DiffEquations[Unit 2]

Differential Equations (MATH313)

Equations of Order One Page 1

EQUATIONS OF ORDER ONE

In this chapter we study several elementary methods for solving first-order differential equations. We begin

studying an equation of the form

Mx, ydx + Nx, ydy = 0 (2.1)

where M and N are both functions of x and y.

2.1 Separation of Variables

Some equations of the form of 2.1 are so simple that they can be put into the form

Axdx + Bydy = 0 (2.2)

that is, the variables can be separated. Then a solution can be written at once.

Example 2.1 Solve the equation dydx =

2yx

Answer:

y = cx

Example 2.2. Obtain the general solution

1 xy = y Answer:

y ln c|1 x| = 1

Example 2.3. Solve the equation

xydx +edy = 0 Answer:

e +y = c



Example 2.4. Solve the equation

dr = bcos dr + r sin d Answer:

r = c1 b cos

Example 2.5. Obtain the particular solution satisfying the initial condition indicated. 2a rdr = r sin d

when = 0 and r = a. Answer:

r lnr/a = r cos a

Drill Problems 2.1 For each of the succeeding problems, obtain the general solution:

1.

y =xy

2.

y = x

y1 +x

3. y +y sin x = 0

4.

y =3x 1

3 + 2y

5. xy = 1 y&/

6.

dydx =

x ey +e'

7.

dydx =

x1 +y

8.

xy + xdx = xy +x +y + 1dy



9.

1 + ln xdx +1 + lnydy = 0

10. xcosydx + tan y dy = 0 For the succeeding equations, find the solutions of the given initial value problem. 11. y = 1 2xy, y0 = 1/6

12.

*+*, =

+, , r1 = 2

13.

y = -.'/ , y0 = 1

14. sin 2x dx + cos 3y dy = 0,y/2 = /3 15. y = 12&34'5 , y0 =

&

2.2 Equations with Homogenous Coefficients

The polynomials

x 3xy + 4y (2.3)

x +y (2.4)

x4y + 7y/ (2.5)

are said to be homogenous polynomials. They are called as such because each term posses the same

degree (power) when all of the powers of each variable are summed.

Example 2.6. The polynomial

x 3xy + 4y (2.3) is homogenous. By inspection, all the terms have a total degree of 2. First term power : x 2; y 0; total 2 Second term power: x 1; y 1; total 2 Third term power: x 0; y 2; total 2 Thus the above polynomial is homogenous of degree 2.



The homogeneity of a function is formally defined as follows

Definition 2.2. The function f(x,y) is said to be homogenous of degree k in x and y if, and only if,

fx, y = ; fx, y Example 2.7. To prove the homogeneity of the polynomial

x 3xy + 4y (2.3) using definition 2.2, we let

fx, y = x 3xy + 4y Then we replace x with x and y with y. Thus

fx, y = x 3xy + 4y

fx, y = x 3xy + 4y

fx, y = x 3xy + 4y

fx, y = 1fx, y3 Since the above polynomial complies with definition 2.2, it follows that it is homogenous whose degree is k =2, as was demonstrated in example 2.6.

Note that k is permitted to be any number. Thus the function



6. x sin yx y sinxy

7. x + 3xyx 2y

8. x/

x + 2y 9. x ln x y ln y 10.

x +y&/x y&/

For homogenous functions, the following theorems hold:

Theorem 2.2.1 If M(x, y) and N(x, y) are both homogenous and of the same degree, the function

Mx, yNx, y

is homogenous of degree zero.

Theorem 2.2.2 If f(x, y) is homogenous of degree zero in x and y, then f(x, y) is a function of y/x alone.

Suppose that a given differential equation is of the form

Mx, ydx + Nx, ydy = 0 (2.1)

where M(x, y) and N(x, y) are both homogenous functions and are of the same degree in x and y. By

theorem 2.2.1, the ratio M/N is homogenous and of degree zero. If we let the ratio M/N = f(x,y), by theorem

2.2.2, the ratio M/N = f(x, y) is a function of y/x alone.. Thus equation 2.1 can be put into form

dydx + g A

yxB = 0 (2.7)

By putting y = vx into equation 2.7, it becomes of the form

x dvdx + v + gv = 0 (2.8)

This equation 2.8 is now variable separable and can be solved by the methods presented in Article 2.1. By

reverting back to the original variables, we can then obtain the solution for equation 2.1.



