Mathematics-III
Solved Exam Paper 2019
Q1 a) If y=A cos (log x) + B sin (log x), Show that x2yn+2 +(2n+1) * yn+1 + (n2 + 1) yn=0 where yn= dny/dxn
If y = Acos(logx) + Bsin(logx) ------------(1)
Differentiating (1) w.r.t x, we get
y1 = -asin(logx)1/x + bcos(logx)1/x
xy1 = -asin(logx) + bcos(logx) ------------(2)
Diff 2 again w.r.t x, then we get
xy2 + y1 = -acos(logx)1/x - bsin(logx)1/x
x2y2 + xy1 = -[acos(logx) + bsin(logx)]
x2y2 + xy1 = -y
y2x2 + y1x + y = 0 ----------------------------(3)
Diff 3 by Leibnitzle theorem n times, we get
[ yn+2x2 + nc1 yn+12x + nc2 yn.2 ] + [ yn+1x + nc1 yn-1] + yn = 0
x2yn+2 + 2nxyn+1 + xyn+1 +n(n-1)yn + nyn + yn = 0
x2yn+2 + (2n+1)xyn+1 + (n2+1)yn = 0
Q2 a) Show that the following function is continuous at the point (0,0):
Answer:
0 < = | | < = + = 0
0 < = + = 2 | x | + 3 | y |
lt(x,y)-->(0,0) 2 | x | + 3 | y | = 0
∴ by the squeez theorem,, we conclude that
lt(x,y)-->(0,0) f(x,y) = (0,0) that is => lt(x,y)-->(0,0) = 0
∴ f is continous at (0,0)
Q2 b) If z(x+y) = x2 + y2 show that
(- )2 = 4(1--)
solution:
z(x+y) = x2 + y2
z =
= =
= =
(- ) = = =
(- )2 = -----------------------------------------------------------1
4(1--) = 4 [ 1 - - ]
= ---------------------------------------------------2
From 1 & 2
(- )2 = 4(1--)
Q3 a)Transform the equation
+ = 0 into polar coordinates.
Solution:
we have x = rcosθ , y = rsinθ
r2 = x2 + y2 , θ = tan-1y/x
= . + . => (cos θ ) -
= ()
= ( cosθ - .) ( cosθ - )
= cosθ( cosθ- ()) - .( cosθ- )
= cosθ( cosθ+ - .) - ( - sinθ+
cosθ-- )
= cos2 θ+ - +-
+ +
= cos2 θ+ 2- 2 + +
----------------------------------------------------------------(1)
= = +
= y/r +
= sinθ +
== ()
= (sinθ + )( sinθ+)
= sinθ(sinθ + ) + ( sinθ+ )
= sinθ[ sinθ- + + ] + [ cosθ [+
sinθ- + ]
= sin2θ- + + () +
() - + ()
= sin2θ- 2+2+ () +
() -----------------------------------------------------------(2)
By adding 1 & 2
+ = (sin2θ+cos2θ)+(sin2θ+cos2θ)1/r+ (sin2θ+cos2θ)1/r2
= (+ 1/r+1/r2) ans
Q4 Find the extreme values of f( x,y ,z) = 2x+3y+z, such that x2+y2=5 and x+z=1
f(x,y,z) = 2x + 3y + z ----------------------(1)
f(x,y) = (x2 + y2) - 5 -----------------(2)
y(x,z) = x+z-1 ------------------(3)
Lagranges Multipliers Equations are
+ λ + m = 0
+ λ + m = 0
+ λ + m = 0
2 + λ(2x) + m(1) = 0 ------------(4)
3 + λ(2y) + m(0) = 0 ------------(5)
1 + λ(0) + m(1) = 0 ------------(6) => m = -1
putting the values of m in (4) and (5), we get
2 + 2λx – 1 = 0 => 2λx = -1 , x = -
3 + 2λy = 0 => 2λy = -3 , y = -
putting the values of x,y in x2 + y2 = 5 , we get
+ = 5 => = 5
2λ2 = 1 => λ = +-
we know that , x = - = +- = +-
y = - = +-+-
From (3) , x+z =1 or z = 1-x
z = 1+-
putting x = , y = and z = 1 - in equation (1), we get
f = + + 1 - = + 1 = 5Ö2 + 1
putting x = -, y = -and z = 1 + in equation (1), we get
f = 2+ 3 + ( 1 + ) = - - +1 +
f = 1 – 6 Ö2 ans
Q5 Evaluate òòs F.n.ds where F = 4xz i^ + y2ĵ + yzk^ and S is surface of the cube bounded by x=0, y=1, z=0 , z=1, by using Gauss divergence theorem
s.no | surface |
| ds |
|
1 | OABC | -k | dxdy | z= 0 |
2 | DEFG | k | dxdy | z= 1 |
3 | OAFG | -ĵ | dxdz | y= 0 |
4 | BCDE | ĵ | dxdz | y=1 |
