18MAB201T-U1-Week2.pdf

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18MAB201T-U1-Week2-Lecture Notes

18MAB201T-Transforms and Boundary Value

Problems

Prepared by

Dr. S. ATHITHAN

Assistant Professor

Department of of Mathematics

Faculty of Engineering and Technology

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Unit-1

PARTIAL DIFFERENTIAL EQUATIONS

TOPICS:

? Formation of partial differential equation by eliminating arbitrary constants

? Formation of partial differential equation by eliminating arbitrary functions

? Formation of partial differential equation by eliminating arbitrary functions of the form φ(u, v) = 0.

? Solution of standard types of first order equations

? Reducible to standard type

? Lagrange’s linear equation: Method of grouping, method of multipliers

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Transforms & Boundary Value Problems S. ATHITHAN

1

Solution of PDE’s by Standard and Derived Types

Standard Types

Type 1: f(p, q) = 0

Type 2: z = px+qy+f(p, q)[Clairaut’s Form]

Type 3: f(z, p, q) = 0

Type 4: f(x, p) = g(y, q)

Derived Types

Type 5: f(xmp, ynq) = 0andf(z, xmp, ynq) = 0

Type 6: f(zmp, zmq) = 0andf(x, zmp) = g(y, zmq)

Example: 1. Solve:p2 +q2 = 5

Hints/Solution:

Givenp2 +q2 = 5 − − − −−> (1). This equation is inType-1.

Let us assume thatz =ax+by +c− − − −−> (2)be a solution of (1). Differentiating (1) partially w.r.t. xandy, we get

∂z

∂x =p= aand ∂z

∂y = q =b.

Put this in (1), we geta2 +b2 = 5 =⇒ b =±p5−a2_.

Now, we have the complete integral as asz = ax±yp5−a2_+c.

Putb = φ(a)in (1),z =ax+φ(a)y+a2−[φ(a)]2 − − − −(2).

Differentiating w.r. toaandyand equating to zero, we get ∂z

∂a =x±

1

2√5−a2(−2a) = 0

and ∂z

∂c = 1 6= 0.

Hence there is no singular integral in this type.

Example: 2. Solve:z =px+qy +p2−q2.

Hints/Solution:

This equation is in Clairaut’s form. Hence complete solution isz = ax +by +a2 −b2 − − − −− > (1)

Differentiating (1) partially w.r.t. aandb, we get a =−x

2 andb =

y

2.

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Putb = φ(a)in (1),z =ax+φ(a)y+a2−[φ(a)]2 − − − −(2).

Differentiating w.r. toa,0 = xφ0(a) + 2a−2φ(a)φ0(a)− − − −(3). Eliminatingabetween (2) and (3) we get the general solution.

Example: 3. Solve:z2 =p2+q2 + 1

Hints/Solution:Givenz2 = p2 +q2+ 1− − − −− > (1)

Letz =f(x+ay)be the solution of (1). Putx+ay = u =⇒ z = f(u).

∴ p = dz

du, q =a dz

du − − − −−> (2)

Substituting (2) in (1) and simplifying and integrating we get

dz du =

√

z2₋₁

√

1 +a2

Z _dz

√

z2 ₋₁ =

Z _du

√

1 +a2

cosh−1z = √ 1

1 +a2u+b

cosh−1z = √ 1

1 +a2(x+ay) +b

Example: 4. Solve:p2 +q2 =x+y

Hints/Solution:From the given equation we have

p2−x =y−q2 =k

.

=⇒ p =±px+k, q =±py−k

z =

Z

p dx+

Z

q dy

= 2

3(x+k)

3/2 ₊ 2

3(y−k)

3/2 _+c

which is the required complete integral (C.I.).

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Transforms & Boundary Value Problems S. ATHITHAN Type 5: f(xmp, ynq) = 0andf(z, xmp, ynq) = 0

Case 1: Ifm6= 1andn 6= 1, then putx1−m = X andy1−n =Y Now, ∂z

∂x =p = ∂z ∂X ·

∂X

∂x = P ·(1−m)x

−m

i.e. xmp = (1−m)P Similarly we have ynq = (1−n)Q

Substituting these in given equation we will havef(P, Q) = 0[Type 1]andf(z, P, Q) = 0

[Type 3]which can be solved by the standard methods.

