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lisabon 2012 [21]
3 years ago
15

(10 pts) (a) (2 pts) What is the difference between an ordinary differential equation and an initial value problem? (b) (2 pts)

What is the difference between a particular solution and a general solution? (c) (3 pts) Given an example of a second-order linear ODE and a second-order nonlinear ODE that is not in the text, Chapter 1. (d) (3 pts) Give an example of a nonlinear 4th-order ordinary differential equation.
Mathematics
1 answer:
laiz [17]3 years ago
4 0

Answer:

Step-by-step explanation:

(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.

(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.

(C) Example of a second order linear ODE:

M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)

The equation will be homogeneous if K(t)=0 and heterogeneous if K(t)\neq 0

Example of a second order nonlinear ODE:

Y=-3K(Y){2}

(D) Example of a nonlinear fourth order ODE:

K^4(x) - \beta f [x, k(x)] = 0

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Answer:

Step-by-step explanation:

Average speed is distance divided by time.

s=(101km)/(1hr)

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4 0
3 years ago
Debra plans to invest $2,250 for 10 years. She can invest in a savings account that pays 4% simple intrest or a savings account
Diano4ka-milaya [45]

Answer:

\$180.55

Step-by-step explanation:

step 1

<u><em>Simple interest</em></u>

we know that

The simple interest formula is equal to

A=P(1+rt)

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

t=10\ years\\ P=\$2,250\\r=4\%=4/100=0.04

substitute in the formula above

A=2,250(1+0.04*10)

A=2,250(1.4)

A=\$3,150

step 2

<u><em>Interest compounded annually</em></u>

we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

t=10\ years\\ P=\$2,250\\r=4\%=4/100=0.04\\n=1

substitute in the formula above

A=2,250(1+\frac{0.04}{1})^{1*10}  

A=2,250(1.04)^{10}  

A=\$3,330.55

step 3

Find the differences between the two final amounts

A=\$3,330.55-\$3,150=\$180.55

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As you can see, however, I inserted the "less than" symbol rather than the equal sign because values greater than a certain value can also equal "x". Solve it like a normal equation and you will arrive at:
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