5 | ABEF | i^ | dydz | x=1 |
6 | OCDG | -i^ | dydz | x=0 |
òòs F¯ n^ ds = òòOABC F¯ n^ ds + òòDEFG F¯ n^ ds + òòOAFG F¯ n^ ds + òòBCDE F¯ n^ ds + òòABEF F¯ n^ ds +
òòOCDG F¯ n^ ds -----------------------------------(1)
òòOABC F¯ n^ ds = òòOABC (4xz i^ + y2ĵ + yzk^)(-k)dxdy
= (as z = 0)
òòDEFG (4xz i^ + y2ĵ + yzk^)(k)dxdy = òòDEFG yzdxdy
= = [x]0-1 = 1/2
òòOAFG (4xz i^ + y2ĵ + yzk^)(-j^)dxdz = òòOAFG y2dxdz = 6 (as y =0 )
òòBCDE(4xz i^ + y2ĵ + yzk^)(j^)dxdz = òòBCDE (-y2)dxdz
-= (x)0-1(z)0-1 = -1 ( as y = 1)
òòABEF (4xz i^ + y2ĵ + yzk^).i^ dydz = òò4xzdydz
==> 4(y)0-1 = 4 (1) = 2
òòOCDG(4xz i^ + y2ĵ + yzk^)(-i^) dydz = = 0 ( as x = 0 )
on putting these values in (1), we get
òòs F¯ n^ ds = 0 + 1/2 + 6 -1 + 2 + 0 => 3/2
Q 6 (a) Evaluate {A*{ B * C }}
i^ ĵ k^
(B*C) = cosq -sinq -3
2 3 -1
= i^ (sinq + 9) – ĵ (-cosq + 6) + k^ (3cosq + 2sinq)
i^ ĵ k^
A*( B * C ) = sinq cosq Q
(sinq + 9) (cosq - 6) (3cosq + 2sinq)
= i^[cosq(3cosq + 2sinq) - q(cosq – 6)]
- ĵ[sinq(3cosq + 2sinq) - q(sinq + 9)]
+ k^[sinq(cosq – 6) - cosq (sinq + 9)]
at q = 0
=> i^[cos0(3cos0 + 2sin0) - 0(cos0 – 6)]
- ĵ[sin0(3cos0 + 2sin0) - 0(sin0 + 9)]
- ĵ[sin0(3cos0 + 2sin0) - 0(sin0 + 9)]
=> 3i^ - 9k^ ans
Q 6 A particle moves along the curve x=t3 + 1, y=t2, z= 2t+5 where t is the time. Find the components of the velocity and acceleration at t=1 in the direction i+j+3k
x = t3 + 1 , y = t3 , z = 2t + 5
¯r = xi^ + yĵ + zk^
¯r = (t3 + 1)i^ + (t3)ĵ + (2t + 5)k^
velocity = = 3t2 i^ + 2tĵ + 2k^
when t = 1, we have , = 3 i^ + 2ĵ + 2k^
unit velocity along ( i^ + ĵ + 3k^) = ( i^ + ĵ + 3k^) / Ö (1 + 1 + 9)
= ( i^ + ĵ + 3k^)
component of velocity (3 i^ + 2ĵ + 2k^) along ( i^ + ĵ + 3k^)
= (3 i^ + 2ĵ + 2k^) . ( i^ + ĵ + 3k^)
= (3+2+6) = = Ö11 ans
Q7 (a) Solve by the method of variation of parameters
+ n2y = sec nx
solution: - (D2 + n2)y = sec nx
m2 + n2 = 0 => m = +- ni
cf = c1cosnx + c2sinnx
y = A'cosnx + B'sinnx -----------------(I)
by diff. Equation (1)
(-nA'sinnx + B'ncosnx = secnx) -------------(II)
Now multiplying by nsinx in equation (I) & multiplying by cosx in equation (II)
nsinnx(A'cosnx + B'sinnx = 0)
cosnx(-nA'sinnx + B'ncosnx = secnx)
A' nsinnx cosnx + B' nsin2 nx = 0
- A' nsinnx cosnx + B' ncos2 nx = 1
-----------------------------------------------------
nB' (sin2nx + cos2nx) = 1
B' =
=
dB =
B = + C |
A'cosnx + B'sinnx = 0
A'cosnx + sinnx = 0
A' = - = - tan nx
òdA = -ò tan nx dx + C2
A = -log sec nx + C2
y = [-log sec nx + C2 ] cos nx + ( + C1)sin nx
Y = C1 sin nx + C2 cos nx + sin nx - cos nx. log sec nx |
Q 7 (b) solve - 2 tanx + 5y = sec x. ex
- 2 tanx + 5y = sin x. ex
y'' – 2 tan xy' + 5y = sin x. ex
compare with y'' + 8y' + qy = R
p = -2tanx , q = 6, R = sin x. ex
for C.F µ = e-1/2òpdx
q1 = q - -
R1 =
µ = e-1/2ò-2tanx dx = elog secx = sec x
q1 = 5 -(-2tanx)-
= 5 + sec2x – tan2x = 6
R1 =
+ q1v = R1
(D2 + 6)v =
A.E = m2 + 6 =0 => m2 = -6 => m = +- 6i
CF = (C1cos6x + C2sin6x)
P.I = => (sin x. ex)
= sin x
= sin x => sin x
= sin x => sin x
= sin x => sin x
= (D-3)sin x => (Dsin x – 3 sin x)
= (cos x – 3 sin x)
C.S = (C1 cos 6x + C2 sin6x) -(cos x – 3 sin x