Case 2: Ifm=n = 1, then putlogx= X andlogy = Y Now,

∂z

∂x =p = ∂z ∂X ·

∂X

∂x =P ·

1

x

i.e. xp = P Similarly we have yq = Q

Substituting these in given equation we will havef(P, Q) = 0[Type 1]andf(z, P, Q) = 0

[Type 3]which can be solved by the standard methods.

Example: 5. Solve:p2x+q2y =z.

Hints/Solution:Herem =n = 1/2.

By using the simplification given above this example, our equation reduced to

P2+Q2 = 4z which is ofType 3

By usual solving, we get2√z = √ 2

1 +a2(x

1/2 _+ay1/2_{) +b}

Type 6: f(zmp, zmq) = 0andf(x, zmp) = g(y, zmq)

Case 1: Ifm6=−1, then putZ = z1+mNow,

∂Z

∂x =P = ∂Z

∂z · ∂z

∂x = (1 +m)z m _·p

i.e. zmp = P (1 +m)

Similarly we have zmq = Q (1 +m)

Substituting these in given equation we will have f(P, Q) = 0[Type 1] and f(x, P) =

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Case 2: Ifm=−1, then putZ = logzNow,

∂Z

∂x =P = ∂Z

∂z · ∂z ∂x =

1

z ·p

i.e. z−1p = P Similarly we have z−1q = Q

Substituting these in given equation we will have f(P, Q) = 0[Type 1] and f(x, P) =

g(y, Q)[Type 4]which can be solved by the standard methods.

Example: 6. Solve:(zp+x)2 + (zq+y)2 = 1.

Hints/Solution:Herem = 1.

By using the simplification given above this example, our equation reduced to

_P

2 +x

2

+

_Q

2 +y

2

= 1 which is ofType 4

By usual solving, we getz2 = 2ax−x2+ 2(p1−a2_)y−y2 _+b

2

Solution of Langrange’s Linear Equations

Example: 7. Find the general solution ofx(z2−y2)p+y(x2−z2)q =z(y2−x2).

Hints/Solution:

The given equation is of the form Lagrange’s linear equationP p+Qq =R.

∴The subsidiary equations are

dx

x(z2 _−y2₎ =

dy

y(x2_−z2₎ =

dz

z(y2 _−x2₎ − − − −(1)

Using the multipliersx, y, z and 1 x,

1

y,

1

z in (1), we get

xdx+ydy+zdz = 0and 1 xdx+

1

ydy+

1

zdz = 0Integrating, we getx

2_+y2_+z2 _=a

andxyz =b.

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Transforms & Boundary Value Problems S. ATHITHAN

3

Solution of Homogeneous Equations

3.1

Linear Partial Differential Equations of Second Order with Constant

Co-efficients

An equation of the form

a0 ∂n_z ∂xn +a1

∂n_z

∂xn−1_∂y +a2

∂n_z

∂xn−2_∂y2 +· · ·+an ∂n_z

∂yn = F(x, y)

wherea0, a1, . . . , an are constants, is said to be aLINEAR PARTIAL DIFFERENTIAL EQUA

-TIONof degree ‘n’ with constant coefficients.

Let ∂

∂x =D, ∂2

∂x2 = D

2_{, . . . ,} ∂ n

∂xn = D n _and

∂

∂y =D

0

, ∂ 2

∂y2 =D

02

, . . . , ∂ n

∂yn =D

0n .

Then the above equation can be written as

(a0Dn+a1Dn−1D0 +a2Dn−2D02+· · ·+an−1DD0n−1 +anD0n)z = F(x, y)

φ(D, D0)z =F(x, y) (1) ThegeneralorComplete solutionof (1) consists of two parts namely theComplementary Function(C.F.) andParticular Integral(P.I.). i.e.The General Solution is

z = C.F.+P.I. =zc+zp

3.1.1 Complementary Function

Definition 3.1(Complimentary Function). The general solution ofφ(D, D0) = 0is called as Complementary Function and it id denoted byzc.

Definition 3.2 (Auxiliary Equation). An equation of the form φ(m) = 0 is called as an Auxiliary Equation.

