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(ISO : 9001-2014 Certified)
UNIVERSITY OF PETROLEUM & ENERGY STUDIES
ASSIGNMENT-3
B.TECH-II SEMESTER, 2014-2015
Course: B. Tech (All Branches) Subject: Mathematics-II
Topic: Differential Equations Code: MATH-102
1. Consider the second order ODE:
π¦β²β² + ππ¦β² + ππ¦ = 0. β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (A)
where π and π are constants. The substitution of π¦ = πππ₯ in (A) yields the
auxiliary equation:
π2 + ππ + π = 0.
The two roots of the above quadratic equation are given by
π =βπΒ±βπ2β4π
2
For π2 β 4π = 0 , we obtain only one solution as π¦ = πππ₯ with π = βπ/2 . Prove
that the second linearly independent solution is given by π₯πβππ₯/2 .
2. Consider the boundary value problem:
π¦β²β² + π¦ = 1 , π¦(0) = π¦(2π) = 0
Show that it has more than one solution. How many of them are linearly
independent ?
3. Let π¦(π₯) be the solution of the ODE : π2π¦
ππ₯2 + 2ππ¦
ππ₯+ π΅π¦ = 0. Find limπ₯ββ π¦(π₯) for the
following choices of constant π΅.
(a) 0 < π΅ < 1
(b) π΅ > 1
4. Consider the second order ODE:
π2π¦
ππ₯2 + π΄ππ¦
ππ₯+ π΅π¦ = 0. (π΄ > 0, π΅ > 0)
Find the condition for which the given ODE always admits two linearly independent
solutions that are products of exponential and trigonometric functions.
5. Let π(π₯) and π₯π(π₯) be the particular solutions of the differential equation:
π¦β²β² + π (π₯)π¦β² + π(π₯)π¦ = 0
Then find the solution of the differential equation π¦β²β² + π (π₯)π¦β² + π(π₯)π¦ = π(π₯).
6. Let π¦: β β β be a solution of the ODE
π2π¦
ππ₯2 β π¦ = πβπ₯, π₯ β β ; π¦(0) =ππ¦
ππ₯(0) = 0
Prove that:
(a) π¦ attains its minimum on β .
(b) limπ₯ββ πβπ₯π¦(π₯) =1
4 .
7. Suppose π¦1(π₯) be the known solution of the differential equation:
π2π¦
ππ₯2 + π(π₯)ππ¦
ππ₯+ π(π₯)π¦ = 0.
Show that if π¦2(π₯) = π£(π₯)π¦1(π₯) is the another linearly independent solution then
π£(π₯) = β«πβ β« π ππ₯
π¦12(π₯)
.
8. If π¦1 and π¦2 are two linearly independent solutions of the homogeneous equation:
π¦β²β² + π(π₯)π¦β² + π(π₯)π¦ = 0.
then show that π(π₯) = βπ¦1π¦2
β²β²βπ¦2π¦1β²β²
π(π¦1,π¦2) and π(π₯) =
π¦1β²π¦2
β²β²βπ¦2β²π¦1
β²β²
π(π¦1,π¦2)
where π(π¦1, π¦2) = | π¦1 π¦2
π¦1β² π¦2
β² | is the Wronskian determinant.
9. Find the general solution of π¦β²β² β π₯π(π₯)π¦β² + π(π₯)π¦ = 0 if it is known that the line
bisecting the positive quadrant of π₯π¦ βplane is its solution.
10. Determine all real numbers πΏ > 1 so that the boundary value problem:
π₯2π¦β²β²(π₯) + π¦(π₯) = 0 , 1 < π₯ < πΏ ; π¦(1) = π¦(πΏ) = 0
has a non-trivial solution.
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