Midsem sol
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This document provides the solutions to questions on a mid-semester exam for a differential equations course. The questions cover determining if functions satisfy initial value problems, solving initial value problems, analyzing behaviors of solutions, and solving exact and homogeneous differential equations.
![JL Sem2_2013/2014
TMS2033 Differential Equations
Mid-Semester Exam SOLUTION
Semester 2 2013/2014
31st
March 2014 9:00 ā 11:00am
Answer all questions.
1. Determine whether any of the functions
1 2 3
1
( ) sin 2 , ( ) , ( ) sin 2
2
y x x y x x y x xļ½ ļ½ ļ½
is a solution to the initial-value problem
4 0; (0) 0, (0) 1y y y yļ¢ļ¢ ļ¢ļ« ļ½ ļ½ ļ½ .
(8 marks)
.2sin4,2cos2 11 xyxy ļļ½ļ¢ļ¢ļ½ļ¢ Hence, )(1 xy is a solution to the differential equation.
Also, it satisfies the first initial condition, 0)0(1 ļ½y . However, )(1 xy does not satisfy
the second condition since 10)0(1 ļ¹ļ½ļ¢y . This means that )(1 xy is not a solution to the
initial value problem.
[3 marks]
.0,1 22 ļ½ļ¢ļ¢ļ½ļ¢ yy )(2 xy satisfy both the initial conditions but is not a solution to the
differential equation, hence, )(2 xy is not a solution to the initial value problem.
[2 marks]
.2sin2,2cos 33 xyxy ļļ½ļ¢ļ¢ļ½ļ¢ upon substitution into the differential equation we have
02sin
2
1
42sin2 ļ½ļ·
ļø
ļ¶
ļ§
ļØ
ļ¦
ļ«ļ xx , which shows that )(3 xy satisfy the differential equation.
Also, 0)0sin(
2
1
)0(3 ļ½ļ½y and 1)0(2cos)0(3 ļ½ļ½ļ¢y , shows )(3 xy satisfy both initial
conditions. This means that )(3 xy is a solution to the initial value problem.
[3 marks]
2. Find the solution to the initial value problem, and specify the interval where the
solution is defined.
.100)2(;4
210
2
ļ½ļ½
ļ«
ļ« QQ
tdt
dQ
(15 marks)
Solve using Method of integrating factor where ļ²ļ½
dttp
e
)(
ļ .
Here, .,
210
2
)( 210
2
ļ²
ļ½ļ
ļ«
ļ½ ļ«
dt
t
e
t
tp ļ](https://image.slidesharecdn.com/midsemsol-140429074744-phpapp02/85/Midsem-sol-1-320.jpg)
Midsem sol
- 1. JL Sem2_2013/2014 TMS2033 Differential Equations Mid-Semester Exam SOLUTION Semester 2 2013/2014 31st March 2014 9:00 ā 11:00am Answer all questions. 