special mathematics

Dorian Popa Constantin-Cosmin Todea

SPECIAL MATHEMATICS. PROBLEMS

U.T. PRESS

CLUJ-NAPOCA

2014

Editura U.T.PRESS

Str.Observatorului nr. 34

C.P.42, O.P. 2, 400775 Cluj-Napoca

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Director: Prof. dr. Daniela Manea

Consilier editorial: ing. Calin D. Campean

Copyright c2014 Editura U.T.PRESSReproducerea integrala sau partiala a textului sau ilustratiilor din aceasta carte

este posibila numai cu acordul prealabil scris al editurii U.T.PRESS.

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ISBN 978-973-662-988-4

Contents

1 Differential equations effectively integrable 1

1.1 Differential equations with separable variables . . . . . . . . . . . . . 1

1.2 Homogenous equations . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Linear equations of order one . . . . . . . . . . . . . . . . . . . . . . 4

1.4 Bernoullis equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.5 Riccatis equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.6 Exact differential equations. Integrant factor . . . . . . . . . . . . . . 9

1.7 Equations of Clairaut and Lagrange . . . . . . . . . . . . . . . . . . . 13

1.8 Higher order differential equations . . . . . . . . . . . . . . . . . . . . 15

1.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Bibliography 20

i

Chapter 1

Differential equations effectivelyintegrable

In this chapter we present the most relevant types of differential equations of order

one and some basic and elementary techniques to solve them. We end this chapter

with a section regarding some higher order differential equations.

1.1 Differential equations with separable variables

A differential equation of the form

y = f(x)g(y) (1.1.1)

where f C(I), g C(J), I, J R are intervals is called equation with separablevariables. For y J1 J , J1 interval with g(y) 6= 0, the equation (1.1.1) is equivalentto

dy

g(y)= f(x)dx

and the solution follows by integrationdy

g(y)=

f(x)dx.

If y0 J and g(y0) = 0 then the equation (1.1.1) admits the singular solution y(x) =y0, for all x I.

Example 1.1.1. Integrate xy(1 + x2)y = 1 + y2.

1

2Solution. We have xy(1 + x2) dydx

= 1 + y2, hence ydy1+y2

= dxx(1+x2)

. We integrate to

obtain

ydy1+y2

=

dxx(1+x2)

. Equivalently

ydy1+y2

= (

1x x

1+x2

)dx, hence

1

2ln(y2 + 1) = ln |x| 1

2ln(1 + x2) + C.

It follows that ln(y2 + 1) = lnx2 ln(1 + x2) + C2. For C2 = ln k, k 1 we obtainthe solution

y2 + 1 =kx2

1 + x2.

Example 1.1.2. Integrate y = (x+ y + 1)2.

Solution. Let x+ y + 1 = z, z = z(x). Then y = z x 1 and y = z 1 hencethe equation becomes z = 1 + z2, that is dz

1+z2= dx. Integrating the previous relation

it follows that dz

1 + z2=

dx

or

arctan z = x+ C

so z = tan(x+ C), hence the solution is

y(x) = tan(x+ C) x 1.

1.2 Homogenous equations


y = f(y

x) (1.2.1)

where f C(I), with I R an interval where f(u) 6= u for any u I, is calledhomogenous equation.

Let z : I R be the function defined by z(x) = y(x)x, x I. Then y = xz and

y = xz + z hence the equation becomes xz = f(z) z, i.e.dz

f(z) z =dx

x

which is an equation with separable variables.

3Example 1.2.1. Integrate 2x3y = y(3x2y + y2), x (0,).Solution. We have y = 3x

2+y3

2x3, that is y = 3

2yx

+ 12( yx)3. The substitution y

x=

z, z = z(x) leads to

xz + z =3

2z +

1

2z3.

Equivalently, to xz = 12z + 1

2z3. The new equation becomes 2dz

z(1+z2)= dx

xand by

integration we get

2

(1

z z

1 + z2

)dz =

dx

x

hence

ln z2 ln(1 + z2) = ln x+ lnC,C 0.Finally by replacing z = y

xin z

2

1+z2= Cx, we get y2 = Cx

3

1Cx .

