Free Essay

Differential Equations

In: Other Topics

Submitted By tarungehlot
Words 2717
Pages 11
Differential Equations by tarungehlots

Concepts of Differential Equation Consider a variable that might denote the per capita capital stock level in an economy. When convenient, we can recognize that the level of capital depends upon time by addition the time argument. Doing so, we would write the per capita capital stock level as , rather than just writing . If we think of time as unfolding continuously, we can also think of capital as being a continuous function of time, and we can assume that the function has derivate . This derivative is the instantaneous change in per capita capital. In many presentations, is written as , but we will here use the more familiar notation . By writing the derivate as , so that we include the time argument, we are emphasizing that the value of the derivative may change with time. Because it is typically cumbersome to repeatedly write down the argument, we can also write the derivative as just , while remembering that its value may change with time. A differential equation is an equation that relates the time derivative of a variable to its level. An example is the equation (1) . The variable is called a state variable because it gives the state of the system at any given point in time. In our example, gives the state or level of per capita capital stock.

2. A Dynamic System

The basic dynamic principle is the idea that “the way things are determines the way things change.” A differential equation is one way of modeling the basic dynamic principle. The state variable indicates the way things are, while the time derivative indicates how things change. The differential equation itself is what relates the two. More precisely, a differential equation typically presents a functional relationship, showing how depends upon . When we want to emphasize the functional relationship, we can write the differential equation as , where the general functional form is the rule that tells us how the value of the time derivative is determined from the level of the state variable .

A difference equation can also be used to represent the basic dynamic principle. When a difference equation is used, time unfolds as sequential time periods, rather than unfolding continuously. We will focus on using the differential equation here.

A solution to a differential equation is a function that satisfies the differential equation for all points in time that are of interest. Some differential equations can be solved. However, most differential equations that are of interest in economics cannot be solved. Nonetheless, we can still learn about the path followed by the state variable over time by using a phase diagram technique.

Assuming the function in equation (1) is nonlinear, we can describe equation (1) as a single state variable system, and equation (1) is a nonlinear and non-autonomous differential equation. As a dynamic system, equation (1) is a single state variable system because there is only one time derivative present, which is . If the system were a two state variable system, you would see two time derivatives and two different state variables. The equation is labeled nonlinear because the functional relationship between and is not linear. Typically, nonlinear differential equations cannot be solved. However, the presence of the general functional form in equation (1) assures us that we cannot solve it. Finally, equation (1) is labeled non-autonomous because it includes variables other than state variables and time derivatives. The variables , , , and in equation (1) are exogenous variables, which are variables that describe the environment impacting the system. Usually exogenous variables are constants, which is the case in equation (1).

Classifying the variables of a differential equation system helps clarify how you the model builder believe the system works. Because our system (1) is a single equation system, we can only have one endogenous variable. The endogenous variables a differential equation system are always the time derivatives. So, for our system, is endogenous. The state variables of a system are always classified as predetermined, so is predetermined for our system. The level of the variable is predetermined at point in time because the instantaneous change in the level of we just determined by the model in the previous instant of time. Non state variables are exogenous, so , , , and are exogenous. The last value that needs to be specified for the system is the initial condition for the state variable , which is the value of the variable at the initial point in time . Thus, we can summarize the classification of variables as follows:

Classification of Variables
Endogenous (1): Exogenous (4):, , , Predetermined (1): Initial Conditions (1): It is common that differential equation models contain auxiliary equations, which are equations that endogenized (i.e., determine) the values for other variables by relating them to the state variable. Using standard techniques for analyzing a differential equation of this type, we can learn how the path for state variable will evolve over time. Once we know the path followed by the variable , we can use the following “auxiliary” equations to determine the paths followed by the other variables of interest. To illustrate, let us add the endogenous variables , , , and to our system, and assume that they are determined from the following four equations: (2) (3) (4) (5)

The variables, , and are directly related to the variable , and only related to . Thus, if we know the path followed by , we can use equations (2)-(4) to determine the paths followed by , , and . Equation (5) indicates that we need to know the path followed by , along with the value of the exogenous variable , to determine the path followed by . So we can determine the path followed by by first using equation (2) to determine the path followed by . This discussion indicates that we can classify the variables of the dynamic system (1)-(5) as follows.

