§Differential Equation

§Differential Equation

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December 5, 2001

§ Chapter 1 Introduction

§Ordinary differential equation

Def. A functional equation containing a func- tion and its derivatives is called a D.E.

Ex.y

+ 2xy = 0 : a D.E. (y = f (x))

Def. The ”order” of a D.E. is the highest derivative order of the function which appear in D.E.

Ex.x

dx²

+ y

− y

x = 0 The order is 2.

Ex.y

+ y

= 0 It’s a 3rd order D.E.

Def. The ”degree” of a D.E. is the highest de- gree of the function with highest order deriva- tive which appears in the D.E.

Ex. (y

)

= 1 + y

= ⇒ y

= (1 + y

)

It’s a 2nd order 3rd degree D.E.

Ex. 1 + x

= sin(y

)

It’s a 1st order. 1st degree D.E.

? Varieties of D.E.s

(1) Ordinary D.E. (O.D.E.):

a D.E. containing single independent variable.

(2) Partial D.E. (P.D.E.):

a D.E. containing at least 2 independent vari-

ables.

? Solution (Integral) of a D.E.

Def. A function satisfying a given D.E. is called its ”solution”.

There are 3 kinds of solutions:(for O.D.E.):

general solution (G.S.) particular solution (P.S.) singular solution (S.S.)

Note: A solution of an n-th order D.E. con- taining n arbitrary constants is called the G.S.

of the D.E.

Ex. y

+ y = 0, y = f (x) ∈ C

. Sol. 2y

y

= −2yy

⇒ y

= −y

+ C

⇒ y

+ y

= C, C ≥ 0

⇒ y

=

= ± q

C e

− y

, e C

= C

⇒ ±

Ce²−y²

= dx

i .e. y = e C cos(x + ˆ C)

or y = e C cos(x) cos ˆ C − e C sin(x) sin( ˆ C)

= C

cos(x) + C

sin(x) where C

, C

: arb. constant.

∴ it is a G.S. of y

+ y = 0

Def. A function which is obtained by assign- ing some fixed no.s in the G.S. of a D.E. is called a P.S. of the D.E.

For example, let

C

= 4, C

= 3, y = 4 cos(x) + 3 sin(x) is a P.S. of y

+ y = 0.

(Hence G.S. is a set of P.S.s)

Def. A function which is a solution of a D.E.

but not a P.S. of the D.E. is called a S.S. of the D.E.

Ex. y

(1 + y

) = 1

Sol. y = ±1 satisfy the eq. y

(1 + y

) = 1.

But by solving it, we find y

+ (yy

)

= 1;

y

=

y²

⇒ y

=

= ±

√

⇒ √

1−y²

= ±dx ⇒ − p

1 − y

= ±(x − c) Thus (x − c)

+ y

= 1 is the G.S. of y

(1 + y

) = 1, but y = ±1 can’t be obtained by given proper c.

i.e. y = ±1: S.S. of the D.E.

§Linear and Non-linear D.E.s

Every term of a D.E. contains at most 1 degree

of a fun. or its derivatives with the coefficient

as the fun. of independent variables is called

a linear D.E., Otherwise the D.E. is called the

non-linear D.E.

Ex. P

(x)y

+ P

(x)y

+ P

(x)y = δ(x) it’s a linear D.E.

Ex. x

y

∂y²

+ (x

− z

)

= 0 It’s a non-linear D.E.

Ex. x

y

∂y²

+ (x

− y

)

= 0 It’s a linear D.E.

Note: Linear D.E. usually does not contain

S.S.

§Primitive and D.E.

Def. A functional equation containing n ar- bitrary (linearly independent) constants, say

f (x, y, c

, c

, · · · c

) = 0, · · · (1)

which can be reduced to a D.E. Then (1) is called the ”primitive” of the D.E.

Note: When we are given a D.E., the result after solving the D.E. is called ”solution”.