Example 1.2.2. Prove that the differential equation

y = f(a1x+ b1y + c1a2x+ b2y + c2

)f C(I), ak, bk, ck R, k {1, 2} becomes an homogenous equation if the system ofequations a1x+ b1y + c1 = 0a2x+ b2y + c2 = 0admits a unique solution (x0, y0).

Solution. Consider the change of variables

{x = t+ x0

y = u+ y0, and let u = u(t) be the

unknown function. Then

a1x+ b1y + c1 = a1(t+ x0) + b1(u+ y0) + c1 = a1t+ b1u,

a2x+ b2y + c2 = a2(t+ x0) + b2(u+ y0) + c2 = a2t+ b2u

and y = dydx

= dudt. Therefore we obtain

du

dt= f

(a1t+ b1u

a2 + b2u

)= f

(a1 + b1

ut

a2 + b2ut

)= g

(ut

),

which is a homogenous equation.

41.3 Linear equations of order one

In this section we present one of the most important example of differential equation,

the linear equation of order one. We will end this section with an example from the

electric circuit theory, which can be solved using differential equation.


y + f(x)y = g(x) (1.3.1)

where f, g C(I), I R is an interval is called a linear differential equation of orderone.

Let F (x) = xx0f(t)dt, x0 I, be an antiderivative of f . Multiplying the equation

(1.3.1) by eF (x) we get

yeF (x) + f(x)eF (x)y = g(x)eF (x)

or

(y eF (x)) = g(x)eF (x)

and by integration with respect to x it follows

y(x) = eF (x)( x

x0

g(t)eF (t)dt+ C

), C R.

The function F : I R is called the integrating factor of the above equation.

Example 1.3.1. Integrate y + 2y = ex, x R.Solution. We compute F (x) = e2x, x R, so we multiply with e2x and the equation

becomes

y(x)e2x + 2e2xy(x) = ex

or (y(x)e2x) = ex , hence by integration we obtain

y(x) = e2x(ex + C), C R.

Example 1.3.2. Let f : [0,) R be a continuous function such that there existslimx

f(x) = l, l R and a > 0. Prove that any solution of the equation

y + ay = f(x)

admits an horizontal asymptote at +.

5Solution. Let y be a solution of the equation. Multiplying with eax the equation

becomes (y(x)eax) = f(x)eax, x [0,). Integrating on [0, x], x > 0 it follows

y(x)eax =

x0

f(t)eatdt+ C,C R

hence

y(x) =

x0f(t)eatdt+ C

eax, x > 0.

Then

limx

y(x) = limx

( x0f(t)eatdt

eax+

C

eax

)= lim

x

( x0f(t)eatdt

)aeax

=

limx

f(x)eax

aeax=l

a.

Thus y = la

is the horizontal asymptote of f at .

Example 1.3.3. Find all continuous functions y : R R satisfying the followingintegral equation

x0

(x s)y(s)ds = x0

y(s)ds+ sinx, x R.

Solution. Since y is a continuous function, the functions from the left and the

right hand are differentiable. First we put the equation in the form

x

x0

y(s)ds x0

sy(s)ds =

x0

y(s)ds+ sinx

and by differentiation with respect to x we get

xy(x) +

x0

y(s)ds xy(x) = y(x) + cos x (?)

A second differentiation leads to y(x) = y(x) sinx, a linear equation of order one,which can be written in the form (y(x)ex) = ex sinx. By integration we get

y(x) = Cex sinx+ cosx2

, x R.Take x = 0 in relation (?) to obtain y(0) = 1 so y(0) = C 1

2= 1, hence C = 1

2.

The solution is

y(x) = ex + sinx+ cosx

2, x R.

6Example 1.3.4. Model a RL-circuit (a circuit with a resistor R and with an inductor

I) and solve the resulting differential equation for the current I(t) A (amperes), where

t is time. Assume that the circuit contains as an EMF, E(t) (electromotive force) a

battery of E = 48 V (volts), which is constant, a resistor of R = 11 (ohms), and

an inductor of L = 0.1 H (henrys), and that the current is initially zero.