Classification of Variables
Endogenous (5): , , , , Exogenous (4):, , , Predetermined (1): Initial Conditions (1):

The core of this model is the path followed by the state variable . Once you understand the core of a model, you can use that knowledge to then understand the auxiliary aspects of the model. Notice that adding the auxiliary equations only changes the number of endogenous variables. We can think of the core of the model, which is equation (1) and the first classification of variables as a subsystem, a system upon which the auxiliary aspects of the model depend.

3. Analyzing the Steady State Finding the Steady States The steady state for the state variable is the state where the variable is not changing, or where . Setting in (1), we know SS1 . There are five variables in equation SS1, the model’s four exogenous variables and the state variable. Equation SS1 determines the steady state values of the state variable as a function of the exogenous variables. Assuming , one value of for which equation SS1 holds is . This steady state is not so interesting in that this result is simply telling us that if no capital initially exists, then none will accumulate over time, (which is because capital is assumed to be essential in the production process). If there is a non zero value for that satisfies SS1, let it be denoted by . To illustrate a situation where there is a unique value for the steady state , consider Figure 2 below. The right side of equation SS1 is linear in , with a zero intercept, as shown in the figure. The assumption implies the left side of equation SS1 also has a zero intercept. The conditions and would ensure that the production function increases at a decreasing rate as increases, as shown in the figure. If it were true that, then the only steady state would be associated with , because the two curves and would only intersect at the point in Figure 2. The Inada condition ensures that the curve initial rises above the curve, and the Inada condition ensures that the curve eventually intersects the curve. Because and for all , we know that this intersection at a capital level only occurs once, which implies is unique, as shown in Figure 2. Figure 2: The steady state for the Solow-Swan growth model

Stability of the Steady State The Inada conditions and also ensure that the steady tate associated with is stable. A steady state is locally stable if the state variable moves toward the steady state value for the variable when it is in the “neighborhood” of the steady state. Define the function so . Notice that, if , then the steady state is locally stable. This is because implies when and when . For our model given by SS1, . So, . Because, our steady state is locally stable if . Notice, in Figure 2, that the slope of the curve, at the steady state , is less than the slope of the curve. That is, . So, we find, meaning our steady state is locally stable. Examining Figure 2, we can also see that the steady state is locally unstable. In the neighborhood of , with positive, notice in Figure 2 that the slope of the curve is greater than the slope of the curve. This implies , which implies . If in the neighborhood of , then is increasing in that neighborhood, meaning it increasing away from the steady state value of zero. Figure 3 is a phase diagram. The traditional phase diagram for a single state variable differential equation is constructed by plotting the state variable along the horizontal axis and plotting the time derivative of the state variable along the vertical axis. So, for our model is plotted horizontally and is plotted vertically. The function shows how depends upon . The Inada conditions tell us that the function is positive and diverging to infinity as approaches zero, and they tell us decreases as increases, eventually becoming negative, as shown in Figure 3. The steady state value for is where equals zero. The arrows along the horizontal axis are presented to emphasize is increasing when but decreasing when . Figure 3: Phase diagram for the Solow-Swan growth model

To summarize, we know that the steady state is locally unstable and the steady state is locally stable. The Inada conditions indicate that the path taken over time by the state variable is as shown in Figure 3. If the initial capital stock level is between zero and , then increases over time and approaches . Alternatively, if is greater than , then decreases over time and approaches . Figure 4 shows these two potential time paths for by plotting as it depends upon time.

Figure 4: Two possible time paths for the state variable in the Solow-Swan growth model

4. A System of Two Differential Equations Suppose a dynamic system is represented by the following system of equations and the following classification of variables (A)

(B) Endogenous (2): , Exogenous (2): , ,
Predetermined (2): ,
Initial Conditions (2): ,

How can we characterize the paths followed by the two state variables?