Depends on the roots of the polynomialφ(m) = 0,i.e.the roots of the auxiliary equation, we have the following cases to write the Complimentary Function.

Case 1:

The roots of the auxiliary equation are real and distinct, then we write the roots ofφ(m) = 0

asm1andm2 and the C.F. becomes,zc = f1(y+m1x) +f2(y+m2x).

Generalized form of C.F. in this case: If m1, m2, . . . , mn be the real and distinct roots of the auxiliary equation, then the C.F. becomes, zc = f1(y+m1x) + f2(y+m2x) +

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Case 2:

The roots of the auxiliary equation are real and equal, then we write the roots ofφ(m) = 0as m1 = m2 =mand the C.F. becomes,zc =f1(y+mx) +xf2(y+mx).

Generalized form of C.F. in this case: Ifm1 =m2 = · · · = mn(= m), then the C.F. be-comes,zc =f1(y+mx)+xf2(y+mx)+x2f3(y+mx)+· · ·+x(n−2)fn−1(y+mx)+ x(n−1)fn(y+mnx).

3.1.2 Particular Integral

Definition 3.3(Particular Integral). The general solution ofφ(D, D0)z = F(x, y)is called as Particular Integral and it is denoted byzp.

Based on the function on the RHS of the equation φ(D, D0)z = F(x, y), i.e. based on F(x, y), the following cases can be considered while evaluating the Particular Integral (P.I.).

Let us see some short cut methods of evaluating P.I. whenF(x)is of the following form:

a. F(x, y) = eax+by

b. F(x, y) = sin(ax+by)or cos(ax+by)

c. F(x, y) = xmyn (polynomial function)

d. F(x, y) = eax+byχ(x, y), whereχ(x, y) = xmynor sin(ax+by)or cos(ax+by)

e. F(x, y) = xmynχ(x, y), whereχ(x, y) = sin(ax+by)or cos(ax+by)

Rule 1: F(x, y) = eax+by We know that

zp =

1

φ(D, D0₎F(x, y)

Now,

zp = 1

φ(D, D0₎e

ax+by

= 1

φ(a, b)e

ax+by

, ifφ(a, b)6= 0 Directly replaceDbyaandD0 byb Ifφ(a, b) = 0, useRule 4or use the suitable note given underRule 4..

RULE2: F(x, y) = sin(ax+by)or cos(ax+by) We know that

zp =

1

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Transforms & Boundary Value Problems S. ATHITHAN Let us consider φ(D, D0) = ψ(D2, DD0, D02)(i.e. considering only the quadratic part), then the above equation becomes

yp =

1

ψ(D2_{, DD}0_{, D}02₎sin(ax+by) (or) cos(ax+by)

= 1

ψ(−a2_,−ab,−b2₎sinax(or) cosax

ReplaceD2, DD0, D02 by−a2,−ab,−b2providedψ(−a2,−ab,−b2) 6= 0

Ifψ(−a2,−ab,−b2) = 0, useRule 4or use the suitable note given underRule 4.

RULE3: F(x, y) = xmyn, m, n ∈_Z+ We know that

zp =

1

φ(D, D0₎F(x, y)

= 1

φ(D, D0₎x

m_yn

Now, taking the lowest degree term (may be constant term) and write the the denominator in the form of[1 +φ(D, D0)], then we have

zp =

1

[1 +φ(D, D0_)]F(x, y)

= [1 +φ(D, D0)]−1xmyn

Expanding this term usingBINOMIAL EXPANSIONand hence we get the desiredzp.

Note:

Here 1

D denotes the integration w.r. toxand

1

D0 denotes the integration w.r. toy.

Note:

Inxmyn, ifm < nthen try to writeφ(D, D0)asφ

_D

D0

and ifm > nthen try

to writeφ(D, D0)asφ

_D0

D

.

Few important Binomial Expansions:

1. (1−x)−1 = 1 +x+x2+. . . 2. (1 +x)−1 = 1−x+x2−x3+. . .

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4. (1 +x)−2 = 1−2x+ 3x2 −4x3 +. . .

5. (1−x)−3 = 1 + 3x+ 6x2 +. . . 6. (1 +x)−3 = 1−3x+ 6x2 −. . .