1. Determine whether any of the functions 1 2 3 1 ( ) sin 2 , ( ) , ( ) sin 2 2 y x x y x x y x xļ½ ļ½ ļ½ is a solution to the initial-value problem 4 0; (0) 0, (0) 1y y y yļ¢ļ¢ ļ¢ļ« ļ½ ļ½ ļ½ . (8 marks) .2sin4,2cos2 11 xyxy ļļ½ļ¢ļ¢ļ½ļ¢ Hence, )(1 xy is a solution to the differential equation. Also, it satisfies the first initial condition, 0)0(1 ļ½y . However, )(1 xy does not satisfy the second condition since 10)0(1 ļ¹ļ½ļ¢y . This means that )(1 xy is not a solution to the initial value problem. [3 marks] .0,1 22 ļ½ļ¢ļ¢ļ½ļ¢ yy )(2 xy satisfy both the initial conditions but is not a solution to the differential equation, hence, )(2 xy is not a solution to the initial value problem. [2 marks] .2sin2,2cos 33 xyxy ļļ½ļ¢ļ¢ļ½ļ¢ upon substitution into the differential equation we have 02sin 2 1 42sin2 ļ½ļ· ļø ļ¶ ļ§ ļØ ļ¦ ļ«ļ xx , which shows that )(3 xy satisfy the differential equation. Also, 0)0sin( 2 1 )0(3 ļ½ļ½y and 1)0(2cos)0(3 ļ½ļ½ļ¢y , shows )(3 xy satisfy both initial conditions. This means that )(3 xy is a solution to the initial value problem. [3 marks] 2. Find the solution to the initial value problem, and specify the interval where the solution is defined. .100)2(;4 210 2 ļ½ļ½ ļ« ļ« QQ tdt dQ (15 marks) Solve using Method of integrating factor where ļ²ļ½ dttp e )( ļ . Here, ., 210 2 )( 210 2 ļ² ļ½ļ ļ« ļ½ ļ« dt t e t tp ļ
- 2. JL Sem2_2013/2014 Let tu u du dt t dtdutu 210lnln 210 2 2210 ļ«ļ½ļ½ļ½ ļ« ļļ½ļļ«ļ½ ļ² ļ² te t 210 210ln ļ«ļ½ļ½ļ ļ« ļ Multiplying the differential equation by Ī¼, we obtain ļØ ļ© ļØ ļ©ļ ļ tQt dt d tQ dt dQ t 840210 8402210 ļ«ļ½ļ« ļ«ļ½ļ«ļ« Upon integrating both sides above, we have ļØ ļ© t Ctt tQ CttQt 210 440 )( 440210 2 2 ļ« ļ«ļ« ļ½ļ ļ«ļ«ļ½ļ« Applying the initial condition directly, we have 1304 961400 14001680 100 )2(210 )2(4)2(40 )2( 2 ļ½ļ ļļ½ļ ļ½ļ«ļ«ļ ļ½ ļ« ļ«ļ« ļ½ C C C C Q Thus, t tt tQ 210 1304404 )( 2 ļ« ļ«ļ« ļ½ This solution is defined as long as 50210 ļļ¾ļļ¹ļ« tt 3. Consider the initial value problem 0(4 )/(1 ), (0) 0.y ty y t y yļ¢ ļ½ ļ ļ« ļ½ ļ¾ a. Find the solution to the initial value problem t yty dt dy ļ« ļ ļ½ 1 )4( The differential equation is separable and hence, ļ² ļ² ļ« ļ½ ļ dt t t dy yy 1)4( 1 To assist in integration, we write the integral terms in partial fractions: 4 1 )4(1 4)4( 1 ļ½ļ½ļļ«ļļ½ļ ļ ļ«ļ½ ļ BAByyA y B y A yy 1,1)1( 11 ļļ½ļ½ļļ«ļ«ļ½ļ ļ« ļ«ļ½ ļ« BABtAt t B A t t Thus, our integration now becomes:
- 3. JL Sem2_2013/2014 441ln4 4 )1( 4 1ln4 4 ln 1ln444lnln 1ln4ln 4 1 ln 4 1 1 1 1 )4(4 1 4 1 4 ļļ«ļ«ļ ļ«ļ½ļ½ ļ ļ ļ«ļ«ļļ½ ļ ļ ļ«ļ«ļļ½ļļļ ļ«ļ«ļļ½ļļ«ļ ļ« ļļ½ ļ ļ«ļ² ļ² tKee y y Ctt y y Cttyy cttyy dt t dy yy tCtt Applying 0)0( yy ļ½ , K y y ļ½ ļ 40 0 Therefore the solution for the initial value problem is ļØ ļ©ļØ ļ©4 0 4 0 144 ty ey y y t ļ«ļ ļ½ ļ b. Determine how the