5. Using a Phase Diagram to Characterize the Two Dimensional Differential Equation System

We now characterize the path followed by and diagrammatically in a “phase diagram.” The first step in constructing the phase diagram is to plot the “nullclines” associated with the steady state conditions and . From (B), we can see that implies . The usual restrictions on the production function imply there is a unique value such that . (Notice that we need the Inada conditions and to conclude this. The restrictions and alone are not enough.) When , the assumption implies, so which implies the right side of (B) is negative. Because and are typically assumed for the utility function, we know must be positive for the left side of (B) to be negative. Thus, when , . Using analogous reasoning, when , . This is shown by the arrows in Figure 1.

We obtain the nullcline by setting in condition (A), which implies
. Because , we know when . The Inada conditions and allow us to draw conclusions about the shape of this nullcline consumption level . The assumption ensures that when is near zero, which implies when is near zero. Moreover, ensures increases faster than as increases from a number near zero, which implies is increasing as increases from zero. The assumption ensures that, when is large enough, increases faster than as increases. This implies that, as increases, must reach a peak and then decline thereafter. This nullcline separates the region where is increasing from where it is decreasing. Condition (A) indicates that, when is greater than its nullcline value for a given level , then . Conversely, when is less than its nullcline value for a given level , then . This is shown by the arrows in Figure 1.

Figure 1: Dynamics implied by the Euler Equation and Capital Accumulation Constraint for the Neoclassical Model of Optimal Growth

Starting at any point in the of Figure 1, the arrows tell us the direction of movement. This allows us to trace the path from any starting point . The starting point for is given to us in the problem. The starting point for is a choice. Without additional restrictions, we cannot determine which value of will occur. However, proceeding with some analysis can provide an indication of what those restrictions might be.

Figure 2 is drawn under the assumption that . (After understanding this analysis, it should be apparent how to present the case for .) Path 1 is associated with a relatively high initial consumption choice . Because is in the upper left quadrant of the phase diagram, the direction of movement must be up and to the left ( increasing and decreasing). . Path 2 is still associated with a relatively high initial consumption choice , but one where falls in the lower left quadrant. Initially, the direction of movement must be up and to the right ( increasing and increasing). However, the nullcline is eventually reached, and the direction of movement of is reversed. Thus, the paths of type 2 eventually become like those of type 1. For all paths of type 1 and type 2, must eventually occur. To rule out such paths as equilibrium paths, we need only add the reasonable condition to the problem.

Figure 2: Phase Diagram for the Neoclassical Model of Optimal Growth

3
2
1
4

Path 3 is associated with a relatively low initial consumption choice . Because is in the lower left quadrant of the phase diagram, the direction of movement must initially be up and to the right ( increasing and increasing). However, on this path, the nullcline is reached, and the direction of movement of is reversed. Eventually, such a path must involve a negative consumption level. One way to rule out such a path is to recognize that, at any given point in time , when the capital-consumption combination is , the consumer can discontinuously increase or decrease the consumption level, for such an action is equivalent to the choosing when . On path 3, by choosing to jump paths by discontinuously increase , (by choosing to reduce the saving level ), the consumer can move to a new path that will provide a higher consumption level in every period, for all time, which is clearly preferable. Thus, it is not optimal to choose a consumption level that generates a path of type 3.

Having ruled out paths of types 1 and 2 because they are not feasible, and paths of type 3 because they are clearly not optimal, we can conclude that the unique optimal choice for , given the initial capital level is the level for that generates path 4, or the level shown in Figure 2. On this path, consumption and capital each increase en route to the steady state , never crossing a nullcline.

It is worth noting that, if the consumer believes that the economy would end at a particular point in time , rather than assuming the economy will never end, then it would be rational to choose a consumption path such that . That is, the consumer should choose the consumption path where capital is just used up when the economy terminates. (The Euler equation and capital accumulation equation would be the same; only the “boundary condition” condition would change to .) This implies, rather than choosing path 4 in Figure 2, the consumer would choose a path either of Type 1 or Type 2, so that vertical axis where is eventually reached. Which path is chosen would depend upon the length of time T. The optimal initial consumption level becomes lower as the length of time T becomes longer.…...

