MAHALAKSHMI ENGINEERING...

Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics

MAHALAKSHMI ENGINEERING COLLEGE,TRICHY.

PARTIAL DIFFERENTIAL EQUATIONS-MA2211

PART-A

1. Form the p.d.e from (x-a)2 + (y-b)

2 + z

2 = r

2 (AU May 2013)

Solution:

Given that (x-a) 2

+(y-b) 2

+z2 = r

2 ---------- (1)

d.p.w.r to x,

2(x-a) + 2z

= 0 [ z is a fun of x and y]

(x-a) + zp = 0 ---------- (2)

d.p.w.r. to y, p=

2 (y-b) + 2z

= 0 q=

= (y-b) + zq = 0 ---------- (3)

Eliminating a and b from 1, 2 and 3

2 x – a = -zp

3 y – b = -zq

(1) (-zp) 2

+(-zq) 2

+z2 = r

2

z2p

2 + z

2q

2 + z

2 = r

2

z2 (p

2+q

2+1)=r

2

which is the required p.d.e

2. Find the p.d.e of all spheres having their centres on the z-axis (AU Dec 2012)

Solution:

Let the Centre of the sphere be (0, 0, c) point on the Z –axis and ‘r’ it’s radius.

(x-0) 2

+(y-0) 2

+(z-c) 2

=r2

[Since centre lies on Z axis]

ie, x2+y

2+(z-c)

2 = r

2 ------- 1

d (1) p.w.r. to x, [c&r arbitrary constants]

2x + 2(z-c)

= 0

x + p (z – c) = 0 ---------(2)


d (1) p.w.r. to y,

2y + 2(z-c)

= 0

y + q (z – c) = 0 ---------(3)

From( 2) and (3)

(2) z – c = - x/p

(3) z – c = -y/q

-x/p = - y/q

qx = py, which is the required p.d.e

3. Form the p.d.e by eliminating the constants a and b from z= (x2+a

2) (y

2+b

2)

Solution: AU Dec 2010

G.T. z=(x2+a

2) (y

2+b

2) ----------- (1)

d (1) p w.r to x

= p = 2x (y

2+b

2)

= y

2 + b

2 ----------- (2)

d (1) p w.r to y,

= q = 2y (x

2+a

2)

= x

2 + a

2 ----------- (3)

Substitute

(2) & (3) in (1)

z =

pq = 4xyz

4. Eliminate the arbitrary function ‘f’ from z = f(y/x) and form a p.d.e (AU Dec 2012)

Solution:

Given that z =f(y/x) -----------(1)

d 1 p.w. r to x, p =

= f

1 (y/x) (-y/x

2) ---------- (2)

d 1 p.w.r to y, q =

=f

1(y/x) (1/x) -----------(3)

Now,

=

(

)

(

)

=

=

px = -qy

is, px + qy = 0 is the required p.d.e.


5. Form the p.d.e by eliminating the arbitrary function from z2-xy = f(x/z)

(AU June 2012)

Solution:

G.T z2 – xy = f(x/z) -------- (1)

d (1) p.w.r. to x

2z

-y = f

1 (x/z) ⌈

⌉

2z p-y = f1(x/z) ⌈

⌉ -------------(2)

d (1 ) p.w.r. to y

2z

-x = f

1(x/z) ⌈

⌉

2z q-x = f1(x/z) ⌈

⌉ -------------(3)

(-xq) (2zp-y) = (2zq-x)(z-xp)

-2xzpq + xyq = 2z2q – 2xzpq – xz + x

2p

xyq = 2z2q – xz + x

2p

x2p + 2z

2q – xyq = xz

x2p –(xy – 2z

2)q = xz is the required p.d.e

6. Form the p.d.e of all planes cutting equal intercepts from the x and y axes

(AU Dec 2009)

Solution:

The equation of such plane is

x/a + y/a + z/b = 1 --------- 1 (x and y have equal intercepts)

p.d.w.r. to x

, p= -

-------------------- 2

p.d.w.r. to y


q= -

-------------------- (3)

From (2) and (3)

p = q

p –q = 0 is the required p.d.e.

