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M 2 Paper

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0% found this document useful (0 votes)
35 views4 pages

M 2 Paper

Uploaded by

goswamiusha586
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Roll No. .....................................

Total Pages: 04

2127
B.A. SECOND YEAR EXAMINATION, 2019
MATHEMATICS
Paper – II
Differential Equations
Time: Three Hours
Maximum Marks: 65

PART – A ¼[k.M & v½ [Marks: 20]


Answer all questions (50 words each).
All questions carry equal marks.
lHkh iz’u vfuok;Z gSAa izR;sd iz’u dk mŸkj 50 'kCnksa ls vf/kd u gksA
lHkh iz’uksa ds vad leku gSAa
PART – B ¼[k.M & c½ [Marks: 25]
Answer five questions (250 words each).
Selecting one from each unit. All questions carry equal marks.
izR;sd bdkbZ ls ,d&,d iz’u pqurs gq,] dqy ik¡p iz’u dhft,A
izR;sd iz’u dk mŸkj 250 'kCnksa ls vf/kd u gksA
lHkh iz’uksa ds vad leku gSAa
PART – C ¼[k.M & l½ [Marks: 20]
Answer any two questions (300 words each).
All questions carry equal marks.
dksbZ nks iz’u dhft,A izR;sd iz’u dk mŸkj 300 'kCnksa ls vf/kd u gksA
lHkh iz’uksa ds vad leku gSAa

[2127] Page 1 of 4
PART – A @ [k.M & v
Q.1 (i) Write the necessary and sufficient condition for exact differential equation.
;FkkrFk vody lehdj.k gksus ds vko’;d ,oa i;kZIr izfrca/k fyf[k,A
(ii) Write the necessary conditions of integrability of total differential equation.
lEiw.kZ vody lehdj.k dh lekdyuh;rk dh vko’;d 'krZ fyf[k,A
(iii) Define linear differential equation of second order.
f}rh; dksfV ds jSf[kd vody lehdj.k dks ifjHkkf"kr dhft,A
(iv) Write the condition and complementary function of the differential equation.
ௗమ ௬ ௗ௬
‫ݔ‬ଶ − ሺ‫ ݔ‬ଶ + 2‫ݔ‬ሻ + ሺ‫ ݔ‬+ 2ሻ‫ ݔ = ݕ‬ଷ ݁ ௫
ௗ௫ మ ௗ௫
vody lehdj.k%
݀ଶ ‫ݕ‬ ݀‫ݕ‬
‫ݔ‬ଶ
− ሺ‫ݔ‬ ଶ
+ 2‫ݔ‬ ሻ + ሺ‫ ݔ‬+ 2ሻ‫ ݔ = ݕ‬ଷ ݁ ௫
݀‫ݔ‬ ଶ ݀‫ݔ‬
dk iwjd Qyu rFkk mldh 'krZ fyf[k,A
(v) Form a partial differential equation from the equation:
z = ܽ‫ ݔ‬+ ܽ ଶ ‫ ݕ‬ଶ + ܾ
lehdj.k% ‫ ݔܽ = ݖ‬+ ܽଶ ‫ ݕ‬ଶ + ܾ ls vkaf’kd vody lehdj.k cukb,A
(vi) Write the complete integral of the equation:
(xp)2 + (yq)2 = z2
lehdj.k% (xp)2 + (yq)2 = z2 dk iw.kZ lekdy fyf[k,A
(vii) Write the characteristic equations of Charpit’s method for non-linear differential
equation.
vjSf[kd vody lehdj.k dks gy djus dh pkjfiV fof/k dk vfHkyk{kf.kd lehdj.k fyf[k,A
(viii) Write the subsidiary equation of Rr + Ss +Tt = V
Rr + Ss +Tt = V dk lgk;d lehdj.k fyf[k,A
(ix) Write the formulae of Euler’s modified method.
vk;yj vkifjofrZr fof/k dk lw= fyf[k,A
(x) Find first approximation by Picard’s iterative method:
ୢ୷
= 1 + xy, x଴ =2, y଴ = 2
ୢ୶
fidkMZ fof/k }kjk izFke lfUudVu Kkr dhft,%
ୢ୷
ୢ୶
= 1 + xy, x଴ =2, y଴ = 2