If we are given an equation (and NOT a D.E.), but asked to find a D.E. from the eq., then the given eq. is called ”primitive”

.

(integral)solution-

.primitive

Theorem: If a primitive has n arb. con- stants, then it has an n-th order O.D.E. Con- versely, an n-th order O.D.E. has a G.S. with n arb. constants.

eg.f (x, y, c

, c

, · · · , c

) = 0, · · · (i)

D.E. of f: g(x, y, y

, · · · y

) = 0 · · · (ii) (ii) −→ (i): (i) is a G.S. of (ii).

(i) −→ (ii): (i) is a primitive of (ii).

pf. Of the theorem: Let the primitive be

f (x, y, c

, c

, · · · c

) = 0 · · · (a)

where y = h(x). Consider f ∈ C

, then

∂f

∂x

+

y

= 0 · · · (1)

∂²f

∂x²

+ 2

y

+

∂y²

y

+

y” = 0 · · · (2) ...

∂ⁿf

∂xⁿ

+ · · · +

y

= 0 · · · (n) If

6= 0, i.e. (a) really contains y. Then we may eliminate n arb. constants from the n equations to obtain a D.E. of n-th order.

Ex. y = ax

+ bx + c, where a,b and c are arb.

constants. Find the D.E.

Sol. Primitive: y − ax

− bx − c = 0

= ⇒ y

− 2ax − b = 0 · · · (1) y

− 2a = 0 · · · (2) y

= 0 · · · (3) (3) is the desired D.E.

Ex. Given y = ax

+ bx

+ c, where a,b & c are arb. Find the D.E.

Sol. We have −y + ax

+ bx

+ c = 0 · · · (a) then



 

 

4ax

+ 3bx

+ 0 · c − y

= 0 · · · (1)

12ax

+ 6bx + 0 · c − y

= 0 · · · (2)

24ax + 6b + 0 · c − y

= 0 · · · (3)

Combining (a),(1),(2)&(3), we find











x

x

1 y 4x

3x

0 y

12x

6x 0 y

24x 6 0 y











| {z }

A







 a

b c

−1









| {z }

X

=







 0 0 0 0









| {z }

0

i.e. AX = 0, A is a n × n matrix. (By Fredolm Alternative) AX = 0 has only the trivial sol. X = 0 ⇔ detA 6= 0.

Since ∃X = (a, b, c, −1)

6= 0

S.t. AX = 0 i.e. detA = 0. Thus 

x

x

1 y 4x

3x

0 y

12x

6x 0 y

24x 6 0 y

3]

= 0

is the desired D.E.

§ Initial-Value Problems,Boundary-Value Problem, and Basic existence and uniqueness Theorem of the 1st order D.E.

e.g.



 



 

 dy

dx = 2x · · · (1) (a D.E.) y(1) = 4 · · · (2) (a supplementary condition) The G.S. of the D.E. :

(?) · · · x

+ c = y, c: arb. constant.

Substitute x=1 & y=4 into (?), we find c=3.

∴ y = x

+ 3 is a sol. of (1), (2).

Def. A problem involves both a D.E. and one or more supplementary conditions which the solution of the given D.E. must satisfy. If all of the associated supplementary conditions re- lated to one X value, the problem is called an

”initial- value problem.” If the conditions re- lated to two different X values, the problem is called a ”two-point boundary-value problem”

or simply a ”boundary-value problem”.

Ex.

dx³

+ y = 0, y(1) = 3 & y

(1) = 4.

? Sol. The above is an initial-value problem.

Ex.

dx³

+ y = 0, y(0) = 1 & y(

) = 5.

This is a boundary-value problem.

Theorem: (Existence and Uniqueness Theorem) Let the functions f and

be continuous in some rectangle α < x < β, r < y < δ, containing the point (x

, y

).

Then in some interval |x − x

| < h con- tained in α < x < β, ∃! y = φ(x) being the solution of

(?)  y

= f (x, y) y(x

) = y

Ex.