RULE4:F(x, y)is any other function [different function from Rule 1, Rule 2 and Rule 3], then resolveφ(D, D0)into linear factors in the form(D−m1D0)(D−m2D0). . .(D−

mnD0)and theP.I.= zpis given by

zp =

1

φ(D, D0₎F(x, y)

= 1

(D−m1D0)(D−m2D0). . .(D−mnD0)

F(x, y)

= 1

(D−m1D0)(D−m2D0). . .(D−mn−1D0)

Z

F(x, c−mnx)dx,

wherey =c−mnx ..

.

In case of second order, we have

zp =

1

φ(D, D0₎F(x, y)

= 1

(D−m1D0)(D−m2D0)

F(x, y)

= 1

(D−m1D0)

Z

F(x, c−m2x)dx, wherey =c−m2x

.. .

Note 3.4. If the denominator is zero inRule 1andRule 2, then applyRule 4.

Note 3.5. If the denominator is zero inRule 1andRule 2, then find the partial derivative of the denominator (Dr) w.r.toDand multiplying the numerator (Nr) byx. Continue this process till we get the non-zero Dr.

Note 3.6. If the denominator is zero and the factors of Dr is repeated inRule 1 andRule 2, then

zp =

1

(bD−aD0₎nF(x, y)

= x

n

bn_n!F(x, y)

Example: 8. Solve:(D2 −4DD0+ 4D02)z = 0

Hints/Solution:

The auxiliary equation ism2−4m+ 4 = 0 =⇒ m= 2,2.

∴C.F.zc = f1(y+ 2x) +xf2(y+ 2x).

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Transforms & Boundary Value Problems S. ATHITHAN

Example: 9. Solve: ∂

3_z

∂x3 −7 ∂3z ∂x∂y2 −6

∂3z ∂y3 = x

2_{y+ sin(x+ 2y) +e}4x+y

Hints/Solution:

Given equation is of the form(D3−7DD02 −6D03)z =x2y+ sin(x+ 2y) +e4x+y The auxiliary equation ism3−7m−6 = 0 =⇒ m= −1,−2,3.

∴C.F.zc = f1(y−x) +xf2(y−2x) +f3(y+ 3x)

(P.I.)1 =

1

D3 _−7DD02 _−6D03x 2_y

= 1

D3

1−

_7D02 D2 +

6D03 D3

−1

(x2y)

= 1

D3(x

2_{y) =} 1

60(x

5_y)

(P.I.)2 =

1

D3 _−7DD02 _−6D03 sin(x+ 2y)

= 1

−D+ 28D+ 24D0 sin(x+ 2y)

= 1 3 ·

9D−8D0

81D2_−64D02 sin(x+ 2y) = −

1

75cos(x+ 2y).

(P.I.)3 =

1

D3 _−7DD02 _−6D03e (4x+y)

= 1

64−28−6e

(4x+y)

= 1

30e

(4x+y)_.

Hence the complete solution is given by

z = C.F.+P.I.= zc+zp =f1(y−x) +xf2(y−2x) +f3(y+ 3x) + 1 60(x

5 y)−

1

75cos(x+ 2y) + 1

30e

(4x+y) .

Example: 10. Solve:(D3 −D2D0 −DD02 −D03)z = 3 sin(x+y)

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∴C.F.zc = f1(y−x) +xf2(y−x) +f3(y+x)

(P.I.) = 1

D3 _−D2_D0 _−DD02_−D033 sin(x+y) ReplacingD2 by-1,DD0 = D02 = −1

= 1

−D−D0_+D+D03 sin(x+y) =

1

03 sin(x+y) (ReplacingD2 by -1,DD0 =D02 =−1)

Since the denominator become zero in this case we can applyNote3.4underRule 4. ApplyingRule 4, we have following steps.