7. Find the complete integral of p + q – pq (AU May 2013)

Solution:

Given p + q = pq --------------(1)

It is of the form F (p,q) = 0 --------------(2)

Hence the trial saln is

z = ax + by + c ---------------(3)

To get the complete integral of solution 3 we have to eliminate any one of the

arbitrary constants.

Now (3) =>

=> b - ab = - a

=> b(1-a) = -a

=> b = -a / 1-a

Hence the complete soln is

z =

8. Solve : pq = x (AU May 2010)

Solution:

It is of the form f1 (x,p) = f2

(y,q)

Let p/x = 1/q = k

=> p/x = k => p = kx

z = ∫

= ∫ ∫


z=k

9. Solve : (D-D1)3 z = o (AU Dec 2011)

Solution:

The A.E. is (m-1)3=0

m = 1, 1, 1

The C.F = f1 (y+x) + x f2(y+x) + x2f3(y+x)

10. Solve: (D3-2D

2D

1) Z = 0 (AU Dec 2009)

Solution:

The A.E. is m3 – 2m

2 = 0

=>m2 (m-2) =0

= >m =0, 0, 2

z = f1(y+0x) + xf2(y+0x) + f3(y+2x)

= f1(y) + xf2(y) + f3(y+2x)

11. Solve :(D2-7DD

1+6D

12) Z= 0 (AU June 2012)

Solution:

A.E is m2-7m + 6 = 0

=> (m-6) (m-1) = 0

=> m = 6,1

C.F. = f1(y+x) + f2(y+6x)

12. Find the P.I. of (D2-2DD

1+D

12) Z= e

x y (AU Dec 2010)

Solution:

a=1, b= - 1

P.I. =

=

2e

x-y

. =


13. Solve :(D-1) (D-D1+1) z = 0 (AU Dec 2012)

Solution:

It is of the form

(D-m1 D

1-c1) (D-m2D

1-c

2) z=0

Compare with general form the given eqn can be written as

(D-OD1-1) (D-1D

1-(-1))Z=0

Here m1=0, m2=1

C1=1, C2=-1

z = exf1(y+ox)+e

-xf2(y+x)

= exf1(y)+e

-xf2(y+x)

14. Form the p.d.e by eliminating the function from z=f(x+t) + g(x-t)

Solution : (AUN/D 2010)

d.p.w.r. to x, p =

=f

1(x+t)+g

1(x-t) ------------- (1)

d.p.w.r to t, q=

=f

1(x+t)-g

1(x-t) ------------- (2)

=f” (x+t) +g”(x-t) --------------( 3)

=f” (x+t) +g”(x-t) -------------- (4)

15. Form a p.d.e by eliminating the arbitrary

constants a and b from z = (x+a) (y+b) --------(1) (AUM/J 2008)

Solution:

d(1) p.w. r to x,

z/ x = 1(y+b) => p = y+b - ------(2)

d. p.w. r to y

z/ y= (x+a) 1 => q = x+a --------(3)

Sub (2) and (3) in (1), z = qp


16. Solve:(D3-3DD

12+2D

13)z=0 (AU A/M 2010)

Solution:

m3-3m+2 = 0=> m = 1,2

C.F = 1(y+x) + 2 (y+2x)

17. Form the p.d.e by eliminating a and b from z =(x+a)2

+ (y+b)2

(AU A/M 2008)

Solution:

Given that z=(x+a)2+(y+b)

2 --------- (1)

d (1) p.w.r to x z/ x =2(x+a) 1 = p=2(x+a) = P/2 = x+a ---------(2)

d (1) p.w.r. to y z/ y =2(y+b) 1 = q = 2(y+b)= q/2 = y+b --------- (3)

1=> z (p/2)2

+ (q/2)2 = p

2/4 + q

2/4=

=> 4z = p2+q

2

18. Form a p.d.e by eliminating the arbitrary Constants a and b from the equation

(x2+a

2)(y

2+b

2) = z (AU Dec 2010)

Solution:

z=(x2+a

2)(y

2+b

2) -------(1)

d.p.w.r to x

=p=2x(y

2+b

2) --------(2)

d.p.w.r to y,

=q=2y(x

2+a

2) ------- (3)

(2) = >

= y

2+b

2 -- ------(4)

( 3) = >

= x

2+a

2 - - - - (5)

Sub (4) and (5) in (1) we have

z =

z =

4xyz = pq.