(

dx

= x

+ y

y(1) = 3 · · · (A)

Sol. Let f (x, y) = x

+ y

, &

= 2y

∵ f &

are continuous in every rectangle α < x < β, r < y < δ containing (1,3)

∴ By theorem, ∃! y = φ(x) satisfies (A).

Ex.

(

dx

=

y(1) = 2 · · · (1)

(

dx

=

y(0) = 2 · · · (2) Sol. Set f (x, y) =

x

&

=

Both funs. are conti. except for x = 0.

Hence the existence and uniqueness theorem can

only be applied in (1), but not in (2).

i.e. in (1), ∃! y = φ(x) being the solution of

(1); but in (2), inconclusive.

Remark: The knowledge that (?) has a unique sol. y = φ(x) is our hunting license to go hunting for it. We can apply any method (in- cluding guessing) to find its solution.

Ex. y

= −xy & y(0) = 0 · · · (1) Sol. Set

f (x, y) = −xy, ∂f

∂y = −x f &

:conti. in x-y plane.

∴ By thm.: ∃! y = φ(x) satisfies (1).

Since y ≡ 0 is a sol. of (1).

Hence y ≡ 0 is the unique sol. of (1).

§Method of Isoclines Procedure:

Graphical Construction of Integral Curves of 1st Order D.E.s

Def. Consider the 1st order D.E.:

dy

dx = f (x, y) · · · (A).

(A) is a curve along which the slope f(x,y) has a constant C is called an isocline of the D.E.(A), i.e. the isoclines of (A) are the curves f(x,y)=C for different values of C. (C is con- sidered as a parameter.)

Def. At each pt. in the xy-plane where f(x,y)

is defined,(A) provides a value for y

, which

can be thought of as the slope of a line segment

through that pt. The totality of all such line

segments form the ”direction field” for (A).

? Method of isoclines procedure:

(i) From (A), determine the family of isoclines f (x, y) = C · · · (B) and carefully construct several members of this family.

(ii) Consider a particular isocline

f (x, y) = C

of this family (B).

At all points (x,y) on the isocline the line seg- ments have the same slope C

and have the same inclination: α

= tan

C

.

(iii) Repeat the step (ii) for each of the iso- cline of (B).

(iv) Draw the approximate integral curves in-

dicated by the line segments obtained in Step

(iii)

ex.

Employ the method of isoclines to sketch the approximate integral curves

= x

+y

· · · (1).

Sol. i) The isoclines (1) are the curves x

+ y

= c

ii) Let c=1, x

+ y

= 1 (i.e.

= 1),etc.

Homework:

(1) Given the initial value problem (?)

(

dx

=

y(0) = 2

a) Sketch the direction field.

b) Employ the method of isocline to sketch the approximate integral curve of (?).

c) Solve (?), then sketch the integral curves.

d) Compare with those obtained in (b),(c).

§Solution of the First Order D.E.s I. D.E.s of separable variables

Def. If a D.E.

f (x, y) = y

· · · (A) can be written as M (x)dx + N (y)dy = 0 · · · (1),

then the D.E. is called a ”separable D.E.”

Theorem: The G.S. of (A) is Z

M (x)dx + Z

N (y)dy = c, where c is an arb. const.

pf. ∵ M (x)dx = −N (y)dy · · · (2)

Let H

and H

by any funs. of x, y,

respectively, s.t. H

(x) = M (x), H

(y) = N (y).

∴ H

(x)dx + H

(y)dy = 0 · · · (3)

If y = φ(x) is a differentiable fun. of x and a sol. of (1). By Chain-Rule:

H

(y) dy

dx = d

dx [H

(φ(x))]

From (2), we have H

(x) + H

(y)

= 0, i.e. H

(x) +

[H

(φ(x))] = 0 · · · (3)

⇐⇒

[H

(x) + H

(φ(x))] = 0

Integrating (3)

with respect to (w, r, t) x, we obtain that H

(x) + H

(φ(x)) = C, where C is an arbitrary const.

i.e.