Similar Documents

Premium Essay

Differential Equation

...Exercise 1 – Find the first derivative and the second derivative of the following functions Answer: Applying constant function and power function rule (A) Y = 3 + 10X + 5X2 dY/dX = 0 + 1.10.X1-1 +2.5.X2-1 dY/dX = 10 + 10X d2Y/ dX2 = 0 + 1.10.X1-1 d2Y/ dX2 = 10 (B) Y = 2X (4 + X3 ) Y = 8X + 2X4 dY/dX = 1.8.X1-1 + 4.2.X4-1 dY/dX = 8 + 8X3 d2Y/ dX2 = 0 + 3.8.X3-1 d2Y/ dX2 = 24X2 (C) Y = 3 /X2 Y = 3X-2 dY/dX = -2.3.X-2-1 dY/dX = -6X-3 dY/dX = -6/X3 d2Y/ dX2 = -3.-6X-3-1 d2Y/ dX2 = 18X-4 d2Y/ dX2 = 18/X4 (D) Y = 18T – 2T2 dY/dT = 1.18.T1-1 – 2.2.T2-1 dY/dT = 18 – 4T d2Y/ dT2 = 0 – 1.4.T1-1 d2Y/ dT2 = - 4 Exercise 2 - Find the partial Derivative of Y with respect to X Answer (A) Y = 10 + 3Z + 2X ∂Y/∂X = 0 + 0 + 1.2.X1-1 ∂Y/∂X = 2 (B) Y= 18Z + X2 + Z.X ∂Y/∂X = 0 + 2.1.X2-1 + Z ∂Y/∂X = 2X + Z Application - The nursing home industry is growing rapidly because the aging of American population. According to the study of an economist, the average cost per patient day of a nursing home can be approximated by C = A – 0.16B + 0.002B2 Where, B is the nursing home’s number of patient days per year ( in thousands) and A is the number that depends on the location and other factors but not on B. Based on the information , how big must a nursing home be ( in terms of patient – days) to minimize the cost per patient day ? Answer – C= ƒ (A, B) Where C is Avg. Cost per patient day A is variable depends on location and other......

Words: 404 - Pages: 2

Premium Essay

Accounting Equation

... Accounting Equation Franklin Weatherspoon ACC/300 OCTOBER 26, 2013 Dr. M Moczynski, CPA, CGFM Accounting Equation In the business world an individual cannot understand a balance sheet, income statement nor transaction recordings within a general ledger, until he or she understands the basic accounting equation: Assets = Liabilities + Owners Equity (Barclay, 2013). Following is a discussion on the relationship of accounting equation and the components of a balance sheet; along with examples showing the affect of each (University of Phoenix, 2013). Relation in Components The accounting formula is the way double-entry bookkeeping is formulated. The accounting formula, which is known as the balance sheet equation, signifies the connection between the assets, liabilities, and owner's equity of a small company (Peeler, 2013). The accounting formula basically expresses a company’s assets, which is obtained either by liabilities or by the company’s capital. The equation has to balance since the company’s entirety, which is known as assets are brought with something like liability or the company’s capital (Peeler, 2013). The effect of one another The company’s equity is affected by capital, functioning as stock. The company has income, which is revenue minus expenses, and gains minus loses, and sometimes extra capital and withdrawals are known as dividends. When the closing of the month approaches, each item will......

Words: 377 - Pages: 2

Premium Essay

Differential Pricing

...associated with long-term or fixed volume/capacity agreements; collateral benefits provided by the access seeker (for example through stimulating extra demand for the access 20 21 11 However, differential pricing can reduce efficient competition. Preferential access pricing between a limited group of network operators can have the effect of discouraging entry of more efficient operators.22 Differential pricing can also discourage investment. In an industry where assets often have little alternative use, there is scope for an access provider to appropriate the commercial returns to the assets 23 of access seekers through high access prices. There appears to be even greater scope for differential pricing to reduce efficient competition where an access provider provides preferential pricing to its own vertically-integrated operations or to its subsidiaries or associates. The incentive for the access provider to discriminate against competitors can inhibit efficient entry and competition in those markets. The Commission expects that in most undertakings the same menu of offerings will be available to all access seekers on a non-discriminatory basis. Where an undertaking provides scope for differential pricing not based on costs the Commission must be satisfied that such differential pricing will promote competition and will enhance the efficient use of, and investment in, infrastructure. As discussed below, when arbitrating disputes the Commission will use a......