Physical Laws. A current I in the circuit causes a voltage drop RI across the

resistor (Ohms law) and a voltage drop LI = LdIdt

across the conductor, and the

sum of these two voltage drops equals the EMF (Kirchhoffs Voltage Law).

Solution. According to these laws the model of RL-circuit is LI +RI = E(t), inthe standard form

I +R

LI =

E(t)

L(1.3.2)

Since R,L are constant and

RLdt = R

Lt+C we multiply the above equation with e

RtL

to obtain

(IeRtL ) =

E(t)

LeRtL .

Therefore

I = eRtl

(E

L

e(R/L)t

R/L+ C

)=E

R+ Ce

RtL .

In our case, RL

= 110.1

= 110 and E(t) = 480.1

= 480; thus,

I =48

11+ Ce110t.

The initial value I(0) = 0 gives I(0) = ER

+ C = 0, C = ER

and the particular

solution

I =E

R(1 eRtL ),

thus

I =48

11(1 e110t).

71.4 Bernoullis equations


y + f(x)y = g(x)y, R \ {0, 1}

where f, g C(I), I R interval, is called Bernoullis equation. For > 0 theequation admits the solution y(x) = 0, x I. On an interval I1 I where y(x) 6=0, x I1 the substitution z(x) = y1(x), x I leads to

z + (1 )f(x)z = (1 )g(x)

which is a linear equation.

Example 1.4.1. Integrate the equation xy2y = x2 + y3.

Solution. We can divide with y2 to obtain the rigorous form of this Bernoulli

equation, or directly we make the substitution z = y3 and notice that z = 3y2y. Byreplacing in the equation we obtain

x

3z = x2 + z,

next we multiply by 3x

to obtain the linear equation z 3xz = 3x. We multiply it by x3

and we have (x3z) = 3x2, hence z = 3x3x2dx. Equivalently z = 3x3(C x1).

Finally we obtain y3 = 3Cx3 3x2.

Example 1.4.2. Integrate the equation y + xy = xy2.

Solution. We use the substitution z(x) = y1(x), then z(x) = y2(x)y(x). Wemultiply the equation by y2 and we obtain

y2y + xy1 = x.

From the above substitutions the linear equation in z is z + xz = x, that is

z xz = x.

Since the integrating factor is F (x) = x22

, we multiply this equation by ex2

2 to

obtain

ex2

2 z xex2

2 z = xex2

2 .

8Equivalently we have (ex2

2 z) = xex2

2 , thus ex2

2 z = xex22 dx. Integrating wehave

z = ex2

2 (ex2

2 C) = 1 Cex2

2 .

We recall that z = y1 to conclude that

y(x) = (1 Cex2

2 )1.

1.5 Riccatis equations


y = f(x)y2 + g(x)y + h(x)

where f, g, h C(I), I R interval, is called Riccatis equation. Generally Riccatisequations cannot be effectively integrated. But if y0 is a particular solution of it, then

the substitution y = y0 +1z

gives the linear differential equation

z + (2f(x)y0(x) + g(x))z + f(x) = 0.

Example 1.5.1. Integrate the equation y = y2 1xy 1

x2, if it admits the particular

solution y0(x) = 1x .Solution. The substitution y = 1

x+ 1

zleads to the equation

(1x

+1

z) = (1

x+

1

z)2 1

x(1x

+1

z) 1

x2,

equivalent to1

x2 z

z2=

1

x2 2xz

+1

z2+

1

x2 1xz 1x2.

We obtain the linear equation in z:

z

z2= 3

xz+

1

z2,

next we multiply by xz2 to obtain xz = 3z+x or z 3xz = 1. We multiply this

equation by x3 to obtain zx3 3x4 = x3, that is (zx3) = x3. Integratingwe have z = x3 x3dx, hence z(x) = x3(x2

2+ C), with the final solution

y = 1x

+1

x3(x22

+ C)=

1 + 2Cx2

x 2Cx3 , C R.