Example 2.8 Solve the equation x xy +ydx xydy = 0

Answer

y x exp AyxB = c

Example 2.9 Solve the equation

xydx +x +ydy = 0 Answer

y2x +y = c

Example 2.10 Find the particular solution of the equation y + 7xy + 16xdx +xdy = 0

when x = 1 and y = 1. Answer

5y + 4x ln x = x y

Drill Problems 2.2b In the problems 1 thru 10, verify if the coefficients of the differentials are homogenous, identify the degree of the coefficients, then obtain a family of solutions.

1.

33x +ydx 2xydy = 0

2.

x 2ydx +2x + ydy = 0

3.

22x +ydx xydy = 0

4.

xydx x + 3ydy = 0

5.

xy = 4x + 7xy + 2y

6.

3xydx + x + ydy = 0

7. x y ln y + y ln xdx + xlny ln xdy = 0



8.

ydy = xxdy ydxe/'

9.

ydx = Ax +



2.3 Exact Equations

In Article 2.1, we presented that when an equation can be put into the form

Axdx + Bydy = 0 (2.2)

a set of solutions can be determined by integration.

In this section, we extend that idea to some equations of the form

Mx, ydx + Nx, ydy = 0 (2.1)

in which separation of variables may not be possible. Suppose that a function F(x,y) can be found that has

for its differential the expression M dx + N dy that is

dF = Mdx + Ndy (2.9)

Then certainly,

Fx, y = c (2.10)

defines implicitly a set of solutions for 2.1. Differentiating 2.10, we obtain

dF = 0 (2.11)

and equating this to 2.9, we have

Mdx + Ndy = 0 (2.11)

which conforms with the form of 2.1 (the form of 2.11 is the simplified form of 2.1). At this point, two things

are then needed:

To find out under what conditions on M and N a function F exists such that its total differential is

exactly M dx + N dy;

If the above conditions are satisfied, how the function F will be determined.

The condition pointed by the first thing mentioned above is stipulated in the following theorem:



Theorem 2.3.1 If M, N, HIH' and

HJH are continuous functions of x and y, then a necessary and sufficient

condition that

Mdx + Ndy = 0 (2.11)

be exactly a differential of a function F, or, in other words, 2.11 be an exact equation, is that

My =

Nx (2.12)

If an equation is exact, the following method can be used to solve the equation:

Since its solution will be F = c, the partial derivative of F with respect to x is the one multiplied to

dx, and the partial derivative of F with respect to y is the one multiplied to dy, that is

Fx = M (2.13)

Fy = N (2.14)

Integrate 2.13 with respect to x, holding y to be constant. Thus a function, say , in terms of x and

y will result. The arbitrary constant that will arise from this integration is any function of y, say T(y).

Thus,

F = LFxM'NOPQRSTQS = x, y + Ty (2.15)

Differentiate 2.15 with respect to y.

dFdy =

Wx, yXy + Ty (2.16)

Since the equation is exact, 2.16 should be equal to 2.14. Equate the two then solve for the

unknown function in terms of y. Then, substitute this to 2.15 and a family of solutions will have the

form

Fx, y = c (2.10)




3xxy 2dx +x + 2ydy = 0 Answer:

xy 3x +y = c


2x xy 2y + 3dx xy + 2xdy = 0 Answer:

xy + 4xy x4 6x = c


y 2xy + 6xdx x 2xy + 2dy = 0 for x = 0, y = 1. Answer:

xy xy + 3x 2y = 2

Drill Problems 2.3 In the problems 1-10, test each for exactness and obtain a general solution

1.

x + ydx +x ydy = 0

2.

6x +ydx + y2x 3ydy = 0

3.

2xy 3xdx +x + ydy = 0

4.

x 2ydx + 2y xdy = 0

5.

2x 3ydx +2y 3xdy = 0



6.

v2uv 3du +3uv 3u 4vdv = 0

7.

cos2y 3xydx +cos2y 2x sin 2y 2xydy = 0

8.

1 +y + xydx +xy + y + 2xydy = 0

9.

w + wz zdw +z +wz wdz = 0

10.

2xy tan ydx +x xsec ydy = 0

2.4 Non-exact Equations; Integrating Factors

The equation of the form

Mx, ydx + Nx, ydy = 0 (2.1)

is non-exact if it fails the test of exactness of Theorem 2.3.1. However, a suitable factor, called the

integrating factor, can be multiplied to the expression of 2.1 to make it exact. Let that factor be (x,y).

Multiplying 2.1 with the integrating factor,

x, y Mx, ydx + x, y Nx, ydy = 0 (2.16)

To ensure that 2.16 is exact, and that (x,y) is indeed an integrating factor, we require that 2.16 pass the

test of exactness of Theorem 2.3.1, that is

My =

Nx

(2.17)

To find an appropriate integrating factor for 2.1, we can solve 2.17 for . Thus,

My + My = N

x +

Nx (2.18)

My Nx + >

My

Nx? = 0 (2.19)



The partial differential equation of 2.19 may have more than one solution; therefore for a given equation of

2.1, a number of possible integrating factors may exist.