solution behaves as t ļ® ļ„ As t ļ® ļ„, the right-hand-side of the solution found in part a. is also ļ® ļ„. This means that ļ„ļ® ļ 4y y And so, 4 04 ļ®ļ ļ®ļ y y c. If y0 = 2, find the implicit function of time T at which the solution first reaches the value 3.99 ļØ ļ© ļØ ļ© ļØ ļ© T T T eT T e T e 44 4 4 4 4 1399 1 399 12 2 01.0 99.3 ļ½ļ«ļ ļ« ļ½ļ ļ«ļ ļ½ ļ (20 marks)
- 4. JL Sem2_2013/2014 4. Consider the differential equation 222 3 yxyx dx dy x ļ«ļ«ļ½ a. Show that the equation is homogeneous Writing the differential equation in ),( yxf dx dy ļ½ form, we have 2 22 3 ),( x yxyx yxf ļ«ļ« ļ½ . ),( 33 ),( 2 22 22 22222 yxf x yxyx xt ytxytxt tytxf ļ½ ļ«ļ« ļ½ ļ«ļ« ļ½ Hence, the equation is homogeneous. b. Solve the differential equation Let y = xv then dx dv xv dx dy ļ«ļ½ . Substitute these into the differential equation we obtain 2 2 2 2222 21 31 3 vv dx dv x vv x vxvxx dx dv xv ļ«ļ«ļ½ļ ļ«ļ«ļ½ ļ«ļ« ļ½ļ« This can be solved using method of separable variables: ļØ ļ© Cx v x dx v dv x dx vv dv ļ«ļ½ ļ« ļ ļ ļ½ ļ« ļ ļ½ ļ«ļ« ļ²ļ² ln 1 1 1 21 2 2 Substitute back x y v ļ½ , then we have )(ln)(ln ))(ln( ln 1 1 CxxxCxy xCxxy Cx x y ļ«ļļļ½ļ«ļ ļļ½ļ«ļ«ļ ļ«ļ½ ļ« ļ c. For a given initial condition as y(1) = 2, determine the solution for this initial value problem. Applying y(1) = 2 into the general solution obtained earlier, to find C
- 5. JL Sem2_2013/2014 3 1 12 ļ ļ½ļ ļļļ½ C CC Then substitute this back into the general solution to obtain the solution to the initial value problem xxxy ļļ½ļļ« ) 3 1 )(ln( . (18 marks) 5. Consider the differential equation .022 ļ½ļ«ļ« dx dy bxexye xyxy a. Find the value of b for which the given equation is exact. Writing the DE in Mdx + Ndy = 0 form, we have xyeM xy ļ«ļ½ 2 and xy bxeN 2 ļ½ For exact DE, we need xy NM ļ½ Thus, 1 )12()12( )2(0)2( 22 2222 ļ½ļ ļ«ļ½ļ«ļ ļ«ļ½ļ½ļ«ļ«ļ½ b xybexye beeybxNeexyM xyxy xyxy x xyxy y b. With the value of b found in part b., find the explicit solution to the differential equation. Let the general solution of the DE be cyx ļ½),(ļ¹ , where xy y xeN 2 ļ½ļ½ļ¹ Integrate this wrt y, )1()( 2 )( 2 )( 2 2 2 ļxh e xh x xe dyxe xy xy xy ļ«ļ½ļ ļ«ļ½ ļ½ ļ² ļ¹ ļ¹ Also, )2(2 ļxyeM xy x ļ«ļ½ļ½ļ¹ Differentiate (1) wrt x and equate with (2), we have
- 6. JL Sem2_2013/2014 2 )( )( 2 2 2 2 2 x h xxh xyexh ye xy xy x ļ½ļ ļ½ļ¢ļ ļ«ļ½ļ¢ļ«ļ½ļ¹ Substitute h(x) above into (1), we obtain 22 22 xe xy ļ«ļ½ļ¹ Hence, the explicit solution to the differential equation is Cxe xy ļ½ļ« 22 (19 marks)