9Example 1.5.2. Integrate (1+x3)yy2x2y2x = 0, x (1,), if the equationadmits a particular solution of the form y0(x) = ax

n, a R, n N.Solution. Replacing y0 in the equation we get

(1 + x3)naxn1 a2x2n axn+2 2x = 0

or

a2x2n + (na a)xn+2 + naxn1 2x = 0, x > 1.It follows n = 2 and a2x4 +ax4 + 2ax 2x = 0 for all x > 1, hence a = 1. The

particular solution is y0(x) = x2. The substitution is y = x2 + 1

zand leads to

z(1 + x3) + 3x2z = 1 or (z(1 + x3)) = 1.So z(1 + x3) = x+ C hence z = x+C

1+x3. Finally

y = x2 +1 + x2

x+ C =Cx2 + 1

C x , C R.

1.6 Exact differential equations. Integrant factor

Let D R be a rectangle D = [a, b] [c, d], a b, c d and P,Q C1(D). Adifferential equation of the form

P (x, y)dx+Q(x, y)dy = 0 (1.6.1)

where Py

(x, y) = Qx

(x, y) for all (x, y) D is called an exact differential equation.Under the previous conditions there exists a function F C2(D) given by the relation

F (x, y) =

xx0

P (t, y)dt+

yy0

Q(x0, t)dt, (x0, y0) D,

such that

dF (x, y) = P (x, y)dx+Q(x, y)dy, (x, y) D.Since the exact differential equation is equivalent to dF (x, y) = 0 the solutions are

implicitly defined by

F (x, y) = C,C R.

10

The function F is called an antiderivative (primitive) of the differential form Pdx +

Qdy.

If an equation of the form (1.6.1) is not an exact equation then a function C1(D) with the property that the equation

(x, y)P (x, y)dx+ (x, y)Q(x, y)dy = 0 (1.6.2)

is an exact differential equation is called integrant factor. Denoting

P1(x, y) = (x, y)P (x, y), Q1(x, y) = (x, y)Q(x, y), (x, y) Dthe equation (1.6.2) is an exact equation if

P1(x, y)

y=Q1(x, y)

x, (x, y) D (1.6.3)

The equation (1.6.3) is equivalent to the equation of integrant factor

Q

x P

y=

(P

y Qx

). (1.6.4)

In practice, usually we are looking for integrant factors of the form = (x) or

= (y). If the equation

Q(x) =(P

y Qx

)(x)

depends only on x. then there exists = (x). If the equation

P(y) =(P

y Qx

)(y)

depends only on y, then there exists = (y).

Example 1.6.1. Integrate the equation eydx (2y + xey)dy = 0.Solution.It is easy to check that P (x,y)

y= Q(x,y)

x= ey, where

P (x, y) = ey, Q(x, y) = 2y xey.We apply the integral formula for F , for x0 = 0 to obtain

F (x, y) =

x0

eydt+ y0

(2t)dt = eyt|x0 2[t2

2

]y0

=

11

xey y2.The solution of the equation is implicitly defined by the equation xey y2 = C.

Example 1.6.2. Find the integrant factor = (y), depending on y, for the equation

(2xy2 3y3)dx+ (7 3xy2)dy = 0

and integrate the equation.

Solution. Let P (x, y) = 2xy2 3y3, Q(x, y) = 7 3xy2, (x, y) R2. We haveP (x, y)

y= 4xy 9y2, Q(x, y)

x= 3y2.

Since depends on y, = (y) we apply the above formula to obtain

(2xy2 3y3)(y) = (4xy 9y2 + 3y2)(y)

We divide by y to obtain y(2x 3y)(y) = 2(2x 3y)(y). Next we divide by2x 3y to obtain an equation with separable variables

yddy

= 2 henced

= 2dy

y.

Integrating we have ln || = 2 ln |y|+ C, thus (y) = Cy2.Multiplying the equation by (y) = 1

y2it follows (2x 3y)dx + ( 7

y2 3x)dy = 0

which is an exact equation. Let P (x, y) = 2x3y,Q(x, y) = 7y23x. For x0 = 0, y0 =

1 we get

F (x, y) =

x0

(2t 3y)dt+ y1

7

t2dt = x2 3xy 7

y + 7.