Unfortunately, the equation of 2.19 which determines the integrating factor is at least as difficult to solve as

the original equation. While in principle, the introduction of an integrating factor generalizes the solution of

differential equations, in practice, they can be only found in special cases.

One such special case is when the form of the original equation gives a suggestion as to what the

integrating factor should be. Thus, the integrating factor can be found by inspection. Table 2.4.1 gives a

suggestion as to what possible integrating factors are applicable for a given group of terms.

Table 2.4.1 Possible Integrating Factors



Example 2.14 Obtain the general solution

y = 3xy

x + 2y4 Answer

xy& 23 y = c

Example 2.15 Obtain the general solution

x + y +ydx xdy = 0 Answer

x tan& AyxB = c


yx ydx xx + ydy = 0 Answer

x + 2y = cxy

Another special case in which an integrating factor can be readily found is by going back to equation 2.19

My Nx + >

My

Nx? = 0 (2.19)

In this equation, if is a function of x or y alone, we can determine conditions that would allow us to find an

integrating factor systematically. For example, if is a function of x alone, 2.19 becomes

>My Nx? = N

x (2.20a)

=

1N>

My

Nx? x (2.20b)



Since is a function of x alone, it should follow that the right hand side of 2.20b is a function of x alone.

Thus, we let

gx = 1N>My

Nx? (2.21)

Therefore

= gxx

(2.22a)

Integrating 2.22a to solve for ,

L = Lgxdx

(2.22b)

ln = Lgxdx

(2.22c)

= exp >Lgxdx? (2.23)

Thus a necessary condition for to be a function of x alone is that

1N >

My

Nx?

is a function of x alone. If that is satisfied, the integrating factor is found by the integral of 2.23.

In a similar fashion, we find that if

1M>

My

Nx?

is a function of y alone, we can find the integrating factor by

= exp >Lhydy? (2.24)



where

hy = 1M>My

Nx? (2.25)


3xy +ydx +x + xydy = 0 Answer

xy +12 xy = c


3xy + 2xy +ydx +x +ydy = 0 Answer

3xy +ye = c

Drill Problems 2.4

Obtain the general solution

1.

y2xy + 1dx xdy = 0

2.

yy xdx + xy + xdy = 0

3.

xy + 1dx +x4ydy = 0

4.

2tds + s2 +stdt = 0

5.

yx4 ydx + xx4 +ydy = 0

6.

x +y + 1dx + xx 2ydy = 0

7.

2yx y + xdx +x 2ydy = 0



8.

y4x + ydx 2x 2ydy = 0

9.

xy + 1dx + xx + 4y 2dy = 0

10.

yy + 2x 2dx 2x + ydy = 0

2.5 The Linear Equation of Order One

In article 2.4, two methods of solving a non-exact equation were illustrated. Another special case worth

mentioning here is that if the equation

Mx, ydx + Nx, ydy = 0 (2.1)

can be put into the form

dydx + Pxy = Qx (2.26)

a general method of solution can be developed. Equation 2.26 is the form of a differential equation of order

one that is linear in y. Now, 2.26 is non-exact, and an integrating factor (x) can be found such that, when

multiplied to 2.26, makes it an exact equation. That integrating factor can be determined using the integral

x = exp >LPdx? (2.27)

If such integrating factor is found, the general solution of the order one linear equation of 2.26 is

y = &LQdx + c& (2.28)


2y 4xdx + xdy = 0 Answer:

xy = 2x4 + c




ydx +3x xy + 2dy = 0 Answer:

xy = 2y + 4y + 4 + ce'


y + 1dx +4x ydy = 0 Answer

20x = 4y 1 + cy + 14

Drill Problem 2.5 Obtain the solution as indicated

1.

y = x 2y

2.

y = x 4xy

3.

x/ + 3ydx xdy = 0

4.

b = csc x + y cot x

5.

b = csc x y cot x

6.

udx +1 3uxdu = 3uecdu

7.

y cosxdx + cos x dy = 0

8.

y x + xy cot xdx + xdy = 0

9.

22xy + 4y 3dx + x + 2dy = 0



10.

2xy +x +x4dx 1 +xdy = 0

11.