= 1

(D−D0_)(D+D0_)(D+D0₎3 sin(x+y)

= 3 1

(D−D0_)(D+D0₎

Z

sin(x+c1 +x)dx [Replacingybyc1 +x]

= 3 1

(D−D0_)(D+D0₎

−cos(2x+c1)

2

= −3

2

1

(D−D0_)(D+D0₎cos(x+y) [By puttingc1 =y−x]

= −3

2 1

(D−D0₎

Z

cos(x+c2 +x)dx [Replacingybyc2 +x]

= −3

2 1

(D−D0₎

_sin(2x+c

2)

2

= −3

4 1

(D−D0₎sin(x+y) [By puttingc2 =y−x]

= −3

4

Z

sin(x+c3−x)dx [Replacingybyc3 −x]

= −3x

4 sin(x+y) [By puttingc3 = y+x]

Aliter:

Since the denominator become zero in this case we can applyNote3.5underRule 4.

(P.I.) = 1

D3 _−D2_D0 _−DD02_−D033 sin(x+y) ReplacingD2 by-1,DD0 = D02 = −1

= 1

−D−D0_+D+D03 sin(x+y) =

1

03 sin(x+y) (ReplacingD2 by -1,DD0 =D02 =−1)

Since the denominator become zero in this case we can applyNote3.5underRule 4.

= 1

3D2 _{+ 2DD}0 _−D023 sin(x+y)

(Again replacingD2by -1,DD0 =D02 =−1, we get) = −3x

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Transforms & Boundary Value Problems S. ATHITHAN Example: 11. Solve:(D3 −D2D0 −DD02 −D03)z = 3 sin(x+y)

Hints/Solution:The auxiliary equation ism3 −m2−m−1 = 0 =⇒ m=−1,1,1.

∴C.F.zc = f1(y−x) +xf2(y−x) +f3(y+x)

(P.I.) = 1

D3 _−D2_D0 _−DD02_−D033 sin(x+y) ReplacingD2 by-1,DD0 = D02 = −1

= 1

−D−D0_+D+D03 sin(x+y) =

1

03 sin(x+y) (ReplacingD2 by -1,DD0 =D02 =−1)

Since the denominator become zero in this case we can applyNote3.4underRule 4. ApplyingRule 4, we have following steps.

= 1

(D−D0_)(D+D0_)(D+D0₎3 sin(x+y)

= 3 1

(D−D0_)(D+D0₎

Z

sin(x+c1 +x)dx [Replacingybyc1 +x]

(P.I.) = 3 1

(D−D0_)(D+D0₎

−cos(2x+c1)

2

= −3

2

1

(D−D0_)(D+D0₎cos(x+y) [By puttingc1 =y−x]

= −3

2 1

(D−D0₎

Z

cos(x+c2 +x)dx [Replacingybyc2 +x]

= −3

2 1

(D−D0₎

_sin(2x+c

2)

2

(P.I.) = −3

4 1

(D−D0₎sin(x+y) [By puttingc2 =y−x]

= −3

4

Z

sin(x+c3−x)dx [Replacingybyc3 −x]

= −3x

4 sin(x+y) [By puttingc3 = y+x]

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(P.I.) = 1

D3 _−D2_D0 _−DD02_−D033 sin(x+y) ReplacingD2 by-1,DD0 = D02 = −1

= 1

−D−D0_+D+D03 sin(x+y) =

1

03 sin(x+y) (ReplacingD2 by -1,DD0 =D02 =−1)

Since the denominator become zero in this case we can applyNote3.5underRule 4.

= 1

3D2 _{+ 2DD}0 _−D023 sin(x+y)

(Again replacingD2by -1,DD0 =D02 =−1, we get) = −3x

4 sin(x+y)

(P.I.)2 =

1

D3 _−7DD02 _−6D03 sin(x+ 2y)

= 1

−D+ 28D+ 24D0 sin(x+ 2y)

= 1 3 ·

9D−8D0

81D2_−64D02 sin(x+ 2y) = −

1

75cos(x+ 2y).

(P.I.)3 =

1

D3 _−7DD02 _−6D03e (4x+y)

= 1

64−28−6e

(4x+y)

= 1

30e

(4x+y) .

Hence the complete solution is given by

z = C.F.+P.I.=zc+zp =f1(y−x) +xf2(y−x) +f3(y+x)−

3x

4 sin(x+y).

