and drawing short segments with their slope equal to this value. To draw a solution from a slope field, use the initial condition and work backwards, following and connecting the tangent lines smoothly.

15. Euler's method (for approximating a first-order differential equation)

    Assume y=y(x)y=y(x)y=y(x) , dydx=f(x,y)\frac{dy}{dx}=f(x,y)dxdy​=f(x,y) , x=x0x=x_0x=x0​ , y=y0y=y_0y=y0​ Euler's method starts with the initial conditions (x0,y0)(x_0, y_0)(x0​,y0​) and uses a small arbitrary increment hhh in xxx called the step size. The first approximation is given by y1=y0+hf(x0,y0)y_1=y_0+hf(x_0,y_0)y1​=y0​+hf(x0​,y0​) The second given by y2=y1+hf(x1,y1)y_2=y_1+hf(x_1,y_1)y2​=y1​+hf(x1​,y1​) , with x1=x0+hx_1=x_0+hx1​=x0​+h And so on

16. Logistic model

    The logistic differential equation for a population PPP that follows a logistic model over time ttt is given by dPdt=kP(1−PM)\frac{dP}{dt}=kP(1-\frac{P}{M})dtdP​=kP(1−MP​) , where kkk is the maximum population growth rate and MMM is the carrying capacity of the population. This relation can also be represented as dPdt=kMP(M−P)\frac{dP}{dt}=\frac{k}{M}P(M-P)dtdP​=Mk​P(M−P) , with the solution of both being in the form P(t)=M1+ae−ktP(t)=\frac{M}{1+ae^{-kt}}P(t)=1+ae−ktM​ , where P(t)P(t)P(t) is t…