Question

Place the steps for solving by the substitution method in order.

1st.
2nd.
3rd.
4th.
5th.



1 : Check by adding the equations to see if a variable will eliminate.

2 : Substitute the value into one of the original equations and solve for the other variable.

3 : Write the solution as an ordered pair.

4 : Solve for the remaining variable.

5 : If it doesn't eliminate, try subtraction. Change the terms in the second equation to the opposite and check to see if a variable will eliminate.

Answer 1

Are you certain that the questions asks about the substitution method? The options refer to Addition/Subtraction.

Answer 2

thats what im confused about to

Answer 3

idk how to do this

Answer 4

A full course in differential equations involves applications of derivatives to be studied after two or three semester courses in calculus. A derivative is the rate of change of one quantity with respect to another; for example, the rate at which an object’s velocity changes with respect to time (compare to slope). Such rates of change show up frequently in everyday life. For example, the compound interest law states that the velocity of interest accumulation is proportional to the starting amount of money, given by dy/dt=ky, where y is the sum of money earning compound interest, t is time, and k is a constant (dt is an instantaneous time interval). Although, typically, credit card interest is compounded daily and reported as the APR, annual percentage rate -- yet a differential equation can be solved to give the instantaneous solution y = ce^(kt), where c is an arbitrary constant (the stated interest rate). This article will show you how to solve types of differential equations commonly encountered, especially in mechanics and physics.

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Contents

1 Steps
1.1 The Basics
1.2 Solving First Order Differential Equations
1.3 Solving Second Order Differential Equations
1.4 Solving Higher Order Differential Equations
2 Real Life Applications
3 Tips
4 Warnings
5 Things You'll Need
6 Related wikiHows
7 Sources and Citations

Edit Steps

The Basics

1
Define derivative. Derivative (also called differential quotient; especially British) - the limit of the ratio of the increment of a function (generally y) to the increment of a variable (generally x) in that function, as the latter tends to 0; the instantaneous change of one quantity with respect to another, as velocity, which is the instantaneous change of distance with respect to time. Compare first derivative, and second derivative:[1]

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First derivative – the derivative of a function, example: "Velocity is the first derivative of distance with respect to time."
Second derivative – the derivative of the derivative of a function, example: "Acceleration is the second derivative of distance with respect to time."
2
Figure 1. Example of a differential equation of second-order and third-degree.
Figure 1. Example of a differential equation of second-order and third-degree.
Know the order and degree of the differential equation. The order of a differential equation is determined by the highest order derivative; the degree is determined by the highest power on a variable. For example, the differential equation shown in Figure 1 is of second-order, third-degree.
3
Know the difference between a general, or complete solution versus a particular solution. A complete solution contains a number of arbitrary constants equal to the order the equation. (To solve an nth order differential equation, you have to perform n integrations, and each time you integrate, you have to introduce an arbitrary constant.) For example, in the compound interest law, the differential equation dy/dt=ky is of order 1, and its complete solution y = ce^(kt) has exactly 1 arbitrary constant. A particular solution is obtained by assigning particular values to the constants in the general solution.Solving Higher Order Differential Equations

Higher order differential equations are much harder to solve, except certain special cases, as follows:

1
Figure 8. Solving a special type of higher order differential equation.
Figure 8. Solving a special type of higher order differential equation.
Check to see if the differential equation satisfies the form shown in equation (1) in Figure 5, where f(x) is a function of x alone, or a constant. If so, simply follow the steps outlined in Figure 8.
2
Solving nth order linear differential equations with constant coefficients: Check to see if the differential equation satisfies the form shown in equation (1) in Figure 9. If so, the differential equation can be solved as follows:
Figure 9. Solving a special nth-order linear differential equation.
Figure 9. Solving a special nth-order linear differential equation.
3
To solve a more general nth-order linear differential equation, check to see if the differential equation satisfies the form shown in equation (1) in Figure 10. If so, the differential equation can be solved in a method analogous to that used for solving second order linear differential equations, as follows:
Figure 10. Solving a more general nth-order linear differential equation.
Figure 10. Solving a more general nth-order linear differential equation.

EditReal Life Applications

1
The solution to the compound interest law differential equation is an exponential.
The solution to the compound interest law differential equation is an exponential.
Compound interest law: the rate of interest accumulation is proportional to the starting amount of money. More generally, the rate of change with respect to an independent variable is proportional to the corresponding value of the function. That is, if y = f(t), dy/dt = ky. Solving this using the method of separable variable, we get y = ce^(kt), where y is a sum of money accumulating at compound interest, c is an arbitrary constant, k is the interest rate, for example, the interest in dollars on one dollar for a year, t is time. Time, therefore, is money.
Note that the compound interest law applies to many areas of daily life. For example, suppose you are trying to dilute a salty solution by running water into the solution to decrease its salt concentration. How much water do you need to add, and how does the concentration of the solution change with respect to the rate you run the water?

Let s = quantity of salt in the solution at any time, x = the amount of water which has run through, and v = volume of the solution. The salt concentration of the mixture is given by s/v. Now suppose a volume Δx is leaked out of the solution, so the amount of salt leaked out is (s/v)Δx, hence the change in the amount of salt, Δs, is given by Δs = -(s/v)Δx. Divide both side by Δx, to get Δs/Δx = -(s/v). Take the limit as Δx-->0, and we have ds/dx = -s/v, which is a differential equation in the form of the compound interest law, where y is now s, t is now x, and k is now -1/v.
Newton's law of cooling is yet another variation of the compound interest law. It states that the time-rate of decrease in body temperature in excess of the temperature of the surrounding air is proportional to the body temperature above that of the surrounding air. Let x = body temperature above that of the surrounding air, t = time, we have dx/dt = kx, where k is a constant. The solution to this differential equation is x = ce^(kt), where c is an arbitrary constant, as above. Suppose this excess temperature, x, was at first 80 degrees, and drops to 70 degrees after a minute. What will it be after 2 minutes?