Z

M (x)dx + Z

N (y)dy = C

II. Exact D.E.

Def. A D.E. M (x, y)dx + N (x, y)dy = 0 is

called ”exact” provided that ∃ a differentiable

fun. u(x, y) s.t. du = M (x, y)dx + N (x, y)dy.

Ex. 2xydx + (x

+ sin y)dy = 0 · · · (A)

? Consider u(x, y) = x

y − cos y. So

du = 2xydx+(x

+sin y)dy i.e. (A) is exact.

Theorem: a D.E.

M (x, y)dx + N (x, y)dy = 0 · · · (A), M,N ∈ C

is exact ⇐⇒ M

= N

· · · (B)

pf.( ⇒) Let (A) be exact, we will show that (B) holds. ∵ (A) is exact, ∃ a diff. fun. u s.t.

du = M dx + N dy Since

du = u

dx + u

dy

= M dx + N dy .

we find that u

= M (x, y), u

= N (x, y)

and M

= u

= u

= N

, M, N ∈ C

( ⇐) Let (B) holds, we will show that there’s a fun. u s.t. du = M dx + N dy,

i.e. u

= M, u

= N.

Set u(x, y) = R

x₀

M (t, y)dt + φ(y) Then we find that

N = u

=

( R

x₀

M (t, y)dt + φ(y))

= R

x₀

M

(t, y)dt + φ

(y)

= R

x₀

N

(t, y)dt + φ

(y)

= N (x, y) − N(x

, y) + φ

(y)

∴ φ

(y) = N (x

, y), φ(y) = R

y₀

N (x

, t)dt and u(x, y) = R

x₀

M (t, y)dt + R

y₀

N (x

, η)dη

= C (∵ du=0)

∴ du = M dx + N dy

(Thus u is a G.S. of Mdx + Ndy = 0)

Ex. 2yxdx + (x

+ sin y)dy = 0 · · · (1)

Sol. M

= 2x = N

= 2x ∴ (1) is exact by Thm.

Then the sol. of (1):

∵ u(x, y) = R

x₀

(2ty)dt + R

y₀

(x

+ sin t)dt

= C

∴ u(x, y) = x

y − x

y + x

(y − y

) − cos y + cos y

= C

⇒ x

y − cos y = C

: the G.S. of (1)

(Note that x

, y

can be chosen freely to suit

our need. In this case.x

, y

=

)

§Integrating Factor

Def. If M dx + N dy = 0 · · · (1) is not exact, but µ(x, y)(M dx + N dy) is exact for a fun.

µ ∈ C

. Then µ is called an ”integrating fac- tor”.

Theorem: Given a D.E. M dx+N dy = 0 · · · (A).

Then µ ∈ C

is an integrating factor of (A).

⇔ (µM)

= (µN )

.

Derivation for µ(x, y):

(µM )

= (µN )

⇔ µ

M + µM

= µ

N + µN

· · · (B) i) Suppose that µ is a fun. of x only.

i.e. µ=f(x),then (B) ⇒ µ

= µ(

) · · · (C)

ii) If

is a fun. of x only , too. then (c) ⇒ µ

µ =

, i.e. dµ

µ = (

)dx,

⇒ lnµ = R (

)dx ⇒ µ = exp( R

N

dx).

Ex. ydx − xdy = 0 · · · (1)

Sol. Let M=y, N=x, then M

= 1 6= −1 = N

(1) is not an exact D.E. , Consider M

− N

N = 1 − (−1)

−x = − 2

x = g(x) Let µ(x, y) = exp ( R −

dx) =

x²

· · · (2)

⇒ µ(x, y)(ydx − xdy) = 0

⇒ y

x

dx − 1

x dy = 0 Then ν(x, y)= R

x₀

(

t²

)dt + R

y₀

( −

)dt = C, and set (x

, y

) = (1, 0),we find

ν(x, y) = − y

x + 0 = e C i.e. y

x + C = 0 : G.S. of (1) .