[2127] Page 2 of 4
PART – B @ [k.M & c
UNIT –I@ bdkbZ – I
Q.2 Solve:
gy dhft,%
Dx + Dy – 2y = 2 cos t – 7 sin t
Dx – Dy + 2x = 4 cos t – 3 sin t
Q.3 Solve:
gy dhft,%
ୢ୶ ୢ୷ ୢ୸
= =
୶ ୷ ୸ିୟඥሺx2 ାy2 ାz2 ሻ

UNIT –II@ bdkbZ – II


Q.4 Solve:
gy dhft,%
ୢమ ୷ ୢ୷
xଶ − 2xሺ1 + xሻ + 2ሺ1 + xሻy = x ଷ
ୢ୶మ ୢ୶
Q.5 Solve:
gy dhft,%
ୢమ ୷ ୢ୷
xଶ − 2ሺx ଶ + xሻ + ሺx ଶ + 2x + 2ሻy = 0
ୢ୶మ ୢ୶
UNIT –III@ bdkbZ – III
Q.6 Solve:
gy dhft,%
pq = x ୫ y୬ z ℓ
Q.7 Solve:
gy dhft,%
x ଶ pଶ + qଶ yଶ − zଶ = 0
UNIT –IV@ bdkbZ – IV
Q.8 Solve by Charpit’s method:
pkjfiV fof/k ls gy dhft,%
2xz − px2 − 2qxy + pq = 0
Q.9 Solve:
gy dhft,%
x ଶ r + 2 xys + yଶ t = 0

[2127] Page 3 of 4
UNIT –V@ bdkbZ – V
Q.10 Use Picard’s method to solve:
fidkMZ fof/k ls gy dhft,%
ୢ୷
= x + y, for x = 0.1 and x = 0.2, given when x = x଴ = 0, y = y଴ = 1
ୢ୶
Q.11 Use Euler’s method to solve:
vk;yj fof/k ls gy dhft,%
ୢ୷ y2 −x
= , x = 0 ,y = 1
ୢ୶ y2 +x

PART – C @ [k.M & l


Q.12 Solve:
gy dhft,%
ሺx2 − 2xy − y2 ሻdx − ሺx + yሻଶ dy = 0
Q.13 Solve by method of removal of the first derivatives:
izFke vody dks gVkus dh fof/k ls gy dhft,%
ୢమ ୷ ୢ୷
ୢ୶మ
− 2 tanx ୢ୶
+ 5y = e୶ secx
Q.14 (a) Solve:
gy dhft,%
ଵ ଵ ଵ ଵ ଵ ଵ
ቀ ୸ − ୷ቁ p + ቀ ୶ − ୸ ቁ q = ቀ ୷ − ୶ ቁ
(b) Solve:
gy dhft,%
ሺy − xሻ ሺqy − pxሻ = ሺp − qሻଶ
Q.15 Solve:
gy dhft,%
t – r sec4y = 2q tan y
Q.16 Use Euler’s modified method obtain a solution of the following equation with initial
condition y = 1 at x = 0 for the range 0 ≤ x ≤ 0.6 in steps of 0.2.
dy
= x + หඥyห
dx
vk;yj vkifjofrZr fof/k ls fuEu lehdj.k dk izkjfEHkd fLFkfr x = 0 ij y = 1 ij gy Kkr
dhft, tgk¡ {ks= 0 ≤ x ≤ 0.6 ,oa in 0.2 gSA
dy
= x + หඥyห
dx
-----------------------------------------

[2127] Page 4 of 4

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