PART – B

1. Form the p.d.e by eliminating the arbitrary function from

2 2 2x +y +z ,ax+by+cz 0 (AUC Dec 2010)

Solution:

Given 2 2 2x +y +z ,ax+by+cz =0 ---------1

Let u = 2 2 2x +y +z ---------2

v = ax+by+cz ---------3

Eqn (1) u,v =0 ---------4

Elimination of from (4) gives

u v x x

= 0u v

y y

--------5

u2x+2zp

x

u 2y+2zq

y

v=a+cp

x

v =b+cq

y

--------6

Now (6) in (5)

2x+2zp a+cp

=02y+2zq b+cq

2x+2zp b+cq - a+cp 2y+2zq =0

2 x+zp b+cq - a+cp 2 y+zq =0

2 x+zp b+cq - a+cp y+zq =0

x+zp b+cq - a+cp y+zq =0

bx+cqx+zpb+zcpq-ay-azq-cpy-cpzq=0


p(zb-cy)+q(cx-az)=ay-bx

2. Solve : (mz-ny)p+(nx- x)q=y-mxl (AUC Apr/May 2010)

Solution:

p qP +Q =R

P=mz-ny, Q=nx-lz, R = y-mxl

The Lagrange’s subsidiary equations are dx dy dz

= =P Q R

ie, dx dy dz

= =mz-ny nx-lz 1y-mx

Using lagrangian multipliers as l, m, n each of ratio is equal to

x+mdy+ndz dx+mdy+ndz=

(mz-ny)+m(nx- z)+n( y-mx) 0

ld l

l l l

dx+mdy+ndz=0l

Integrating , 1x+my+nz=Cl

Choosing another set of multipliers x, y, z

xdx+ydy+zdz xdx+ydy+zdz=

x(mz-ny)+y(nx-lz)+z(1y-mx) 0

xdx+ydy+zdz=0

22 22Cyx z+ + =

2Integrati ,

2 2ng

2

2 2 2

2x +y +z =C

The general solution

2 2 2x +y +z , lx+my+nz =0 ( is arbitrary)


3. Solve: 2 2 2x (y-z)p+y z-x q=z x-y (AUC Dec 2010/June 2012)

Solution :

2 2 2x (y-z)p+y (z-x)q = z (x-y) ---------1

Lagrange’s equation is

p qP +Q =R

Here 2 2 2P x y-z ,Q y z-x ,R z x-y

The S.E is dx dy dz

= =P Q R

2 2 2

dx dy dz= =

x y-z y (z-x) z x-y

Choosing 2 2 2

1 1 1, ,

x y zas multipliers each ratio is equal to

22 2 2 22 2

ddy dydxdx dz + +y x y zx z= = =

y-z z-x x-y y-z+z-x+x-y

2 2 2

dydx dz+ + =0x y z

Integrating we have

2 2 2

1 1 1dx dy dz=0

x y z

-2 -2 -2x dx y dy z dz 0

-2 -2+1

1

x +1 y-2+1 z+ + =C

-2+1 -2+1 -2+1

11 1 1- - - =C

x y z

11 1 1+ + =C

x y z 1 1 1u= + +

x y z

Similary choosing 1 1 1+ +x y z

as Lagrange’s multipliers we get

v=x y z

1 1 1+ + , xyz =0x y z


4. Solve : 2 33 2 1 1 1D +D D +4DD +4D z=cos(2x+y) (AUC Jun’2012)

Solution:

The A.E is m3+m

2+4m+4=0

(m+1) (m2+4) =0

m=-1, m = ±2i

synthetic division

1

1 2 3

1 1 4 4

0 -1 0 -4 C.F= y-x y+2ix y-2ix

1 0 4 0

Now,

2 33 2 1 1 1

1P.I=

D +D D +4DD +4cos

D2x+y

1 1

1=

-4D-4D -4D-cos

4D2x+y D

3is D

2.D

=1

1 1- Cos(2x+y)8 D+D

2 12 1Re D 4, 1, 2place D DD

1

1 D Cos (2x+y)-8 D(D+D )

(X and ÷ by D)

2 1

1 D Cos (2x+y)=-

8 D +DD

1 -2 sin (2x+y)=-

8 -4-2

2 sin (2x+y)

48

= 1 sin(2 x y)24

1 2 31z y-x y+2ix y-2ix sin(2x+y)

24


5. Solve the p.d.e. ( ) ( ) ( )x y z p y z x q z x y (AUC Dec 201

Solution :

Lagrange’s type Pp+Qq=R

The S.E is dydx dz= =P Q R

P x(y-z)

Q y(z-x)

R z(x-y)

dx dy dz

= =x(y-z) y(z-x) z(x-y)

Choosing 1,1,1 as lagrange’s multipliers, each of above ratio is equal

to

dx+dy+dz dx+dy+dz

=xy-xz+yz-yx+zx-zy 0

dx+dy+dz=0

Integrating, 1d(x+y+z)=0 x+y+z =c u=x+y+z

choosing 1 1 1

, ,x y z

as Lagrange’s multipliers

1 1 1 1 1 1dx+ dy+ dz dx+ dy+ dz

x y z x y z=

y-z+z-x+x-y 0

dydx dz+ + =0x y z

Integrating, 2log x +log y +log z =log C

2

2

log(xyz)=logC

(xyz)=C v=xyz

x+y+z, xyz =0


6. Solve: x-2z + 2z-y =p q y-x

Solution:

Pp+Qq=R

The equation is of the form Pp+Qq=R

P = x-2z, Q = 2z-y, R=y-x

The S.E dx dy dz

= =P Q R

ie , dx dy dz

= =x-2z 2z-y y-x

Using multipliers as 1,1,1

Each ratio = dx+dy+dz dx+dy+dy

=x-2z+2z-y+y-x 0

ie, dx+dy+dx=0 Integrating 1x+y+z =C u = x+y+z

Next, using multipliers as y, x, 2z

Each ratio = ydx+xdy+2zdz

yx-2yz+2xz-xy+2yz-2xz

ie ydx+xdy+2zdz

0

ie, ydx+xdy+2zdz=0

d(xy)+2zdz =0

integrating , 2

22zxy+ =C

2

2 2

2xy+z =C v = z xy+

2(x+y+z, xy+z )=0


7. 2 2 2x yz p y zx q z xy (AUC May 2010)

Solution:

The equation is of the form Pp+Qq=R

P = 2 2 2x -yz,Q=y -zx,R=z -xy

Lagrange’s subsidiary equations are

dydx dz= =P Q R

ie,2 2 2

dx dy dz= =

x -yz y -zx z -xy

Using lagrange’s multipliers x,y,z we have

3 3 3 2 2 2

xdx+ydy+zdz dx+dy+dz=

x +y +z -3xyz x +y +z -xy-yz-zx

2 2 2

xdx+ydy+zdz

x+y+z (x +y +z -xy-yz-zx)

2 2 2

dx+dy+dz

x +y +z -xy-yz-zx

xdx+ydy+zdz dx+dy+dz

=x+y+z 1

xdx+ydy+zdz=(x+y+z)(dx+dy+dz)

Integrating

222 2 x+y+zyx z+ + =

2 2 2 2

2 2 2 2 2 2x +y +z =x +y +z +2xy+2yz+2zx

2(xy+yz+zx)=0

1u=xy+yz+zx=C

u (x,y,z) = xy+yz+zx -------------1


Now,

2 2 2 2

dx-dy dy-dz=

x -yz - y -zx y -zx - z -xy

2 2

d y-zdx-dy=

y+z y-z +x y-zx -y +z(x-y)

d(x-y) d(y-z)

=x-y (x+y+z) (y-z)(x+y+z)

d(x-y) d(y-z)

x-y y-z

2log(x-y)=log(y-z)+logC

2

x-ylog =logC

y-z

2

x-y=C

y-z, ie,

x-yv=

y-z

The general solution is , 0x y

xy yz zxy z

8. Solve : 2 2 2 2 2 2x(y -z )p+y(z -x )q=z x -y (AUC May 2013)

Solution:

It is of the form Pp+Qq=R

Here 2 2 2 2 2 2P=x y -z , Q=y z -x , R= z (x -y )

The S.E is dydx dz= =p Q R

2 2 2 2 2 2

dx dy dz= =

x(y -z ) y z -x z x -y

Use lagrange’s multipliers x,y,z

Each ratio 2 2 2 2 2 2 2 2 2

xdx+ydy+zdz xdx+ydy+zdz= =

0x (y -z )+y (z -x )+z x -y


ie, xdx +ydy+zdz =0 integrating, 22 2y ax z+ + =

2 2 2 2

2 2 2x +y +z =a ie, 2 2 2u= x +y +z

Similarly, taking 1 1 1

, ,x y z

as L.M we get

Each ratio = 2 2 2 2 2 2

1 1 1dx dy dz

x y z

y z z x x y

=

1 1 1

0

dx dy dzx y z

ie, 1 1 1

0dx dy dzx y z

, integrating, logx+logy+logz = log b

ie, log (xyz) = logb b=xyz ie, v=xyz

2 2 2(x +y +z ,xyz)=0

9. (D3-2D

2D

1)z = 2e

2x+3x

2y (AUC Dec 2011)

Solution:

The A.E is m3-2m

2 =0 m

2 (m-2) =0

m=0,0, 2

C.F = 1 2 3(y+ox) x (y+ox) (y+2x)

2x+oy

1 3 2 1

2eP.I =

D -2D D

= 2x1 2e

8-0 a=2, b=0

=2xe

4 Replace

1

2

0

D

D

2

2 3 2 1

1P.I = 3x y

D -2D D


2

3 2 13

3

1= .3x y

D -2D DD

D

2

13

3.x

21

yD

DD

-11

2

3

3 2D= 1- x y

D D

1 122

3 1

3 2D 4D= 1+ + +.... x y

D D D

22

3

3 2xx y+

D D

2 2

3 4

3 1= x y +6 (x )

D D

5 6

. 660 360

x xy

5 6x y x= +

60 60

2x 5 6

1 2 3

e x y xz=f y+ox +xf y+ox +f y+2x + + +

4 60 60

10. Solve 3 3

x+2y

3 3

z z2 e 4sin ( )

x x yx y

Solution:

The given equation can be written as

(D3-2D

2D

1) z=e

x+2y+4sin (x+y)

The A.E is m3-2m

2=0 m

2(m-2)=0

m=0,0 (or) m=2

C.F = f1(y) + xf2(y)+f3(y+2x)


P.I= x+2y

3 2 1 3 2 1

1 1e + 4sin (x+y)

D -2D D D -2D D a=1, b=2 m=1, n=1

x+2y

1

1 1= e + 4sin(x+y)

1-4 -D+2D Replace

1

1

2

D

D

2

2

1

1

1

1

1

D

D

DD

x+2y

1

1 4sin(x+y)=- e +

3 -D+2D

x+2y

2 1

1 4D(sin(x+y))=- e +

3 -D +2DD

x+2y1 4cos(x+y)=- e +

3 1-2

x+2y1=- e -4cos(x+y)

3

x+2y

1 2 3

1z=f (y)+xf (y)+f (y+2x)- e -4cos(x+y)

3

11. Solve: (D2+DD

1-6D

12

) z=y cosx (AUC May 2013)

Solution: m2+m-6=0 m =2, -3

C.F= 1 2(y+2x) (y-3x)

P.I = 22 1 12

1 12

2 2

1 1ycosx = ycosx

D +DD -6D DD 6DD 1+ -

D D

=2

-1-11 12 1 1

2 2 2 2 2

1 DD 6D 1 D 6D1+ - ycosx= 1+ - ycosx

D D D D D D

=

21 1

2 2 2

1 D 6D1- - +.... ycosx

D D D

21 1

2 2

1 D 6D= ycosx - ycosx + ycosx

D D D

2 2

1 1 6= ycosx - cosx + (0)

D D D


2

1= ycosx -sinx

D

1 1 1

= (ycosx -sinx) = ysinx+cosxD D D

= -y cos x + sin x

C.F+P.Iz

12. Solve: P (1+q) = qz

Solution:

Give that p(1+q) = qz -------------- 1

It is of the form f (z,p,q) =0

Let u = x+ay

Now u u

=1, =ax y

dz dz

p= =,q=adu du

dz dz dz(1) 1+a =a .z

du du du

dz

1+a =azdu

dz=az-1

dua

dz az-1=

du a

adu =dz az-1

adzdu=

az-1

Integrating on b.s

u = log (az-1) +logc


Hence the complete solution is (since the number of a.c= no.of. I.V)

x+ay=log [c(az-1)]

13. Solve 2 2z=px+qy+ p +q +1 (AUC Dec 2011, May 2013)

Solution:

Given that 2 2z=px+qy+ p +q +1

It is of the form z = px+qy+f(p,q) (Clairaut’s form)

Hence the complete integral is 2 2z=ax+by+ a +b +1

(a and b are arbitrary constants)

To find singular solution:

2 2z=ax+by+ a +b +1 --------1

d (1) p.w.r. to a,

2 2 2 2

a -ao=x+ x=

a +b +1 a +b +1 --------2

d (1) p .w.r. to b,

2 2 2 2

b -bo=y+ y=

a +b +1 a +b +1 --------3

2 22 2

2 2

a +bx +y =

1+a +b

2 2

2 2

2 2

a +b1- x +y =1-

1+a +b

2 2

2 2

11-x -y =

1+a +b

2 2

2 2

11-x -y =

1+a +b ---------i

2 2

2 2

11+a +b =

1-x -y --------ii


(2) 2 2x=-a 1-x -y by (i)

2 2(3) y=-b 1-x -y by (ii)

Now 2 2 2 2

-x -ya= , b=

1-x -y 1-x -y

Substitute in (1)

2 2

2 2 2 2 2 2

-x y 1z= - +

1-x -y 1-x -y 1-x -y by(ii)

= 2 2

2 2

1-x -y

1-x -y

2 2 2 2 2z= 1-x -y z =1-x -y

2 2 2x +y +z =1 is the singular solution

To find the general integral

Put b = (a) in (1),

22z=ax+ (a)y+ 1+a + (a) -------------4

d. (4) p.w.r to ‘a’

1

22

2a+2 (a) (a)o=x+ '(a)y+

2 1+a + (a)

-------------5)\

Eliminate ‘a’ between (4) and (5) we get the general solution

14. Solve : p2+q

2 = x

2 +y

2

Solution :

Give that p2+q

2 = x

2 +y

2

It is of the form F1(x,p) = F2 (y,q)

Let p2-x

2 = y

2-q

2 =a

2 constant

2 2 2 2 2 2 2 2p -x =a p =a +x p= x +a ------------1


2 2 2 2 2 y q a q y a -------------2

pdx qdyz -----------3

Substitute (1) and (2) in (3)

2 2 2 2z= x a dx y -a dy

2 22 2

2 2-1 -1y y -ax x +a ya ax= sin h + + - cos h +b

2 a 2 a2 2

Which is the complete integral

15. Solve :p (1-q2) = q(1-z) (AU Nov/Dec 2009)

Solution:

Given p(1-q2) = q(1-z) -------------1

It is of the form F (z,p,q) = 0

Let z = f (x+ay) be the solution of (1) --------------2

Put x+ay = u u u

1, ax y

(*)

Z = f(u)

z dz u dzp= = . = .1

x du x du

z dz u dzq= = . = .a

y du y du

Using (*) 3

Substituting (3) in (1)

2

2dz dz dz1-a =a 1-z

du du du


2

2

2

2

dz1-a =a(1-z)

du

dza = 1-a+az

du

dz 1 = az+(1-a)du a

dz 1= du

aaz+(1-a)

-1

2 1az+(1-a dz= u+b

a

-1 +12az+ 1-a 1

= x+ay +b-1 a+1 a

2

12az+ 1-a 1

= x+ay +b1 aa

2

2 1az+(1-a)= (x+ay)+b

a a which is complete integral

15.Solve: 2

2 22

2 1 12 2

z(D -5DD +6D )z = ysinx (or) -5 +6 =ysinx

x x

zzy y

Solution:

m2-5m+6=0 (m-3) (m-2) =0 m=3,2

1 2C.F= (y+2x)+ (y+3x)

P.I. = 22 1 1

1ysinx

D -5DD +6D

= 21 1

2

2 2

1ysinx

5DD 6DD 1- +

D D

= 2

-11 1

2 2

1 5D 6D1- - ysinx

D D D


=

21 1

2 2

1 5 61 ... y sinx

D D

D D D

=

21 1

2 2

1 5D 6Dysinx+ (ysinx)- ysinx

D D D

=2 2

1 5 6ysinx+ sinx- (0)

D D D

21

1D ysin x = sin x

0ysinxD

2

1= ysinx+5(-cosx)

D

1

= -cosx.y-5sinxD

= -ysinx+5cosx

The general solution is

Z =C.F +P.I

17. Solve 2 33 1 1D -7DD -6D z=sin(2x+y)

(AUC May 2013)

Solution:

m3-7 m-6=0

m=-1, -2, 3

1 2 3C.F= (y-x)+ (y-2x)+ (y+3x)

2 31 3 1 1

1P.I = sin(x+2y)

D -7DD -6D D

2by -1

2 =-1

2 32 1 1

sin(x+2y)=

D(D )-7DD -6D

21D by-22 =-4

1 1

sin(x+2y)=

D(-1)-7(-2)D -6(-4)D DD

1by-(1)(2)=-2

1 1

sin(x+2y)=

-D+14D +24D

1

sin(x+2y)=

-D+38D


2

1

2 1

-D-38D= sin(x+2y)

D -1444D

1-D-38D sin(x+2y)=

-1-1444(-4)

1

= -cos(x+2y)-38cos(x+2y).25775

1

=- cos(x+2y)+76cos(x+2y)5775

77=- cos(x+2y)

5775

1=- cos(x+2y)

75

18. Find the singular integral of z=px+qy+p2+pq+q

2 (AUC Dec 2012)

Solution:

Z =px+qy+p2+pq+q

2

It is of the form z=px+qy+f(p,q) (clairaut’s form)

Hence the complete soln is z =ax+by+a2+ab+b

2 -------------1

(since the number of a.c = number of I.V)

To find the singular integral

d.(1). p.w.r.to ‘a’ and ‘b’ we get

0 = x+2a+b -------------2

0 = y+2b+a -------------3

(2) x 1 2a+b+x =0 -------------4

(3)x2 2a+4b+2y=0 -------------5

(4) –(5) -3b+x-2y =0


3b =x-2y b = 2

3

x y

Similarly,

2y(3) 0=y+2 x- +a

3

2x-y y-2x a=- a=

3 3

Substitute a,,b in (1)

2 2

y-2x x-2y y-2x y-2x x-2y x-2yz= x+ y+ + +

3 3 3 3 3 3

Simplifying,

2 2 2 213z= -x -y +xy z= -x -y +xy9 3

2 2 2 23z=-x -y +xy 3z+x +y -xy=0 which is the singular integral.

19.22 1 1 1 y(2D -DD -D +6D+3D )z=xe (AU June 2012’)

Solution:

Given 22 1 1 1(2D -DD -D +6D+3D )=0

1 11D+ D D-D +3 =0

2

The solution of 1 1

1 1 2 2D-m D -α D-m D -α z=0 is

1 2α x α x

1 1 2 2z=e f (y+m x)+e f y+m x

Here 1 1

2 2

1α =0, m = -

2

α =-3, m = 1

ox -3x

1 2

1C.F=e f y- x +e f y+x

2

1 1

D D+O

D D 1

2

y

2 1 1 1

1P.I= xe

2D -DD -D +6D+3D


ox+y

2 1 1 2 1

1=e x

2(D+0) -(D+0)(D +1)-(D +1) +6(D+0)+3(D +1)

2

y

2 1 1 1 1

1=e x

2D -DD -D-D -1-2D +6D+3D +3)

2

y

2 1 1 1

1=e x

2D -DD +5D+D -D +2

2

y

2 1 1 1

1=e x

2D -DD +5D+D -D2 1+

2

2 -12 1 1 1

y 2D -DD +5D+D -De= 1+ .x2 2

22 1 1 1y 2D -DD +5D+D -De= 1- +... x

2 2

ye 5= x-2 2

z= C.F +P.I

20. Solve : 22 1 1 1 2x-y(D -2DD +D -3D+3D +2)z=e (AU Dec 2011)

Solution:

The given equation can be written as

1 1 2x-yD-D -1 D-D -2 z=e

C.F:

1 1

1 1 2 2D-m D -α D-m D -α z=0 is

1 2α x α x

1 1 2 2z=e f (y+m x)+e f (y+m x)

1 2 1 2Here α =1, α =2, m =1, m =1

x 2x

1 2C.F=e f (y+x)+e f (y+x) 1

Replace D 2

D 1


2x-y

1 1

1P.I= e

D-D -1 D-D -2

2x-y1

= e2+1-1 (2+1-2)

= 2x-ye

2

z = C.F +P.I

=

2x-yx 2x

1 2

ee f (y+x)+e f (y+x)+

2

21. 22 1 13 4 cos(2 )D DD D z x y xy

(AU Nov/Dec 2012)

Solution:

Give that

22 1 1D +3DD -4D z=0

The A.E is m2+3m-4=0 (m-1) (m+4) =0

1, -4m

1 2C.F (y-4x) (y+x)

21 2 1 1

1P.I = cos(2x+y)

D +3DD -4D

1cos(2x+y)

-4+3(-2)-4(-1)

1=- cos(2x+y)

6

2 22 2 1 1 1 12

2

1 1P.I = xy= xy

D +3DD -4D 3DD -4DD 1+

D


21 12

2

1= xy

3DD 4DD 1+ -

D D

21 12

2

1= xy

3D 4DD 1+ -

D D

2-1

1 1

2 2

1 3D 4D= 1+ - xy

D D D

21 1

2 2

1 3D 4D= 1- - +... xy

D D D

1

2

1 3D= xy- xy

D D

2

1 3= xy- x

D D

2

2

1 3x= xy-

D 2

21 1 3x= xy-

D D 2

2 3 2 31 x 3x 1 x y x= y- = -

D 2 6 D 2 2

= 3 4x x

y-6 8

22. Solve: (D2-DD

1+2D)z = e

2x+y+4 (AU Dec 2012)

Solution:

Here m1 = 0, c1 =0

m2 =1, c2 = -2

C.F. =f1(y)+e-2x

f2(y+x)


P.I1 = 2x+y 2x+y

2x+y

2 1

e ee = =D -DD +2D 4-2+4 6

ox+oy

2 2 1

1P.I = 4e

D -DD +2D

ox+oy

1

1= 4 e

2D-D +2

ox+oyx= 4 e

2

= 2 x

z= C.F+P.I

-2x 2x+y

1 2

1= f (y)+e f (y+x)+ e +2x

6

23.Find the p.d.e of all planes which are at a Constant distance from the

origin (AU May 2010)

Solution:

The equation of a plane which is at a distance ‘k’ from the origin is

x+my+nz=kl where 2 2 2+m +n =1l ----------------1

Let l =a ; m =b ; n=c we get

ax+by+cz=k ---------------2

2 2 2 2 2 2

(1)

a +b +c =1 c =1-a -b

2 2c= 1-a -b ---------------3

Substitute (3) in (2)

2 2ax+by+ 1-a -b z=k ---------------4

d.p.w.r to ‘x’ 2 2 za+ 1-a -b 0

x


2 2

2 2

-a1-a -b p p=

1-a -ba

d (4) p.w.r to ‘y’

2 2 zb+ 1-a -b 0

y

2 2

-bq=

1-a -b

Now,

2 2a b= =- 1-a -b (say)

p q

2 2a=p ; b=q and 1-a -b =-λ

2 2 2 21-p λ -q λ =-λ

Squaring on b.s

2 2 2 2 21-p λ -q λ =λ

2 2 2 21- p +q λ =λ

2 2 2 2 2 2 21=λ +λ (p +q )=λ 1+p +q

2

2 2 2 2

1 1λ = =±

1+p +q 1+p +q

2 21- = 1+P +q λis-veλ

(*)

(4) pλx+qλy-λz=k

2 2

(px+qy-z)λ=k

kpx+qy-z=

λ

kz=px+qy-λ

=px+qy+k 1+p +q

[by (*)]

*****

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