Remark: We may also assume that µ is a fun. of y only. i.e. µ

M + µM

= µ

N + µN

⇒ µ

= ( N

− M

M )µ

If (

) ia also only a fun. of y, then µ = exp ( R

M

dy) :an integ. factor of (A).

§Isobrasic D.E. and Homogeneous D.E.

I. Def. A D.E. f (x, y, y

, · · · , y

) = 0 · · · (1) is called ” isobrasic of weight γ ”

provided that

f (tx, t

y, t

y

, · · · , t

y

)

= t

f (x, y, y

, · · · , y

) · · · (2)

Ex. (x + 2x

y)y

+ 2y + 3xy

= 0 · · · (A) Sol. Check (A) to see if (A) is isobrasic.

Let

f (x, y, y

) = xy

+ 2x

yy

+ 2y + 3xy

f (tx, t

y, t

y

) = (tx)(t

y

)

+2(tx)

(t

y)(t

y

)+2(t

y)+3(tx)(t

y)

= t

xy

+ t

2x

yy

+ t

2y + t

3xy

= t

f (x, y, y

) · · · (B)

If (B) holds, t

= t

= t

i.e. γ = m = −1

Thus (A) is isobrasic D.E. of weight -1 .

Note. If we set t =

in an isobrasic D.E. , (2) becomes

f (1,

,

x^m−1

, · · · ,

)

=

f (x, y, y

, · · · , y

) = 0 i.e. f (1,

,

x^m−1

, · · · ,

) = 0 · · · (3)

From (3) ,

f = φ(

,

x^m−1

, · · · ,

) = 0

⇒

= ψ(

,

x^m−1

, · · · ,

) Hence for a 1st order isobrasic D.E. , we may have a standard form

y⁰

x^m−1

= ψ(

) · · · (4)

Let u =

⇒ y = x

u , and y

= mx

u + x

u

i.e.

x^m−1

= mu + xu

= ψ(u) i) If ψ(u) − mu 6= 0 :

du

dx

= u

=

⇔

=

i.e. ln |cx| = R

ψ(u)−mu

du

x = ˜ c exp( R

ψ(u)−mu

du) ii) If ψ(u) − mu = 0 : xu

= 0 ⇒ u = c

or roots of ψ(u) − mu = 0 are sol.s of u =

∴ y = x

u

Ex. We know that

(x + 2x

y)y

+ 2y + 3xy

= 0 · · · (1) is an isobrasic D.E. of weight -1 .

Find the sol. of (1) . Sol.

Set t =

∵ f (tx, t

y, t

y

) = t

f (x, y, y

)

with m = γ = −1

∴ f (1, xy, x

y) = xf (x, y, y

)

= (x

+ 2x

y)y

+ 2xy + 3(x

y

) = 0

= φ(xy, x

y)

Assume that u = xy , we have y

=

=

u

x x

Substituting the above into (1) , (1 + 2u)(u

−

) + 2(

) +

= 0

⇔ (1 + 2u)u

+

= 0

⇔ (1 + 2u)

= −

i) u 6= 0 and u 6= −1

Then R

u(u+1)

du = R

u

+ R

u+1

= ln |u(u + 1)| = − R

x

= − ln |cx|

i.e. u(u + 1) =

u = xy =

q

1+_cx⁴ 2

y =

q

1+_cx⁴ 2x

ii) u = 0 or u = −1 α)u = 0 ⇔ xy = 0 i.e. y = 0

β)u = −1 ⇔ xy = −1 i.e. y = −

Thus the sol.s of (1) are : y =

q

1+_cx⁴ 2x

y = 0 or y = −

Homework: Solve the following D.E.s:

(1)2x

y

− x

y

+ 2xy + 1 = 0

(2)x

y

− x

y + y

= 0

II. Homogeneous D.E.s

Def. A D.E. M (x, y)dx+N (x, y)dy = 0 · · · (1) is said to be ”homogeneous” provided that

M , N are of the same weight γ in x and y.