Words: 13617 - Pages: 55

Premium Essay

Differential Pricing in Telecom Sector

...Differential Pricing: Many important industries involve technologies that exhibit increasing returns to scale, large fixed and sunk costs, and significant economies of scope. Two important examples of such industries are telecommunications services and information services. In each of these cases the relevant technologies involve high fixed costs, significant joint costs and low, or even zero, marginal costs. Setting prices equal to marginal cost will generally not recoup sufficient revenue to cover the fixed costs and the standard economic recommendation of "price at marginal cost" is not economically viable. Some other mechanism for achieving efficient allocation of resources must be found. The outcome of this investigation is that (i) efficient pricing in such environments will typically involve prices that differ across consumers and type of service; (ii) producers will want to engage in product and service differentiation in order for this differential pricing to be feasible; and, (iii)differential pricing will arise naturally as a result of profit seeking by firms. It follows that differential pricing can generally be expected to contribute to economic efficiency Thus differential pricing is “the practice of selling the same product to different customers at different prices even though the cost of sale is the same to each of them. More precisely, it is selling at a price or prices such that the ratio of price to marginal costs is different in different sales” ...

Words: 2467 - Pages: 10

Premium Essay

Differential Staining

...MLT1 Experiment 5/Task 6 Differential Staining There are many different ways to stain bacteria for viewing under a microscope. For the most part these are categorized as “either simple, nonspecific or differential (specific) (LabPaq, p.128). The most common staining method used is Gram staining. Gram staining is a differential or specific method of staining. Gram staining is a way to tell the difference between gram positive and gram negative bacteria. The cell wall of the gram positive bacteria is made up of several “layers of peptidoglygan,” and “techoic acids,” (LabPaq, p. 128). This peptidoglycan is what absorbs the color part of the crystal violet stain causing the gram positive bacterial cell to appear violet colored when viewed with a microscope. In contrast, the cell wall of the gram negative bacterial cell is not as thick as that of the gram positive cell wall. Peptidoglygans are on the inside of the cell rather than in outer layers. The outer part of the gram negative cell is made up of phospholipids and lipipoly-saccharides (Betsy and Keough, 2005). The outer cell wall does not hold onto the violet color of the crystal violet and appears pink in color when viewed under a microscope. The purpose of iodine staining is to aid the bacteria to keep the stain by creating an iodine-crystal violet mixture that will not dissolve. Iodine is also known as a mordant in this case. A mordant is usually an inorganic oxide that when mixed with...

Words: 417 - Pages: 2

Premium Essay

Action Equation

...Action Equation Michele Riazzi Grand Canyon University/LDR645 May 27, 2014 Action Equation Plan Do: What do you need employees to do? * Employees must enroll a minimum of 5 students * It is preferable for students to begin closer to the beginning of the month * Enrollment counselors should be listening by asking open ended questions * Direct students to take an action by beginning the application * Ask for referrals each month with new and existing students * Try to have all students for the month registered by the 15th to begin the next month Know: What do employees need to know to take action? * Know what to do and why we do it * Enrollment Counselors need to understand how the revenue stream is affected by monthly expectations. * Database management is critical to create the most opportunities with the inquiries provided. * Enrollment Counselors need to set clear expectations with student from the initial conversation. They, students and enrollment counselors, must follow through on the enrollment process. * Enrollment counselors must ask all students and potential students for referrals. This may be accomplished by speaking with current students on a weekly basis. Feel: What do employees need to feel to take action? * Enrollment counselors are usually the first impression potential students have of Grand Canyon University. This is a valuable role and must be taken seriously. * The enrollment counselor’s...