2x + 3y = y +2x + 3&/whenx=-1andy=0

12.

b = x 2xywhenx=1andy=1

13.

e didt + Ri = E;L,RandEareconstants,whent=0,i=0

14.

e didt + Ri = E sint ;L,RandEareconstants,whent=0,i=0

15.

y = 22x y;thesolutioncurvepassesthroughthepoint0,-1

2.6 Bernoullis Equation

If the equation can be translated into the form

dydx + Pxy = QxyQ (2.28)

a substitution can be made so that it conforms with the form of equation 2.26, and hence can be solved

using the methods developed in the preceding section. Such equations are called Bernoullis equation.

Take note that if n = 1, 2.28 is equal to 2.26, therefore the foregoing technique applies only to the case

when n 1.

We first multiply the equation by 1 nyQ

kdydx + Pxy = QxyQl W1 nyQX (2.29a)

1 nyQ dydx + 1 nPxy&Q = 1 nQx (2.29b)



By putting z = y&Q, and hence dz = 1 nyQ, 2.29a becomes dzdx + 1 nPxz = 1 nQx (2.29c)

Notice that 2.29c is now linear in z. The method presented for solving linear equations of order one in the

preceding section can now be applied.


y6y x 1dx + 2xdy = 0 Answer:

x = y6 + ce


2xy = yy + 3x Answer:

x = yc x

2.7 Substitutions Suggested by the Differential Equation

In some differential equations, the equation itself suggests a method by which it can be solved.

Substitutions can be done when a certain group of terms is repeated in other terms of the equation.


x + 2y 1dx + 3x + 2ydy = 0 Answer:

x + 3y + c = 3 ln|x + 2y + 2|

Substitutions can also be done when a term and its differential appears in the equation.


1 + 3x sin ydx x cos y dy = 0 Answer:

4x sin y = cx4 1



If the differential equation

Mx, ydx + Nx, ydy = 0 (2.1)

has its coefficients linear in two variables, that is

a&x +b&y +c&dx + ax +by +cdy = 0 (2.30)

a substitution can be done depending on the conditions of the coefficients.

One special case of 2.30 is when c1 and c2 are both zero. Equation 2.30 then has homogenous coefficients

of degree one. We have already described a method of solution for such cases (see Section 2.2).

The equation will be linear in one of the variables if a1 and b1 or a2 and b2 are zero.

In connection with 2.30, we consider the lines

a&x +b&y +c& = 0ax +by +c = 0

(2.31)

Three cases are possible here: the lines may intersect (has a single unique solution), may be parallel

(inconsistent equations, has no solution), or actually lie on the same set of points (equivalent equations,

has infinitely many solutions). We tackle each of the cases here.

If the system of equation of 2.30 has a single solution, let that solution be (h,k). Then the translation

x = u + hy = v + k

(2.32)

will change the equations 2.31 into equations of lines through the origin of the uv-coordinate system,

namely

a&u +b&v = 0au +bv = 0

(2.33)

Therefore, since dx = du and dy = dv, the change of variables of 2.32 will transform the differential equation

of 2.30 into

a&u +b&vdu + au +bvdv = 0 (2.34)




x + 2y 4dx 2x + y 5dy = 0 Answer:

x y 1 = cx + y 3

If the lines are parallel, then a1 = a2 and b1 = b2. Therefore, there is a recurring group of term in the

equation. Thus we can substitute a single variable to these recurring terms as we did before.

If the lines are equal, then there exists a constant k such that

ax +by = ka&x +b&y (2.35)

Thus, 2.30 can be written as

a&x +b&y +c&dx + Wka&x +b&y +cXdy = 0 (2.36)

Again, a recurring group of terms arises. We can do a substitution similar to what we did before.


2x + 3y 1dx 2x + 3y + 2dy = 0 Answer

2x + 2y + c = 6 ln|2x + 3y 7|

Drill Problems 2.7 Solve each of the given equations accordingly

1.

3x 2y + 1dx +3x 2y + 3dy = 0

2.

x + sin y sin y dx + 2x cos y dy = 0

3.

dydx = 9x + 4y + 1

4.

y = y xye



5.

dydx = sinx + y

6.

xydx +x 3ydy = 0

7.

3 sin y 5xdx + 2x cot y dy = 0

8.

y = 1 + 6x expx y

9.

x 3y + 2dx + 3x + 3y 4dy = 0

10.

6x 3y + 2dx 2x y 1dy = 0

11.

9x 4y + 4dx 2x y + 1dy = 0

12.

2x 3y + 4dx + 3x 1dy = 0whenx = 3, y = 2

13.

x + y 4dx 3x y 4dy = 0whenx = 4, y = 1

14.

43x + y 2dx 3x + ydy = 0whenx = 1, y = 0

15.

y = 23x + y 1whenx = 0, y = 1

DiffEquations[Unit 2]

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