(m=1)

i.e. M (tx, ty) = t

M (x, y) N (tx, ty) = t

N (x, y)

)

· · · (2) Note. From (1), we have

dy

dx

= −

= −

= −

· · · (3)

Let

= t , and substitute it into (3), we obtain that

dy

dx

= −

y x)

N (1,^y_x)

= φ(

)

Hence y

= φ(

) is the standard form for homogeneous D.E.s

Ex. Solve (y + p

x

+ y

)dx − xdy = 0 · · · (1) Sol.

Let M (x, y) = y + p

x

+ y

N (x, y) = −x

⇒ M(tx, ty) = ty + p

(tx)

+ (ty)

= t(y + p

x

+ y

) N (tx, ty) = −tx = t(−x)

So (1) is homogegeous.

Again, from(1), we have

dy

dx

=

√

x

=

+ q

1 + (

)

= φ(

)

Set u =

y = xu and y

= u + xu

Then (1) becomes u + xu

= u + √

1 + u

· · · (2)

⇔ x

= √

1 + u

⇔

1+u²

=

Set u = tan θ θ = tan

u

⇔ R

sec θ

dθ = R sec θdθ = ln |cx|

i.e. ln | sec θ + tan θ| = ln |cx|

√ u

+ 1 + u = q

(

)

+ 1 +

= cx · · · (3) Thus

q

(

)

+ 1 +

= cx is the G.S. of (1)

Homework. Solve the D.E.s:

(1) y

=

x²

(2) (x + y)dy = (x − y)dx

Remark. A homogeneous D.E. is an isobra- sic D.E. of weight γ with m equals to 1.

§Linear D.E. of First order and its expansions.

I. 1st order linear D.E.

P

(x)y

+P

(x)y = Q(x) , P

, P

, Q ∈ C

· · · (1)

e.g. f (x, y, y

) = 3x + 5x

y + 6xy

= 0 · · · (A)

Def. A D.E. with the form of (1) is called

the 1st order linear D.E. (P

6= 0)

Derivations.

From (1), we have

y

+ p(x)y = f (x) p, f ∈ C

· · · (2)

⇔

+ [p(x)y − f(x)] = 0

⇔ [p(x)y − f(x)]dx + dy = 0 · · · (3) Let M (x, y) = p(x)y − f(x)

N (x, y) = 1

Then since M

= p(x) 6= 0 = N

(3) is not exact.

Assume that µM dx + µN dy = 0 is exact , i.e. (µM )

= (µN )

Consider µ = µ(x) ,then µ

=

= (

)µ

and then µ = exp( R

N

dx)

Now

= p(x)

µ = e

: integrating factor of (3) and (3) becomes

e

y

+ e

p(x)y = e

f (x)

⇔

(e

y) = e

f (x)

⇔ e

y = R e

f (x)dx + c

⇔ y = e

[ R

e

f (u)du+c] · · · (4) which is a G.S. of (1) .

Ex. (?)  y

+ 2y = e

y(0) = 3 Sol.

Since (?) is a 1st order linear D.E.,

we find that

y = e

[ R

e

· e

dt + c]

= e

[ R

e

· e

dt + c]

= e

(e

+ c)

= e

+ ce

y(0) = 1 + c = 3

∴ c = 2

∴ y(x) = e

+ 2e

II. Bernoulli D.E.

Def. First order D.E. of the form y

+ P (x)y = Q(x)y

· · · (1)

is called a Bernoulli D.E.