Let t = time in minutes, x = excess temperature in degrees, we have 80 = ce^(k*0) = c. Also, 70 = ce^(k*1) = 80e^k, so k = ln(7/8). So x = 70e^(ln(7/8)t) is a particular solution to this problem. Now plug in t = 2, we have x = 70e^(ln(7/8)*2) = 53.59 degrees after 2 minutes.
Various strata of the atmosphere as height increases from sea level
Various strata of the atmosphere as height increases from sea level
In atmospheric thermodynamics, atmospheric pressure p above sea level changes in proportion to the altitude h above sea level--yet another variation of the compound interest law. The differential equation here is dp/dh = kh, where k is constant.
In chemistry, the velocity of a chemical reaction in which x is the amount transformed in time t is the time-rate of change of x. Let a = concentration at the beginning of the reaction, then dx/dt = k(a-x), where k is the velocity constant. This is another variation of the compound interest law where (a-x) is now the dependent variable. See that d(a-x)/dt = -k(a-x), so d(a-x)/(a-x) = -kdt. Integrate, to get ln(a-x) = -kt + a, since a-x = a at time t = 0. Rearranging, we see that the velocity constant k = (1/t)ln(a/(a-x)).
In electromagnetism, given an electric circuit with voltage V and current i (amperes), the voltage V is consumed in overcoming the resistance R (ohms) of the circuit and the inductance L, as governed by the equation V=iR + L(di/dt), or di/dt = (V - iR)/L. This is another variation of the compound interest law, where V - iR is now the dependent variable.
2
The modes of a vibrating string are harmonics.
The modes of a vibrating string are harmonics.
In acoustics, simple harmonic vibration has acceleration being directly proportional to the negative of distance. Recall that acceleration is the second derivative of distance, so d2s/dt2 + k2s = 0, where s = distance, t = time, and k2 is the magnitude of acceleration at unit distance. This is the simple harmonic equation, a second order linear differential equation with constant coefficients, as solved in Figure 6, equations (9) and (10). The solution is s = c1cos kt + c2sin kt.

This can be further simplified by setting c1 = b sin A, c2 = b cos A. Substitute these in, to get b sin A cos kt + b cos A sin kt. Recall from trigonometry, that sin (x+y) = sin x cos y + cos x sin y, so the expression reduces to s = b sin (kt + A). The waveform obeying the simple harmonic equation oscillates between b and -b, with period 2π/k.
Vibrating spring: take an object, with mass m, on a vibrating spring. By Hooke's Law,[3] when the spring is stretched or compressed s units from its natural length (or equilibrium position), it exerts a restoring force F proportional to s, or F = -k2s. By Newton's Second Law (force equals mass times acceleration),[4] we have m d2s/dt2 = -k2s, or m d2s/dt2 + k2s = 0, which is an expression of the simple harmonic equation.
Rear shock absorber and spring of a BMW R75/5 motorcycle
Rear shock absorber and spring of a BMW R75/5 motorcycle
Damped vibrations: consider the vibrating spring as above, with a damping force. A damping force is any effect, such as friction, that tends to reduce the amplitude of oscillations in an oscillator. For example, a damping force could be supplied by a shock absorber in an automobile. In most cases, the damping force, Fd, is approximately proportional to the velocity of the object,[5] or Fd = -c2 ds/dt, where c2 is a constant. Combining the damping force with the restoring force, we have -k2s - c2 ds/dt = m d2s/dt2, by Newton's second law. Or, m d2s/dt2 + c2 ds/dt + k2s = 0. This differential equation is a second-order linear equation that can be solved by solving the auxillary equation mr2 + c2r + k2 = 0, after substituting s = e^(rt).
Solving this by the quadratic formula, we get r1 = (-c2 + sqrt(c4 - 4mk2)) / 2m; r2 = (-c2 - sqrt(c4 - 4mk2)) / 2m.
Overdamping: If c4 - 4mk2 > 0, r1 and r2 are real and distinct. The solution is s = c1e^(r1t) + c2e^(r2t). Since c2, m, and k2 are all positive, sqrt(c4 - 4mk2) must be less than c2, which implies that both roots, r1 and r2, are negative, and the function is in exponential decay. In this case, oscillation does not occur. A strong damping force, for instance, could be supplied by high-viscosity oil or grease.
Criticial damping: If c4 - 4mk2 = 0, r1 = r2 = -c2 / 2m. The solution is s = (c1 + c2t)e^((-c2/2m)t). This is still exponential decay, with no oscillation. However, the slightest decrease in damping force will cause the object to oscillate past the equilibrium point.
Underdamping: If c4 - 4mk2 < 0, the roots are complex, given by -c/2m +/- ωi, where ω = sqrt(4mk2 - c4)) / 2m. The solution is s = e^(-(c2/2m)t) (c1 cos ωt + c2 sin ωt). This is an oscillation damped by the factor e^(-(c2/2m)t. Since both c2 and m are positive, e^(-(c2/2m)t) will go to zero as t approaches infinity. So eventually the motion will decay to zero.

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Edit Tips

Note: the converse of differential calculus is called integral calculus, which deals with summation of the effects of continuously changing quantities; for example, computing the distance (compare to d = rt) covered by an object when its instantaneous rates (velocities) over a time interval are known.
Many differential equations simply cannot be solved by the above methods. The methods above, however, suffice to solve many important differential equations commonly encountered.
Substitute your solution back into the original differential equation, to see whether the equation is satisfied. This will verify that you have solved the differential equation correctly.
HERE U GO TO HELP :) :)

Answer 5

lol thanks

Answer 6

:D