Words: 956 - Pages: 4

Free Essay

Math Equation

...Chapter 2: Equations and Inequalities MAT 1103: Fundamentals of Mathematics 2.1 Equations 1) Equation Statement indicating that 2 quantities are true. Example: Solution set: 3x − 2 = 10 values of variable that satisfy equations. 2) Restricted values a. Fraction: 1 , x≠a x−a b. Radical: c. Logarithmic: x−a, x≥a log( x − a) , x > a 3) Solve Linear equation in 1 variable ax + b = 0 Example 1: Solve a. (a and b are real numbers and a ≠ 0) 2x + 3 = 0 b. 3( x + 2) = 5 x + 2 c. 3y 2 1 − = y 2 3 5 1 Chapter 2: Equations and Inequalities MAT 1103: Fundamentals of Mathematics 4) Solve Rational Equations Example 2: Solve a. 3 7 + =2 5 x+2 b. 3x 5 − =3 x −1 x + 3 2.2 Applications of Linear Equations 1) English-mathematics vocabulary Mathematical operator + x English words More, greater, add, sum, exceeds, increase, higher, total, extra Less, difference, lower, minus, decrease, fewer Times, multiple Mathematical ratio 2x 3x 1/3 x ¼x English words Double Triple One third One quarter 2) General guideline for solving word problem a. Read the problem. b. Read the problem again. c. Draw a picture / table / flow chart. d. Find and label the unknowns that you are looking for. e. Find and label the known quantities. f. Write down all the formulas and relations between the known and unknown. g. Solve the problem. h. Check the answer & reply in words. 2 Chapter 2: Equations and Inequalities MAT 1103:......

Words: 773 - Pages: 4

Free Essay

Gravity Equation

...NBER WORKING PAPER SERIES THE GRAVITY EQUATION IN INTERNATIONAL TRADE: AN EXPLANATION Thomas Chaney Working Paper 19285 http://www.nber.org/papers/w19285 NATIONAL BUREAU OF ECONOMIC RESEARCH 1050 Massachusetts Avenue Cambridge, MA 02138 August 2013 I want to thank Fernando Alvarez, Michal Fabinger, Xavier Gabaix, Sam Kortum, Bob Lucas, Jim Tybout, Jon Vogel and seminar participants in Berkeley, Bilkent, Bocconi, Boston University, Chicago, Erasmus, Hitotsubashi, LBS, Louvain-CORE, LSE, the NY Fed, Oxford, Princeton, Rochester, Sciences Po, Toulouse, UBC Vancouver, Yale and Zurich for helpful discussions, and NSF grant SES-1061622 for financial support. I am indebted to Jong Hyun Chung, Stefano Mosso and Adriaan Ten Kate for their research assistance. During the last year, I have received compensation for teaching activities from the Toulouse School of Economics, as well a research grant from the National Science Foundation (SES-1061622), in excess of $10,000. The views expressed herein are those of the author and do not necessarily reflect the views of the National Bureau of Economic Research. NBER working papers are circulated for discussion and comment purposes. They have not been peerreviewed or been subject to the review by the NBER Board of Directors that accompanies official NBER publications. © 2013 by Thomas Chaney. All rights reserved. Short sections of text, not to exceed two paragraphs, may be quoted without explicit permission provided that full......

Words: 18767 - Pages: 76

Premium Essay

Differential

...LRBI Checklist Differential Reinforcement Differential Reinforcement ifferential reinforcement is the reinforcement of one form of behavior and not another, or the reinforcement of a response under one condition but not another. Differential reinforcement uses positive reinforcement to differentiate or separate appropriate student behavior from inappropriate behavior by increasing one while decreasing the other. Definition D Things to Do  Select behaviors to be decreased or increased.  Select alternative, incompa tible or communicative behavio , rs to be taught.  Determine time interval.  Set criterion.  Ignore inappropriate behav iors.  Monitor the student’s performance. Level 1: Positive Interaction Procedures 1 LRBI Checklist Differential Reinforcement There are six strategies that comprise or make up differential reinforcement: Select behaviors to be decreased or increased. Select behaviors to be decreased or increased. These behaviors should be objectively defined and must be observable and measurable. For DRO, DRI, DRA, and DRC, there will be a specific behavior to decrease, such as talk-outs, hitting, not following directions, out of seat, disrupting other students, or tantrums. For DRH, there will be a specific behavior to increase, such as contributing in class, positive social interactions, getting to class on time, or task completion. D R H RL D problem behaviors Level 1: Positive Interaction Procedures ...