The solutions are given by x2 3xy 7y

= C,C R.

Example 1.6.3. Find the integrating factor = (x+ y2), depending on x+ y2 for

the equation (3y2 x)dx+ (2y3 6xy)dy = 0.Solution. We have P (x,y)

y= 6y, Q(x,y)

x= 6y. For shortness we denote by t the

expression x+ y2, hence t = t(x, y) = x+ y2. We apply formula 1.6.3 to obtain

Q(t)t

x P(t) t

y=

(P

y Qx

)(t).

12

Since tx

= 2x and ty

= 1 we obtain

(t)[(2y3 6xy)1 (3y2 x)2y] = [6y (6y)](t),equivalently (t)(2y3 6xy 6y3 + 2xy) = 12y(t). Divide by y and we get(t)(4y2 4x) = 12(t). Divide also by 4 and we obtain the equation

(x+ y2)(t) = 3(t)that is td

dt= 3 which is a separable variable equation, hence

d

= 3dt

t.

We integrate and obtain ln || = 3 ln |t| + ln |C|, thus (t) = Ct3. We concludethat an integrant factor can be chosen (x, y) = (x+ y2)3.

Example 1.6.4. Integrate the equation

(4x3 + 2xy)dx+ (3y2 + x2)dy = 0

and prove that it admits a unique solution y : R R satisfying the equation y(0) = 1.Solution. Let P (x, y) = 4x3 + 2xy, Q(x, y) = 3y2 + x2, (x, y) R2. Since

P (x,y)y

= Q(x,y)x

= 2x the equation is exact. For x0 = y0 = 0 we obtain

F (x, y) =

x0

P (t, y)dt+

y0

Q(0, t)dt = x4 + x2y + y3

so the solutions of the equations are implicitly defined by

x4 + x2y + y3 = C, C R.For x = 0, y = 1 we find C = 1, the solution satisfying the condition y(0) = 1 is

implicitly given by

x4 + x2y + y3 = 1.

Now let f : R R, f(y) = y3+x2y+x41 for a fixed x R. Since f (y) = 3y2+x2 0for all y R and lim

yf(y) = +, lim

yf(y) = it follows the equation f(y) = 0

admits a unique solution for all x R. From the equation we get

y = 4x3 + 2xy

3y2 + x2

so the function y is differentiable on R.

13

1.7 Equations of Clairaut and Lagrange


y = xy + g(y)

where g C1(I), I R interval, is called Clairauts equation. We make the substitu-tion y = p to obtain the equation y = xp+ g(p). By differentiating we get

p = p+ xp + g(p)p

hence (x + g(p))p = 0. We obtain the singular solution (given by parametric equa-tions)

x = g(p), y = pg(p) + g(p)and from p = 0 we obtain p = C and the general solution is

y = Cx+ g(C), C R.


y = xf(y) + g(y)

where f, g C1(I), I R interval, with f(u) 6= u, u I , is called Lagrangesequation. The same substitution y = p leads us to y = xf(p) + g(p) hence, bydifferentiating we get

p = f(p) + xf (p)p + g(p)p.

We obtain the linear equation

(p f(p) = (xf (p) g(p))p,

with the unknown x = x(p), which has a general solution x = h(p, C). We obtain the

general parametric solution of the Lagranges equation

x = h(p, C), y = h(p, C)f(p) + g(p), C R.

Example 1.7.1. Integrate the equation y = xy +

1 + y2.

14

Solution. Let y = p, so y = xp+

1 + p2. Differentiate to obtain

p = p+ xp +2pp

2

1 + p2

hence

0 = p(x+p

1 + p2)

First we have p = 0 that is p = C and the general solution

y = xC +

1 + C2.

Next we obtain the singular solution

x = p1 + p2

, y = p2

1 + p2+

1 + p2, p R.

Example 1.7.2. Integrate the equation y = x(1 + y) + y2.