Discussions :

i) n = 1 :

(1) ⇒ y

+ P (x)y = Q(x)y i.e. y

+ (P (x) − Q(x))y = 0

⇒

= −[P (x) − Q(x)]dx

⇒ y(x) = ce

(C: arb. )

ii) n 6= 1 :

(1) ⇒ y

y

+ P (x)y

= Q(x)

Let u = y

,then u

= (1 − n)y

y

Substitute the above into (1),

we find

u⁰

1−n

+ P (x)u = Q(x)

⇔ u

+ (1 − n)P (x)u = (1 − n)Q(x)

Use the formula of 1st order linear D.E.,

we have

u(x) = e

[ R

f (t)e

dt + c]

where p(x) = (1 ˜ − n)P (x) f (x) = (1 − n)Q(x)

And finally, y = u

: G.S. of (1).

Ex. y

+ 2xy = xe

y

· · · (A) Sol.

(A) is a Bernoulli D.E. with n = 3 i.e. y

y

+ 2xy

= xe

· · · (B) Set u = y

⇒ u

= −2y

y

⇒ (B) becomes: −

+ 2xu = xe

· · · (C)

⇒ u

− 4xu = −2xe

· · · (C)

∴ u(x) = e

[2 R

( −te

)e

dt +c]

= 2e

( R

−te

e

dt + c)

=

e

( R

e

d(3t

) + c)

=

e

(e

+ c)

=

e

+

e

And then y = u

= [

(e

+ c)]

III.Riccati D.E.

y

= P (x) + Q(x)y + R(x)y

P, Q, R  C

· · · (1)

Def.The 1st order D.E. with the form of eq. (1) is called a Riccati D.E. .

Disscussions

i) If one P.S. y

of (1) is known .

i.e. y

= P (x) + Q(x)y

+ R(x)y

· · · (2) (1)-(2): y

− y

= Q(x)(y − y

)

+ R(x)(y

− y

) · · · (2)

Let y − y

= u, y = u + y

⇒ u

= Q(x)u + R(x)(u

+ 2uy

)

⇒ u

= (Q(x) + 2R(x)y

)u + R(x)u

· · · (3) (3) is a Bernoulli D.E. with n = 2 for u.

Set f (x) = Q(x) + 2y

R(x), then u

u

− f(x)u

= R(x) · · · (4)

Let v = u

, v

= −u

u

· · · (4)

then becomes

v

+ f (x)v = −R(x)

∴ v(x) =

= e

[ R

−R(t)e

dt + c]

= g(x, c)

and y(x) = y

(x) +

g(x,c)

ii) If 2 P.S.s y

and y

of (1) are known.

y

= P (x) + Q(x)y

+ R(x)y

· · · (5) (1)-(5): y

− y

= Q(x)(y − y

)

+R(x)(y

− y

) · · · (6) From (2)

:

1

= Q(x)+R(x)(y+y

) · · · (7) and from (6):

2

= Q(x)+R(x)(y+y

) · · · (8)

Set y − y

= Y

, y − y

= Y

, then

(7)-(8):

1

−

= (y

− y

)R(x) = f (x), i.e.

1

−

= f (x),

1

−

= f (x)dx · · · (10) Integrating (10) w.r.t. x, we have

ln |Y

| − ln |Y

| = R

f (t)dt + ln |˜c|

∴ ln |

| = R

f (t)dt

y−y1

y−y2

= ˜ ce

· · · (11) (11) is a G.S. of (1)

Ex.y

= −

+

+

x³

· · · (1) Find the solution y(x)

Sol.

y = x ⇒ 1 = −

+

+

x³

= 1,

∴ y = x: a P.S. of (1)

y = −x ⇒ −1 = −

+

+

x³

∴ y = −x: also a P.S. of (1) Hence we find

( (y − x)

=

(y − x) +

(y

− x

) (y + x)

=

(y + x) +

x³

(y

− x

)

⇒







(y−x)⁰

y−x

=

+

x³

(y + x)

(y+x)⁰

y+x

=

+

x³

(y − x)

⇒

−

=

x³

(2x)dx

⇒ ln |y − x| − ln |y + x| = −

+ ln |˜c|

⇒

= ce

⇒ y =

1−ce^{− 2x}

x

§D.E. of the form y

= f (

) · · · (∗) Discussions:

i) If

=

= k:const.

f (

) = f (

)

= f (k +

)

= y

= ψ(

) where d = γ − kc

All we have to do is set u = ax + by + c

ii) If

6=

:

Let  X = x + h

Y = y + k s.t.