Example: 12. Solve:zxx−4zxy+ 4zyy = e2x+y

Hints/Solution:

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Transforms & Boundary Value Problems S. ATHITHAN

∴C.F.zc = f1(y+ 2x) +xf2(y+ 2x)

(P.I.) = 1

D2 _−4DD02_{+ 4D}02e 2x+y

ReplacingDby 2,D0by 2, we get

= 1

4−8 + 4e

2x+y

= 1 0e

2x+y

Since the denominator become zero in this case we can applyNote3.4underRule 4. ApplyingRule 4, we have following steps.

= 1

(D−2D0_)(D−2D0₎e

2x+y

= 3 1

(D−2D0₎

Z

e2x+c1−2x_{dx [Replacingybyc}

1−2x]

= 1

(D−2D0₎e

2x+y _·

x dx [By puttingc1 = y+ 2x]

=

Z

e2x+c2−2x _{·x dx [Replacingybyc}

2−2x]

= x

2

2 e

2x+y _{[By puttingc}

2 = y+ 2x]

Aliter-1:

Since the denominator become zero in this case we can applyNote3.5underRule 4.

(P.I.) = 1

D2 _−4DD02_{+ 4D}02e 2x+y

ReplacingDby 2,D0by 2, we get

= 1

4−8 + 4e

2x+y ₌ 1

0e

2x+y

Since the denominator become zero in this case we can applyNote3.5underRule 4.

= x

2D−4D0e

2x+y

(Again replacingDby 2,D0by 2, we get)

= x

0e

2x+y

= x

2

2 e

2x+y

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(P.I.) = 1

D2 _−4DD02_{+ 4D}02e 2x+y

ReplacingDby 2,D0by 2, we get

= 1

4−8 + 4e

2x+y ₌ 1

0e

2x+y

Since the denominator become zero in this case we can applyNote3.6underRule 4. Herea = 2, b= 1andn = 2

= x

n

1n_·n!e 2x+y

= x

2

2 e

2x+y

Hence the complete solution is given by

z = C.F.+P.I.= zc+zp =f1(y+ 2x) +xf2(y+ 2x) + x2

2 e

2x+y_.

Example: 13. Solve:r −s−2t = (y−1)ex

Hints/Solution:Given equation is of the form(D2−DD0 −2D02)z = (y−1)ex The auxiliary equation ism2−m−2 = 0 =⇒ m= −1,2.

∴C.F.zc = f1(y−x) +f2(y+ 2x)

Since RHS is different kind of function which is not comes underRule 1orRule 2orRule 3, here we need to applyRule 4

(P.I.) = 1

D2 _−DD02_−2D02(y−1)e x

ApplyingRule 4, we have following steps.

= 1

(D−2D0_)(D+D0₎(y−1)e

x

= 1

(D−2D0₎

Z

(c1 +x−1)exdx [Replacingybyc1 +x]

= 1

(D−2D0₎(y−2)e

x_{dx [After}

Z n

by puttingc1 =y−x]

=

Z

(c2 −2x−2)exdx [Replacingybyc2−2x]

= yex [After

Z n

by puttingc2 = y+ 2x]

Hence the complete solution is given by

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Transforms & Boundary Value Problems S. ATHITHAN

4

Exercise/Practice/Assignment Problems

1. Solve the following partial differential equations (pde’s)

(a) 3p2 −2q2 = 4pq. Ans: z = ax±a

" −1 + √ 10 2 #

y+c

(b) p+q = pq. Ans:z = ax+ ay

a−1 +b

(c) √p+√q = 1. Ans:z =ax− ay

a+ 1 +c

(d) p(1−q2) = q(1−z). Ans: 2 a

p

az + (1−a) = 1

a(x+ay) +b

(e) p2 +pq = z2. Ans:logz = √ 1

1 +a(x+ay) +b

(f) z =p2 +q2. Ans: C.I.: 4(1 +a2)z = (x+ay+b)2, S.I.: z = 0.

(g) z =px+qy+ p

q −p. Ans: C.I.: z =ax+by+

a

b −a, S.I.:z = y

1−x.

(h) (1−x)p+ (2−y)q = 3−z. Ans: C.I.: z = ax+by−a−2b+ 3, S.I.: z = 3.