Words: 1949 - Pages: 8

Premium Essay

Accounting Equation

...There are numerous aspects to the accounting equation and each has its own set of criteria. In order to maintain proper balance of the account equation, assets equaling liabilities plus shareholder’s equity there are several things to consider which include recordable transactions and financial statements. A transaction is any event that has an impact on the financial statements of the business. In order for a transaction to be recorded it must result in assets equaling liabilities plus shareholder’s equity. Examples of recordable transactions include; the sale of merchandise to a customer, a purchase of supplies or equipment, and borrowing funds from a lender. The aforementioned equation assets = liabilities + shareholder’s equity is the fundamental accounting equation. For an account to have a transaction post to it both sides of the equation must remain equal. You could not for example increase assets without subsequently decreasing liabilities or shareholder’s equity. There are four primary financial statements in accounting; Balance Sheet, Income Statement, Statement of Retained Earnings, and Statement of Cash Flows. A Balance Sheet is a statement that shows all of a business’s assets, liabilities, and equity for a point in time. The function of a balance sheet is to show a company’s liquidity and calculate net worth. An Income Statement is a statement that measures a company’s financial performance over a given period, a year for example. The primary function of......

Words: 372 - Pages: 2

Free Essay

Classification of Differential Equations

...CLASSIFICATION OF FIRST ORDER DIFFERENTIAL EQUATIONS SUPPLIMENTARY PROBLEMS 1. Write the given equation in standard form 3.16. exy'-x=y' solution exy'-x=y' =exy'-y'=x ex-1y'=x y'=xex-1 ANS 3.25 dy+dx=0 Solution. dy+dx=0 dydx+1=0 y'=-1 2. The differential equations are given in both standard and differential form. Determine whether the equation in standard form are homogeneous and/or linear, if not linear, whether they are Bernoulli: determine whether the equations in differential form as given are separable and/or exact. 3.28 y'=xy+1: xy+1dx-dy=0 Solution. y'=xy+1 = y'-xy=1 this is in the form y'+pxy=qx for px=-x , qx1 thus this is linear next xy+1dx-dy=0 is not separable si9nce the variables cannot be separeted. To check for exactness take Mx=xy+1 and Nx=-1 for an exact equation, ∂M∂y=∂N∂x ∂M∂y=x and ∂n∂y=0 thus ∂M∂y≠∂N∂x so it is not exact 3.30 y'=x2y2: -x2dx+y2dy=0 Solution. y'=x2y2 =≫ y'+0y=x21y2 The equation is of the form y'+pxy=qxyn where px=0, qx=x2 and n=-2 so it is a Bernoulli equation NEXT For -x2dx+y2dy=0 it is separable since the variables are separated. To check for exactness take Mx=-x2 and Ny=y2 thus ∂M∂y=0 and ∂N∂x=0 hence its EXACT. 3.35 y'=2xy+x: 2xye-x2+xe-x2dx-e-x2dy=0 Solution. For y'=2xy+x =≫y'-2xy=x This is in the form y'+pxy=qx for px=-2x and qx=x Thus it’s a linear equation NEXT 2xye-x2+xe-x2dx-e-x2dy=0 this is not separable since the variables...

Words: 336 - Pages: 2

Free Essay

Ordinary Differential Equations

...MATH 364A: Ordinary Differential Equations (Midterm 1) Name: Student ID: Signature: Question 1 (40 points) Solve the following initial value problems. (a) y + t3 y = t3 y(0) = 0. (b) y = − (1+x) y y(−1) = 1. Question 2 (40 points) Solve the second-order initial value problem 2y − 3y − 5y = 0 y(0) = 0 2 y (0) = 1. Question 3 (40 points) For each equation below, first determine whether the equation is exact or not exact. If the equation is exact, find the solution. (a) cos(y) + (2y − x sin(y)) dy = 0. dx (b) 3xy 2 + (y 3 + 3x2 ) 3 dy = 0. dx Question 4 (40 points) Consider the following first-order differential equation y = (y 2 − 1)(4 − y 2 ). Find all critical (equilibrium) solutions and classify their stability. 4 Question 5 (40 points) (True/False) In each of the following, determine whether the given function solves the given differential equation or initial value problem. (a) If k > 0 denotes any real constant, then the function y(t) = e−kt solves the initial value problem y = −ky y(0) = 1. (b) The function y(t) = (1 − t)−1 = 1/(1 − t) solves the initial value problem y = y2 y(0) = 1. (c) The function y(t) = e2t solves the second-order initial value problem y = 2y y(0) = 1 y (0) = 2. (d) Any function y(x) defined by the implicit equation x4 + 2x2 y 2 + y 4 = C solves the differential equation x3 + xy 2 + (x2 y + y 3 )y = 0. 5 Question 6 (Extra Credit, 10 points) An object with......