 αx + βy + γ = αX + βY ax + by + c = aX + bY where  −αh − βk + γ = 0

−ah − bk + c = 0

(Note that the solution for h and k definitely exists, since

detA ≡    

α β a b

= αb − aβ 6= 0 )

Then ( ∗) can be written as y

= f (

) = f (

Y X) a+b(_X^Y )

) i.e. y

= φ(

) · · · (1)

Let u =

, Y

(= y

) = u

X + u

(1) ⇒ Y

= φ(u) = u

X + u

∴ either (a) φ(u) − u = 0 or (b) φ(u) − u 6= 0

We may find sol. of u from (a) and (b) .

Ex.(x + y − 1)dy + (5x + 5y + 3)dx = 0 Please solve for y = y(x).

Sol.

dx

= −

= −5 −

Let x + y − 1 = u, then u

= 1 + y

∴ y

= u

− 1 = −5 −

du

dx

=

= −4(

) i)If u + 2 6= 0

−

(

du) = dx

−

(1 −

)du = dx

⇒ x + c = −

(u − 2 ln |u + 2|)

⇒ x + c = −

(x + y − 1 − 2 ln |x + y + 1|) ii)If u + 2 = 0

i.e. u = −2 = x + y − 1

∴ y = −1 − x

Ans: The sol.s of y are

x + c = −

(x + y − 1 − 2 ln |x + y + 1|) or y = −1 − x

Ex. (x − y + 3)y

− (x + y − 7) = 0 Sol.

y

=

Set X = x + h , Y = y + k

and then  −h − k − 7 = 0

−h + k + 3 = 0

We find h = −2 , k = −5. So y

= Y

=

=

Y X

1−_X^Y

Let

= u , Y

= u

X + u

⇒ u

X + u =

, Xu

=

⇒ X

=

= 1 + u

Then either u + 1 = 0 or u + 1 6= 0 i) u + 1 6= 0 :

du

u+1

=

⇒ ln |u + 1| = ln |cX|

⇒ u + 1 = ±cX

⇒ u = −1 ± cX

⇒

= −1 ± c(x − 2)

⇒ y = −(x − 2) ± c(x − 2)

+ 5

= −(x − 7) ± c(x − 2)

· · · (a) ii) u + 1 = 0 :

u = −1

⇔

= −1

⇔ y = −x + 7 . (a P.S. of (a) ) The sol. y(x) = −(x − 7) ± c(x − 2)

§Orthogonal Trajectories

figure.

Def.Let F (x, y, c) = 0 · · · (A) be a given one- parameter family of curves in the xy-plane.

A curve that intersects the curves of the family (A) at right angle is called an ”orthogonal trajectory” of (A) .

F (x, y, c) = 0 ↔ D.E.

= f (x, y)

Then

of g(x, y, ¯ c) = 0 is

= −

(∵ y

· y

= −1), which means

g(x, y) = 0 is orthogonal to F (x, y, c) = 0 Procedure for finding the orthogonal trajectories for a given family of curves

F (x, y, c) = 0 :

i)Find the corresponding D.E. of (A):

dy

dx

= f (x, y)

ii)Replace f (x, y) by −

:

dy

dx

= −

· · · (B)

which is the D.E. of the orthogonal trajectories.

iii)Find the G.S. of (B), which is the desired family of orthogonal trajectories.

Ex.Find the orthogonal trajectories of y

= cx · · · (A)

Sol.

i)Find the D.E. of (A):

∵ y

= cx ⇒ 2yy

= c · · · (B) Plug (B) into (A):

y

= (2yy

)x = 2xyy