Test 1 Prep Flashcards

1
Q

T or F: all models are DE’s

A

false

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2
Q

f’ (c)<0 means what?

A

f is decreasing at x = c

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3
Q

f’ (c)>0 means what?

A

f is increasing at x = c

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4
Q

What is a DE?

A

A DE is an equation involving a dependent variable (unknown function) and one or more of it’s derivaties with respect to the independent variable(s)

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5
Q

what is an ODE (Ordinary DE)?

A

a DE involing ONLY ordinary derivatives w/ respect to a single indepednedent variable

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6
Q

what is PDE (Partial DE)?

A

a DE involing partial derivatives w/ respect to a single indepednedent variable

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7
Q

how do you determine the oder of DE?

A

Look at the highest derivative in the DE. Ex. If the highest is first derivative, the order of the DE, First order.

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8
Q

how do you determine independent and dependent variables.

A

dy/dx. derative of y W/ RESPECT TO X
derivative of (dep var.) W/ RESPECT TO (indep variable)

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9
Q

Linear vs Non Linear DE

A

A DE is linear if the depedent variable and tis’ derivatives appear in additive combinations of their first power. (No Multiplying variables. Not to the power of a the independent variable)

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10
Q

how does implicit differtiation work?

A

Basically chain rule everything. Leave primes except for variable were deriving with respect to

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11
Q

how to check something is a solution?

A

Plug it in and check if LHS = RHS.

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12
Q

Explicit vs implicit functions

A

A function is explicit when we can write is “y = some function of x”. It’s implicit when we don’t see that.

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13
Q

For a DE with dependent varaible y and independent varaible x, an ______ to the DE on an interval I is a function y=f(x) that satifies the equation for all x ∈ I

A

explicit solution

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14
Q

A relation G(x,y) is said to be an _______ to a DE on an interval I provided that there exists at least one function f that satisfies the relation as well as the DE on I.

A

implicit solution

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15
Q

difference b/w interval vs domain

A

intervals have to be a an connect set, domains can have breaks or unions.

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16
Q

What is an initial value problem (IVP)?

A

an IVP for an nth order DE is one in which the n inital conditions on the solution y and it’s derivatives must be satisfied:

y(x_0) = y_0
y’(x_0) = y_1
y^(n-1) (x_0) = y_n-1

17
Q

how to use existence thrm?

A
  1. write DE in proper form: y’ = f(x,y)
  2. Identify all discontinuities in f
  3. make a rectangle containging (x_0, y_0) on the inside that also has none of the points where f is discontinuous. You really want to make sure that there is no discontinuity at the IC.
  4. If you can do step 3 then according to the existence theorem, a solution to the IVP exists
  5. if you can’t do step 3 then no conculsion can be drawn. IVP may or may not have a sol-n.
18
Q

how to use existence & uniqueness thrm

A
  1. write in proper form. y’ = f(x,y) (y’ with a coeffeicent of one)
  2. compute partial derivative of the dependent variable. ∂f/∂y
  3. identify all discontinuities in f and ∂f/∂y
  4. try to construct a rectangle containing (x_0, y_0) on the inside that also has none of the points where f or ∂f/∂y is discontinuous.
  5. if you can do step 4, then according to the thrm a unique solution to the IVP exists.
  6. If you cannot do step 4, then there may or may not be a unique solution to IVP. We know nothing!
19
Q

why do we care abt existence thrm?

A

if the DE model, doesn’t have a solution, then it’s a bad model and we can’t predict the future.

20
Q

why do we care abt the existence & uniqueness thrm?

A

if a model has more than one sol-n, which is when the thrm fails. We don’t know which sol-n correctly predicts the future.

21
Q

how to find singular sol-ns?

A
  1. upon separating, look for any possible values of y (since y is the sol-n) that cause us to divide by zero.
  2. for any such y (ȳ), ask yourself if it is possible to find a constant in the one parameter family that allows you to find a sol-n that sovles the IC for all x_0 in the sol-ns interval.
    a. If you can get ȳ from the your general sol-n then ȳ isn’t a signular solution
    b. If you can’t, y = ȳ is a singular solution.