Vocabulary

1. A list of words to remember:

to consider – рассматривать

to describe - описывать

to determine – определять; syn. to define

a domain – область (определения)

hence – следовательно, поэтому

to obtain – получать

obvious - очевидный

to occur – встречаться, появляться

particular – фиксированный, частный, отдельный, конкретный

a ratio – пропорция, соотношение, частное

to substitute – заменять, подставлять вместо

a substance – вещество

sufficient – достаточный

sufficiently – достаточно

2. Translate the following sentences into English:

1) Достаточно ли это условие, чтобы форма была полным дифференциалом?

2) Для иллюстрации нашей идеи рассмотрим следующую форму.

3) Очевидно, что равенство выполняется и, следовательно, данная форма является полным дифференциалом.

4) Если мы хотим указать, что форма не является полным дифференциалом, мы обозначаем ее символом…..

5) Чтобы получить это равенство, необходимо продифференцировать данную форму по х.

6) Когда две фигуры тождественны, соотношение длин их соответствующих сторон равно.

7) Рассмотрим исключения в этом конкретном случае.

Problem Solving

3. Can you solve the following problems and explain how you obtained the solution?

Example 1.

Find the general solution of each of the following differential equations:

(a)

(b)

(c)

In (b) find the particular solution y=f(x) such that f (0) =-ln 2, and in (c) find the particular solution which has value 5 when x= 0.

Example 2.

Find the general solution of each of the following differential equations:

(a)

(b) (D²+6D+9)y=0

(c ) (D²-6D+10)y=0

Translate the solutions given below into Russian.

Source: Crowell and Slesnick's Calculus with Analytic Geometry The Dartmouth CHANCE Project Version 3.0.3, 5 January 2008

Example 2:

For the first, the characteristic equation is t²-5t+ 6 = 0, which is equivalent to (t-2)(t-3) = 0. Hence the two roots are 2 and 3, and the general solution is given by y=c₁e^2x +c₂ e^3x.

In (b), the characteristic equation is t²+6t+ 9 = 0, which is equivalent to (t+3)²=0. Thus there is only one root, 3. The solutions of the differential equation are therefore all functions y=(c₁x+c₂)e^-3x, where c₁ and c₂ are arbitrary constants.

The characteristic equation for (c) is t²-6t+ 10 = 0 and, since its discriminant is equal to 4, the roots are not real. Solving the quadratic equation, we find that the roots are 3+і and 3-і.

Hence the general solution is y=e^3x(c₁cosx+c₂sinx).

In the example above find the English equivalents for: который, следовательно (3 синонима), корень, произвольный, так как, равен.

Translate the following example with the explanation into Russian in the written form:

*Source: Crowell and Slesnick's Calculus with Analytic Geometry The Dartmouth CHANCE Project Version 3.0.3, 5 January 2008

Example 3.

Find the particular solution of the differential equation D(D-5)y = 0 which has value equal to 2 and derivative equal to -15 when x = 0. The characteristic equation is t(t - 5) = 0, whose roots are obviously 0 and 5. The general solution is therefore

y=c₁e^0x+c₂e^5x=c₁+c₂e^5x

The derivative is y’=5c₂e^5x.

When x = 0, we are told that y = 2 and y’ = -15. Substituting these values in the preceding equations, we obtain

2=c₁+c₂e^5.0=c₁+c₂

-15=5c₂e^5.0=5c₂.

It follows that c₂ = -3 and thence that c₁ = 5. Hence the required solution is

y=5-3e^5x.

It is extremely useful to recognize alternative forms of the general solution of the differential equation (D² + aD + b)y = 0 in the case where the roots of the characteristic equation are the complex numbers α + iβ and α - iβ. In particular, it is easy to verify that the functions

y=ce^αxsin(βx+k), (1)

y=ce^αxcos(βx+k) (2)

where c and k are arbitrary real numbers, are both solutions. To see that this is so, we expand (1) using the trigonometric identity for the sine of the sum of two numbers. The result is

y = ce^αx(sin βx cos k + cos βx sin k)

= e^αx[(c sin k) cosβx + (c cos k) sinβx].

Setting c₁ = csink and c₂ = ccosk, we obtain y = e^αx (c₁ cosβx+c₂ sinβx), which we know to be a solution. The proof for (2) is analogous.

Conversely, any solution y = e^αx (c₁ cosβx+c₂ sinβx) can be written in the forms (1) and (2). For if both c₁ = c₂ = 0, then y = 0, and we need only set c = 0 in (1) and (2). If c₁ and c₂ are not both zero, then , and we can write

y= e^αx

To put this equation in the form of (1), we set c = and observe that, since = 1

it follows from our definition of the functions sine and cosine that there exists a real number k such that cos k = and sin k= . Hence, we get

y = ce^αx(sin k cos βx + cos k sin βx)

= ce^αxsin (βx +k).

Again, by an analogous argument, the solution can also be written in the form of equation (2).

An advantage in using the forms (1) and (2) for the general solution is that it is easy to see what the graphs of such functions look like. They are all sinusoidal curves lying between the graphs of y = ce^αx and y = - ce^αx.

Give the English equivalents for: получить; производная; равный; подставить эти значения; корни характеристического уравнения; требуемое решение; доказать; доказательство; произвольные вещественные числа; наблюдать; можно записать; кривые.

Grammar Notes: