Date 01.06.

2021 Dept: NAME and ME

Meeting ID: 910 761 8655
Passcode: 647298
Matrix and differential equation
1. Matrix (Any one out of two)
2. Ordinary differential equation (Any two out of three)
3. Partial differential equation (Any three out of Four)

1. An ordinary differential equation (or ODE) is a relation that

contains of only one independent variable and one or more of their
derivatives with respect to that variable for example of ordinary
differential equation,
dy
(i) + y=2 x e x + 4 y
dx
d 3 y d2 y dy
(ii) 3
− 2 −8 +12 y=0
dx dx dx
Given, y=tan −1 x
differentiate with respect to x
dy 1
=
dx 1+ x 2
2 dy
Or, ( 1+ x ) dx =1
Differentiate again with respect to x
( 1+ x 2 ) d dy + dy d (1+ x 2 )=0
( )( )
dx dx dx dx
2
2 d y dy
Or, ( 1+ x ) 2 +2 x dx =0 … … …( i)
dx

dy
Exp-1.Solve the equation, dx
=12 x +8

Or, dy =(12 x+ 8) dx
Now integrating both sides.
Or, ∫ dy=∫(12 x+ 8)dx
x2
Or, y=12 ( )
2
+8 x+ c

Or, y=6 x 2 +8 x +c Ans.

 d2 y dy
1. 2
dx 2
−3 + y=0 p593
dx
3 2
d y d y dy
2. dx 3
− 2 −8 +12 y =0
dx dx
2
d y dy
3. 5 2 −2 +3 y=0
dx dx
dv
Exp-2. dt =a
Or, dv =adt
Or, ∫ dv=∫ adt
Or v=at +u [u is a constant ]
Or v=u+at
dS
Or dt =u+ at
Or, dS=(u+at)dt
Or, ∫ dS=∫ (u+at )dt
t2
Or, S=ut+a ()
2
+c
1 2
Or, S=ut+ 2 a t + c
If t=0 and S=0 then c=0
1
∴ S=ut + a t 2
2
Partial differential equation :

1. Define partial differential equation:

Partial differential equation: An equation containing one or
more partial derivatives of an unknown function of two or
more independent variables is known as a partial differential
equation
for the example of partial differential equation :
δz δz
+ =z+ xy ……. …… (i)
δx δy
δz 2 δ 3 z δz
δx( )+ 3 =2 x
δy δx
……. …….. (ii)
2. Find a partial differential equation by eliminating a and b from
2
z=ax +by + a + b 2
P-1.5, Art-1.10 Ex-1
 Given, z=ax +by + a2+ b2 … … …(i)
Differentiating (i) partially with respect to x we get
δz
=a+ 0+0+0
δx
δz
or, δx
=a

Again, differentiating (i) partially with respect to y we get

δz
=0+ b+0+ 0
δy
δz
Or, δy
=b

Substituting the values of a∧b in equation (i)

δz 2 δz 2
δz δz
z= ( ) ( ) ( )( )
δx
x+
δy
y+
δx
+
δy

which is the required Partial differential equation.

3. Eliminate arbitrary constants a and b from z=(x−a)2 +( y −b)2 form the
partial differential equation P-1.6, Art-1.10 Ex-2.
Given, z=(x−a)2 +( y −b)2 … … … … (i)
Differentiating (i) partially with respect to x we get
δz
=2 ( x−a ) +0
δx
δz
or, δx
=2 ( x−a ) … … …(ii)

Again, Differentiating (i) partially with respect to y we get

δz
=0+ 2 ( y −b )
δy
δz
or, δy
=2 ( y−b ) … … …(iii)

Squaring (ii), (iii) and adding we have

 δz 2 δz 2
( )( )
δx
+
δy
=4 (x−a)2 +4 ( y−b)2

2 2
δz δz
Or, ( δx ) + ( δy ) =4 {( x−a ) +( y−b ) }
2 2

2 2
δz δz
Or, ( δx ) +( δy ) =4 z [using (i)]

which is the required Partial differential equation.

4. Form a partial differential equation by eliminating arbitrary
constant a and b from the following relations P-1.6, Art-1.10 Ex-3
a). z=a ( x + y ) +b
b). z=ax +by + ab
sol : (a) Given, z=a ( x + y ) +b
or, z=ax +ay + b … … …(i)
Differentiating (i) partially with respect to x we get
δz
=a+ 0+0
δx
δz
or, δx
=a … … (ii)

Again, Differentiating (i) partially with respect to y we get,

δz
=0+ a+0
δy
δz
or, δy
=a …… (iii)
δz δz
From (ii) and (iii) we get, =
δx δy

which is the required Partial differential equation

Sol: (b) given, z=ax +by + ab … … …(i)
Differentiating (i) partially with respect to x we get
δz
=a+ 0+0
δx
 δz
or, δx
=a … … … (ii)

Again, differentiating (i) partially with respect to y we get

δz
=0+ b+0
δy
δz
or, δy
=b … … … (iii)

Substituting the values of a∧b in equation (i)

z=x ( δxδz )+( δyδz ) y + δxδz . δyδz
which is the required Partial differential equation.
5. Form partial differential equation eliminating h and k from the
equation ( x−h)2 +( y−k )2 + z 2=ℷ2 P-1.6, Art-1.10 Ex-5(a)
Sol : Given, ( x−h)2 +( y−k )2 + z 2=ℷ2 … … …(i)
Differentiating (i) partially with respect to x we get

( δxδz )=0
2 ( x−h ) +2 z .

δz
Or, 2 ( x−h )=−2 z . ( δx )
δz
Or, ( x−h )=−z . ( δx ) … … … …(ii)

Again, Differentiating (i) partially with respect to y we get,

( δzδy )=0
2 ( y−k )+ 2 z .

δz
Or, 2 ( y−k )=−2 z . ( δy )
δz
Or, ( y−k )=−z . ( δy )… … … … (iii)

Substituting the values of ( x−h)∧( y−k ) in equation (i)

 2 2
δz δz
{ ( )} { ( ) }
−z .
δx
+ −z .
δy
+ z 2=ℷ 2

δz 2 2 δz 2 2 2
Or, z2
δx ( ) ( )
+z
δy
+ z =ℷ

which is the required Partial differential equation

6. Find the partial differential equation of all spheres of radius ℷ
having center in the xy plane P-1.7, Art-1.10 Ex-5(b) Mist-2017
marks -15
Sol: 6. From the coordinate geometry of three dimension of any
spheres of radius ℷ having center (h,k,0) in the xy plane is given by
( x−h)2 +( y−k )2 +(z−0)2 =ℷ2
Or, ( x−h)2 +( y−k )2 + z 2=ℷ2 … … … …(i)
Differentiating (i) partially with respect to x we get

( δxδz )=0
2 ( x−h ) +2 z .

δz
Or, 2 ( x−h )=−2 z . ( δx )
δz
Or, ( x−h )=−z . ( δx ) … … … …(ii)

Again , Differentiating (i) partially with respect to y we get

( δzδy )=0
2 ( y−k )+ 2 z .

δz
Or, 2 ( y−k )=−2 z . ( δy )
δz
Or, ( y−k )=−z . ( δy )… … … … (iii)

Substituting the values of ( x−h)∧( y−k ) in equation (i)

2 2
δz δz
{ ( )} { ( ) }
−z .
δx
+ −z .
δy
+ z 2=ℷ 2
 δz 2 2 δz 2 2 2
Or , z2 ( ) ( )
δx
+z
δy
+ z =ℷ

which is the required Partial differential equation

Date 14.09.2021 Dept: NAME and ME

Origin of partial differential equation

δz δz δ2 z δ2 z δ2 z
=p , =q , =r , =t , =s
δx δy δx 2 δy 2 δx . δy
δu δv δu δv δu δv δu δv δu δv δu δv
P= δy . δz − δz . δy , Q = δz . δx − δx . δz , R = δx . δy − δy . δx

xy=2 … … … (i) differentiate w.r.t x

d d
Or, x dx ( y ) + y dx ( x ) =0

dy
Or, x dx + y .1=0

dy
Or, x dx + y=0

xz=7 y … … …(ii)

Differentiating (ii) partially with respect to x

δz
x. + z=0
δx

Differentiating (ii) partially with respect to y

δz
x. =7
δy

7. Form partial differential equation eliminating a∧b from the equation,

z=(x¿¿ 2+ a)( y¿ ¿2+ b)¿ ¿ P-1.7, Art-1.10 Ex-7
Sol: z=(x¿¿ 2+ a)( y¿ ¿2+ b)… … … (i)¿ ¿
Differentiating (i) partially with respect to x we get
δz
=2 x ( y ¿¿ 2+ b)¿
δx
1 δz
Or, ( y ¿¿ 2+b)= 2 x . δx … … …(ii) ¿
Again, Differentiating (i) partially with respect to y we get
 δz
=2 y ( x ¿¿ 2+ a)¿
δy
1 δz
Or, ( x ¿¿ 2+a)= 2 y δy … … … (iii) ¿
From equation (i) we have,
z=(x¿¿ 2+ a)( y¿ ¿2+ b)¿ ¿
1 δz 1 δz
Or, z= 2 y δy . 2 x . δx
1 δz δz
Or, z= 4 xy . δx . δy
δz δz
Or, 4 xy z= δx . δy
which is the required Partial differential equation.

8. Show that the differential equation of all cones which have their vertex at the
origin is px+ qy=z verify that yz + zx + xy=0 is a surface satisfying the above
equation. P-1.7, Art-1.10 Ex-9
Sol: The equation of any cone with the vertex at the origin is
a x 2+ b y 2+ c z 2 +2 fyz+2 gzx +2 hxy=0 … … …(i)

Differentiating (i) partially with respect to x we get

δz δz δz
( )
Or, 2 ax+ 0+2 cz δx +2 fy δx + 2 gz .1+2 gx δx + 2hy =0
δz
Or, 2 ax+ 2 gz +2 hy+ δx ( 2 cz +2 gx +2 fy )=0
Or, ax + gz+ hy + p ( cz+ gx + fy )=0 … … …(ii)
Differentiating (i) partially with respect to y we get
δz δz δz
0+2 by +2 cz
δy ( )
+2 f y + z +2 gx +2 hx=0
δy δy
δz
Or, by + fz+hx + δy ( cz+ gx + fy )=0

Or, by + fz+hx +q ( cz + gx +fy )=0 … … … (iii)

Multiplying (ii) by x , (iii) by y we have
a x 2+ gxz + hxy+ px ( cz +gx + fy )=0 … …(iv)
b y 2 + fyz+hxy +qy ( cz + gx+ fy ) =0 … …(v)
Now adding (iv) and (v) we get
a x 2+ b y 2+ fyz+ gzx +2 hxy + ( cz + gx +fy ) (px +qy )=0
Or,a x 2+ b y 2+ c z 2 +2 fyz+2 gzx +2 hxy+(−c z 2−fyz−gzx )+ ( cz+ gx + fy ) ( px+ qy)=0
Or,0+(−c z2 −fyz−gzx)+ ( cz + gx +fy ) ( px+ qy)=0
Or,−z (cz + gx+ fy)+( cz + gx +fy ) ( px+ qy)=0
Or, ( cz + gx+ fy ) {−z + ( px+ qy ) }=0
0
Or, px+ qy−z= ( cz + gx+ fy )
Or, px+ qy−z=0
∴ px +qy =z Showed.
2nd part: verify that yz + zx + xy=0 is a surface satisfying the above equation. P-
1.7, Art-1.10 Ex-9
Given surface , yz + zx + xy=0 … …(1)
Differentiating (i) partially with respect to x we get
δz δz
y
δx ( )
+ x + z + y =0
δx

Or, y + z + yp+ xp=0

Or, p ( x+ y ) =− y−z
− y−z
¿ , p= … …(2)
x+ y

Differentiating (i) partially with respect to y we get

( y δzδy + z)+ x δyδz + x=0

Or, x + z+ yq+ xq=0
−x−z
Or, q= x+ y … …(3)

Now, px+ qy=z

∴ px +qy −z= ( −x+y−zy ) x +( −x−z

x+ y )
y−z

−xy−xz− xy− yz−xz − yz

Or, px+ qy−z= x+ y
−2(xy−xz − yz)
Or, px+ qy−z=
x+ y
−2(0)
Or, px+ qy−z=
x+ y

Or, px+ qy−z=0

∴ px +qy =z
 Hence (1) is a surface satisfying by, px+ qy−z=0
9. Find a partial differential equation by eliminating a,b and c from
x2 y 2 z 2
+ + =1 P-1.9, Art-1.10 Ex-13
a2 b 2 c2
x2 y 2 z 2
Sol: Given, 2 + 2 + 2 =1 … …(i)
a b c
Differentiating (i) partially with respect to x we get
2x 2 z δz
+ 0+ 2 =0
c δx
2
a
2 x −2 z δz
or, 2 = 2 δx
a c
2 2 δz
or, c x + a z δx =0 … … …(ii)
Differentiating (i) partially with respect to y we get
2 y 2 z δz
0+ + =0
b 2 c 2 δy
2 2 δz
or, c y+ b z δy =0 … … … (iii)
Differentiating again (ii) partially with respect to x we get
2 δz δz δ2 z
2
(
Or, c +a δx δx + z 2 =0
δx )
δz 2 2 δ 2 z
2
c +a
2
( )
δx
+ a z 2 =0 … … …(iv)
δx
2 2 δz
From (ii) c x + a z δx =0
2
2 −a z δz
or, c = . … …( vi)
x δx
Putting c in equation (iv) we have
2

−a2 z δz 2 δz 2 2 δ 2 z
. +a ( ) + a z 2 =0
x δx δx δx
−z δz δz 2 δ 2 z
Or, x . δx +¿( δx ) + z 2 =0 ¿ [div by a 2]
δx
δ2z δz 2 δz
Or, zx 2 + x( δx ) −z δx =0 … …( vii)
δx
This is the desire partial differential equation
Alternate Method
Differentiating again (iii) partially with respect to y we get
2 2 δz 2 2 δ 2 z
c +b ( ) + b z 2 =0 … … …(v )
δy δy
Similarly from (iii) and (v)
 δ2 z δz 2 δz
Or, zy 2
+ y (
δy
) −z =0 … … (viii )
δy
δy
Differentiating (ii) partially with respect to y we get
2 δz δz δ2 z
0++ a ( . + z . )=0
δy δx δxδy
δz δz δ2 z
Or, . + z . =0 … … ( ix )
δy δx δxδy
( vii ) ,(viii ) and ( ix ) are three possible form of the required partial differential
equation
10. Derive partial differential equation by elimination of arbitrary function ∅
from the equation ∅ ( u , v )=0 where u∧v are functions of x , y ∧z P-1.11, Rule-2,
Art-1.11 Mist-2019
Proof : Given, ∅ ( u , v )=0 … … (i)
we treat z as dependent variable and x,y as independent variable
δz δz δy δx
so that, δx = p , δy =q and δx =0 , δy =0
Differentiating (i) partially with respect to x we get

δ ∅ δu δx δu δy δu δz δ ∅ δv δx δv δy δv δz
( . + . + .
δu δx δx δy δx δz δx
+ )
. + . + .
δv δx δx δy δx δz δx
=0 ( )
δ ∅ δu δu δu δ ∅ δv δv δv
( )
Or, δu δx .1+ δy .0+ δz . p + δv δx .1+ δy .0+ δz . p =0 ( )
δ ∅ δu δu δ ∅ δv δv
( ) (
Or, δu δx + δz . p + δv δx + δz . p =0 … … … … … .(ii) )
δ ∅ δu δu δ ∅ δv δv
( )
Or, δu δx + p δz =−¿ δv δx + p δz ( )
δ∅ δv δv
− +p
δu δx δz
Or, δ ∅ = δu δu … … …(ii)
+p
δv δx δz
Similarly differentiating (i) partially with respect to y we get
δ ∅ δu δu δ ∅ δv δv
(
δu δy
+q
δz
=−¿ ) δv δy
+q(δz )
δ∅ δv δv
− +q
δu δy δz
Or, δ ∅ =¿ δu δu … … …(iii )
+q
δv δy δz
Elimination of arbitrary function ∅ with the help of ( ii )∧(iii)
 δv δv δv δv
− +p − +q
δx δz δy δz
=
δu δu δu δu
+p +q
δx δz δy δz
δv δu δu δv δv δu δv δu δu δv δu δv δu δv δu δv
Or, δx . δy + q δz . δx + p δz . δy + p δz . q δz = δx . δy + δx . q δz + p δz . δy + p δz . q δz
δv δu δu δv δu δv δu δv δu δv δv δu
( ) ( )(
Or, p δz . δy − δz . δy +q δz . δx − δx . δz = δx . δy − δx . δy )
Or, pP+qQ=R
δu δv δu δv δu δv δu δv δu δv δu δv
where, P= δy . δz − δz . δy , Q = δz . δx − δx . δz , R = δx . δy − δy . δx
∴ Pp+Qq=R … … …(iv )
Thus we obtain a linear partial differential equation of first order and first
degree in p∧q