Solution. Let y = p then y = x(1 + p) + p2 and we differentiate to obtainp = 1 + p+ xp + 2pp, hence

0 = 1 + p(x+ 2p),dx

dp= x 2p

that is

dx = (x+ 2p)dpwhich is a linear equation with the unknown x = x(p).We write this equation in the

form x+x = 2p and we multiply this equation by ep to obtain epx+ epx = 2pep,that is (epx) = 2pep. Equivalently we get

x = 2eppepdp = 2ep(pep ep + C) = 2p+ 2 2Cep.

We replace x in the first formula of y to obtain

y = (1 + p)(2p+ 2 2Cep) + p2.

So, the parametric equations are

x = 2p+ 2 2Cep, y = (1 + p)(2p+ 2 2Cep) + p2, C R.

15

1.8 Higher order differential equations

We present some classes of differential equations of order n, n > 1, which can be

reduced to differential equations of order strictly less than n.

1. Equations of the form F (x, y(k), y(k+1), . . . , y(n)) = 0 with the unknown

y Cn(I), y = y(x), I R interval. The substitution z = y(k) leads to the equationof order n k

F (x, z, z, . . . , z(nk)) = 0.

Example 1.8.1. Integrate y + 2y = e2x, x R.

Proof. The substitution z = y leads to the equation z + 2z = e2x. Multiplying by

e2x we get (ze2x) = 1 so z(x) = e2x(x+ C1) and

y(x) =

e2x(x+ C1)dx = 1

2e2x(x+ C1 +

1

2) + C2, C1, C2 R.

2. Equations of the form F (y, y, y, . . . , y(n)) = 0. Let y = p(y) where pbecomes the new unknown of the equation. We have

y = p;

y = p(y)y = pp;

y = p(y)(y)2 + p(y)y = p2p + p(p)2;

. . .

so we get a differential equation of order (n 1) with the unknown p.

Example 1.8.2. Integrate 2yy = (y)2 + 1.

Proof. The substitution y = p, p = p(y) leads to 2ypp = p2 + 1 which can be written

as 2pdpp2+1

= dyy

. It follows

2pdpp2+1

=

dyy

, hence ln(p2 + 1) = ln |Cy| so p = Cy 1.

16

To obtain the solution we have to integrate the equation y = Cy 1, which isequivalent to dy

Cy1 = dx. We get

4(Cy 1) = C2(x+ C1), C, C1 R.

3. Equations of the form F (x, yy, yy, . . . , y

(n)

y) = 0. Remark that these equa-

tions are homogenous with respect to y, y, . . . , y(n). The substitution

z =y

y, z = z(x)

leads to a differential equation of order n 1. Indeedy

y= z2 + z;

y

y= z3 + 3zz + z, . . .

Example 1.8.3. Integrate the equation x2yy = (y xy)2.

Proof. Divide by y2 to obtain x2 yy

= (1 xyy

)2. Take the substitution yy

= z, hence

yy

= z + z2 and replace it to obtain the equation x2(z + z2) = (1 xz)2, that is

x2z + x2z2 = 1 + x2z2 2xz.

We have a linear equation x2z + 2xz = 1 which is equivalent to (x2z) = 1, hence

x2z = x+ C. Since z = yy

we obtain the separable variable equation

y

y=x+ C

x2.

Equivalently we get dyy

= ( Cx2

+ 1x)dx. Integrating, we obtain

ln |y| = C 1x

+ ln |x|+ C1

hence the solution

y = xeCxC1.

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1.9 Problems

Problem 1.9.1. Integrate the following differential equations of order one:

(1) xydx+ (x+ 1)dy = 0;

(2)y2 + 1dx = xydy;

(3) 2x2yy + y2 = 2;

(4) (x y)dx+ (x+ y)dy = 0;

(5) xy y = x tan yx;

(6) x y 1 + (y x 2)y = 0;

(7) xy 2y = 2x4;

(8) y + y tanx = 1cosx

;

(9) y + 2y = y2ex;

(10) xy y2 + (2x+ 1)y = x2 + 2x, if admits the particular solution y0 = x;