Words: 324 - Pages: 2

Free Essay

Differential Gear

...The differential was first invented in China, in the third century,A.D. After turning a car the outside wheel has to turn faster then the inside one in order to cover the greater distance. That is why two wheels are not driven at a same speed. So we need a differential gear. A car differential is a place halfway between the wheels, on either the front or both axes. Main mechanisms: Wheels receive power from the motor via drive shaft. The power receiving wheels which makes the vehicle move fowrord are called as the drive wheels. Wheels receive power from the motor via drive shaft. The main function of the differential gear is to allow the drive wheels to turn at different Rpm while from the engine. The main characterics of differential is to, * To aim the engine power at wheels. * To transmit the power to the wheels while allowing them to rotate at differential speeds. the left wheel has to travel more distance compared to the right wheel. If the wheels where connected using a soil shaft. The wheels where connected using a soil shaft. The wheels would have to slip to accomplish the turn. The mechanism in a differential allows left and right wheels to turn at different Rpm, while transferring power to both the wheels. Differential parts: There are three types of gear available, Pinion drive gear: Transfers power from the drive shaft to the ring gear. Ring gear: Transfers power to the differential case assembly Side/Spider gears: Helps both wheels to......

Words: 656 - Pages: 3

Premium Essay

Accounting Equation

...accounting equation, using 100 to 150 words. There are many different rules, regulations and requirements in accounting. However it does not matter how complex an area of accounting may look because accounting is based on one elemental principal which is Assets = Liability + Owners (shareholder) Equity. The key to remember is in any transaction there is always at least two sides and each side of the equation always stays balanced with each other. An asset is something in value a company owns. Examples would be Assets make up cash, property, office equipment, inventory and accounts receivable. Liabilities are existing debts and obligations that are owed by a company. Examples would be salaries, finaning a purchase, notes payable. Owners equity is the owners right to the assets of an entity after all liabilities are paid. Examples would be retained stock, cash invested in the business and paid in capital. Even though there are many different rules, regulations and requirements in accounting. These are the fundamental principals of accounting. ❖ Complete E1-5 on p. 34 of Financial Accounting Assets Cash Cleaning Equipment Cleaning Supplies Accounts Receivable Liability Notes Payable Salaries Payable Accounts Payable Owner Equity Commerce Stock |CheckPoint 2 |10.0 |10.0......

Words: 251 - Pages: 2

Free Essay

Rent Differentials

... KOFORIDUA POLYTECHNIC RENT DIFFERENTIALS BETWEEN RESIDENTIAL AREAS AND ZONGO COMMUNITIES IN GHANA AND DETERMINANTS OR FACTORS THAT MOTIVATE PEOPLE TO STAY OR HIRE AT SUCH PLACES/ AREAS. A CASE STUDY OF KOFORIDUA OLD ESTATES AND KOFORIDUA ZONGO. BY BANAFO BENJAMIN DUODU OBED FEKOW SARPONG ALFRED ANGMOR SOLOMON K. OPOKU MENSAH A. RICHARD YEBOAH A. DERRICK ANUM AYIVOR ISHMAEL E. OSABUTEY CATHERINE A. OWUSU BISMARK 2012 DEDICTION Our expectation is that this case study will be dedicated to all potential researchers who might be embarking on this same case study and to Mr. Jamal Mohammed paved the way for us to embark on this practical research. ACKNOWLEDGEMENT We are most grateful to the Lord Almighty for strength, knowledge and wisdom granted to us in our area of study. We also acknowledge every member of our group who contributed their effort to this successful research. TABLE OF CONTENT PAGE CHAPTER ONE * BACKGROUND INFORMATION ...

Words: 2429 - Pages: 10