(i) z =px+qy+p2q2. Ans: C.I.: z = ax+by+a2b2, S.I.:z3 = −27 16 x

2_y2_.

(j) yp= 2yx+ logq. Ans: C.I.: z = (x2+kx) + e

ky

k +C

(k) √p+√q = 2x. Ans: C.I.: z = 1

6(2x+a)

3

+ay+c

(l) p+x =qy. Ans: C.I.: z =a(x+ logy)− x

2

2 −c

(m) z2 =xypq. Ans: C.I.: logz = √1

a(logx+alogy) +c

(n) 4z2q2 =y+ 2zp−x. Ans: C.I.: z2 = 2

3(y+a)

2/3₊ 1

2(x+a)

2 _+c

(o) z2(p2 + q2) = x2 + y2. Ans: C.I.: z2 = xpx2 _+a2 _{+ y}p_y2 _+a2 ₊

a2 sinh−1 _x a

−cosh−1 _y

a

+c

(p) (y2+z2)p =xyq−xz. Hint: Use the multipliers sets(0,1,0),(0,0,1)and

(x, y, z) Ans: f

_y

z, x

2

+y2 +z2

= 0

(q) z(xp+yq) = y2 −x2. Hint: Use the multipliers sets(1,0,0),(0,1,0)and

(x, y, z) Ans: f

_x

, x2 −y2 +z2

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(r) x(y2 +z)p− y(x2 +z)q = z(−y2 +x2). Hint: Use the multipliers sets

(x, y,−1)and

₁ x, 1 y, 1 z

Ans: f x2 +y2−2z, xyz = 0

(s) (3z−4y)p+ (4x−2z)q = 2y−3x. Hint: Use the multipliers sets(x, y, z) and(2,3,4) Ans: f x2+y2 +z2,2x+ 3y+ 4z

= 0

(t) (y+z)p+(z+x)q = x+y. Hint: Use the multipliers sets(1,−1,0),(0,1,−1)

and(1,1,1) Ans: f

_y−z

x−y,(x−y)

p

x+y+z

= 0

(u) (mz−ny)p+ (nx−lz)q =ly−mx. Hint: Use the multipliers sets(x, y, z) and(l, m, n) Ans: f x2 +y2 +z2, lx+my +nz

= 0

2. Solve the following pde’s (In all the following problemsDandD0represents the operator ∂

∂x and ∂

∂y respectively)

(a) (D3 −7DD02 − 6D03)z = x2y + sin(x+ 2y) Ans: z = f1(y −x) +

xf2(y−2x) +f3(y+ 3x) +

1

60(x

5_y)− 1

75cos(x+ 2y).

(b) ∂ 2_z

∂x2−5 ∂2z ∂x∂y+ 6

∂2z

∂y2 = (D

2_−5DD0

+ 6D02)z =r−5s+ 6t =e(x+2y)

Ans:z = f1(y+ 2x) +f2(y+ 3x) +

1

2e

(x+2y)_.

(c) (D4−2D3D0+ 2DD03−D04)z =e(2x+3y) Ans:z =f1(y−x) +f2(y+

x) +xf3(y+x) +x2f4(y+x)−

1

5e

(2x+3y)

.

(d) r+t = cos 2xcos 2y Ans: z = f1(y+ix) +f2(y−ix)−

1

16[cos(2x+ 2y) + cos(2x−2y)].

(e) r−2s= x3y+e2x Ans:z =f1(y) +f2(y+ 2x) +

e2x

4 +

(x5_y)

20 +

x6

60. (f) r −6s+ 5t = zxx −6zxy + 5zyy = exsinhy +xy Ans: z = f1(y) +

f2(y+ 2x) +

e2x

4 +

(x5_y)

20 +

x6

60.

(g) (D3 +D2D0 − DD02 − D03)z = 3 sin(x+ y) Ans: z = f1(y −x) +

xf2(y−x) +f3(y+x)−

3x

4 sin(x+y).

(h) (D2 +DD0 −6D02)z = ycosx Ans: z = f1(y−3x) +f2(y + 2x) +

sinx−ycosx.

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Transforms & Boundary Value Problems S. ATHITHAN Acknowledgement:

Some of the portions of this material are derived from various sources. I thank the authors for those books and related materials.