(11) y = y2 2yex + e2x + ex, if admits the particular solution y0 = ex;

(12) y + y2 2y sinx+ sin2 x = cosx, if admits the particular solution y0 = sinx;

(13) y = xy ln y;

(14) y = xy + y2;

(15) y = 2xy + sin y;

(16) y = 2xy + ln y;

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Problem 1.9.2. Integrate the exact differential equations:

(1) (2 9xy2)xdx+ (4y2 6x3)ydy = 0;

(2) yxdx+ (y3 + lnx)dy = 0;

(3) 3x2+y2

y2dx 2x3+5y

y3dy = 0;

(4) ydx+ xdy = 0;

Problem 1.9.3. Find the integrant factor for the equations:

(1) (x+ sinx+ sin y)dx+ cosydy = 0 depending on x, = (x);

(2) (x y)dx+ (y + x2)dy = 0 depending on x2 + y2, = (x2 + y2);

Problem 1.9.4. Integrate the following equations of order higher than one:

(1) x2y = y2;

(2) xy = y + x2;

(3) y2 + 2yy = 0;

(4) y3y = 1;

(5) yy y2 = y2y;

(6) x2yy = (y xy)2;

(7) y + 2xyy = 0;

(8) xyy xy2 = yy.

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Solutions: 1.9.1 (1). y = C(x+ 1)ex; (2). ln | x |= C +y2 + 1; (3). y2 2 =Ce

1x ; (4). ln(x2 + y2) = C 2 arctan y

x; (5). Divide by x to obtain y y

x= tan y

x

and then take the substitution yx

= z to obtain a variable separable equation with

the final solution sin yx

= Cx; (6). Make the substitution z = y x, z = z(x)and obtain the final solution (y x + 2)2 + 2x = C; (7). y = Cx2 + x4; (8).y = sin x + C cosx; (9). y = 0 and y(ex + Ce2x) = 1; (10). Take the substitution

y = z + x which leads to the Bernoulli equation in z,xz z2 + z = 0 and the finalsolution y = 1

Cx+1+x; (11). Take the substitution y = z+ex and get the final solution

y = 1Cx + e

x; (12). y = 1C+x

+ sinx; (13). y = Cx lnC and y = 1 + lnx; (14).y = Cx + C2 and x2 + 4y = 0; (15). x = Cp sin pcos p

p2, y = 2Cp sin p2 cos p

p; (16).

x = Cpp2, y = 2(Cp)p

p+ ln p; 1.9.2 (1). x2 3x3y2 + y4 = C; (2). 4y lnx+ y4 = C;

(3). x + x3

y2+ 5

y= C; (4). xy = C; 1.9.3 (1). = ex; (2). = (x2 + y2)

32 ; 1.9.4 (1).

The substitution is y = z and the final solution

C1x C21y = ln | C1x+ 1 | +C 2;

(2). y = x3

3+ C1

x2

2+ C2; (3). y

= p, y = pp and the final solution is y = C andy3 = C1(x + C2)

2; (4). C1y2 1 = (C1x + C2)2; (5). The equation in p = p(y) is

p 1yp = y with the solution y = C1 and

yy+C2

= C3eCx; (6). Divide by y2 to obtain

the equation x2 yy

= (1 xyy

). Use the substitution yy

= z. to obtain the equation

z = 2zx

+ 1x2

The final solution is y = xeC1x C2; (7). Divide by y

2; (8). y = C2eC1x2 ,

Bibliography

[1] D. Alpay, A Complex Analysis Problem Book, Birkhauser, Springer Bassel, 2011.

[2] E. Kreyszig, Advance Engineering Mathematics 10 edition, John Wiley and Sons,

Inc., Hoboken,New Jersey, 2011.

[3] L. Mejlbro, Complex Functions Examples c-7. Applications of the Calculus of

Residues, online, www.BookBoon.com, 2008.

[4] A. Philippov, Recueil de Problemes Dequations Differentielles, Editions Mir,

Moscou, 1976.

[5] S. Toader and G. Toader, Matematici Speciale I, II, U.T. Press, Cluj-Napoca,

2009.

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