Docsity
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Docsity AI
Docsity AI
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search for your university

Find the specific documents for your university's exams

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Math 2214 Sample Test: Complex Roots & Variation of Parameters, Exams of Differential Equations

Virginia Polytechnic Institute and State University (Virginia Tech)Differential Equations

The solutions to the sample second test of math 2214, focusing on differential equations with complex roots and the application of variation of parameters. Finding the general solution, initial value problems, and the behavior of solutions as t approaches negative and positive infinities.

Typology: Exams

Pre 2010

Uploaded on 02/13/2009

koofers-user-4xt
koofers-user-4xt 🇺🇸

5

(1)

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Math 2214 Monday, October 13
Solutions to Sample Second Test
1. The characteristic equation is r2−4r+13 =0. Using the formula, we find that the
roots are (4±√16 −52)/2, that is 2 ±3i. Therefore the general solution to the
differential equation is y=e2t(C1cos3t+C2sin3t), where C1,C2are arbitrary con-
stants. We now plug in the initial conditions: first t=0, y=0 yields 0 =C1, so
y=C2e2tsin3t. Now we differentiate by product rule to obtain y0=2C2e2tsin3t+
3C2e2tcos 3t. Plugging in t=0, y0=3, we obtain 3 =3C2. Therefore C+2=1 and
we conclude that the solution to the initial value problem is y(t) = e2tsin3t.
As t→ −∞,y(t)→0, while as t→∞,y(t)oscillates with ever increasing amplitude,
and this increasing amplitude tends to infinity.
2. First we find yC, the complementary solution which is the general solution to the cor-
responding homogeneous equation, namely y00 −4y0+4y=0. This has characteristic
equation r2−4r+4=0, which has a repeated root 2. Therefore yC=C1e2t+C2te2t.
To find a particular solution yP, our first try would be Ae2t, where Ais a constant to
be determined. However this does not work, because e2tappears in yC. Therefore we
multiply this by tand try Ate2t. However this still does not work because te2talso
appears in yC. Therefore we try y=At2e2t; this should now work. Differentiating
twice, we obtain y0=A(2t2+2t)e2tand y00 =A(4t2+8t+2). Plugging back into the
original differential equation, we obtain
A(4t2+8t+2)e2t−4A(2t2+2t)e2t+4At2e2t=2e2t.
The terms in t2e2tand te2tcancel and we are left with A2e2t=2e2t. Therefore A=1
and we conclude that the solution to the differential equation is y=yC+yP=C1e2t+
C2te2t+t2e2t.
3. We need to find yPand for this we use variation of parameters; note that undetermined
coefficients will not work here, because the coefficients of the left hand side are not
constants. So we look for a solution in the form y=ut +vt3where u,vare functions
of tto be determined, and set u0t+v0t3=0, that is u0+v0t2=0. Differentiating, we
obtain
y0=u0t+u+v0t3+3t2v=u+3t2v
y00 =u0+6tv +3t2v0.
Plugging back into t2y00 −3ty0+3y=2t3, we obtain
t2(u0+6tv +3t2v0)−3t(u+3t2v) + 3(ut +t3v) = 2t3.
As always happens at this point, the terms in u,vcancel and we are left only with terms
in u0,v0, that is t2u0+3t4v0=2t3, which simplifies to u0+3t2v0=2t. If multiply the
equation u0+v0t2=0 from above by 3 and subtract, we obtain −2u0=2t. Therefore
pf2

Partial preview of the text

Download Math 2214 Sample Test: Complex Roots & Variation of Parameters and more Exams Differential Equations in PDF only on Docsity!

Math 2214 Monday, October 13

Solutions to Sample Second Test

  1. The characteristic equation is r^2 − 4 r + 13 = 0. Using the formula, we find that the roots are ( 4 ±

16 − 52 )/2, that is 2 ± 3 i. Therefore the general solution to the differential equation is y = e^2 t^ (C 1 cos 3t + C 2 sin 3t), where C 1 ,C 2 are arbitrary con- stants. We now plug in the initial conditions: first t = 0, y = 0 yields 0 = C 1 , so y = C 2 e^2 t^ sin 3t. Now we differentiate by product rule to obtain y′^ = 2 C 2 e^2 t^ sin 3t + 3 C 2 e^2 t^ cos 3t. Plugging in t = 0, y′^ = 3, we obtain 3 = 3 C 2. Therefore C + 2 = 1 and we conclude that the solution to the initial value problem is y(t) = e^2 t^ sin 3t. As t → −∞, y(t) → 0, while as t → ∞, y(t) oscillates with ever increasing amplitude, and this increasing amplitude tends to infinity.

  1. First we find yC, the complementary solution which is the general solution to the cor- responding homogeneous equation, namely y′′^ − 4 y′^ + 4 y = 0. This has characteristic equation r^2 − 4 r + 4 = 0, which has a repeated root 2. Therefore yC = C 1 e^2 t^ +C 2 te^2 t^. To find a particular solution yP, our first try would be Ae^2 t^ , where A is a constant to be determined. However this does not work, because e^2 t^ appears in yC. Therefore we multiply this by t and try Ate^2 t^. However this still does not work because te^2 t^ also appears in yC. Therefore we try y = At^2 e^2 t^ ; this should now work. Differentiating twice, we obtain y′^ = A( 2 t^2 + 2 t)e^2 t^ and y′′^ = A( 4 t^2 + 8 t + 2 ). Plugging back into the original differential equation, we obtain

A( 4 t^2 + 8 t + 2 )e^2 t^ − 4 A( 2 t^2 + 2 t)e^2 t^ + 4 At^2 e^2 t^ = 2 e^2 t^.

The terms in t^2 e^2 t^ and te^2 t^ cancel and we are left with A 2 e^2 t^ = 2 e^2 t^. Therefore A = 1 and we conclude that the solution to the differential equation is y = yC + yP = C 1 e^2 t^ + C 2 te^2 t^ + t^2 e^2 t^.

  1. We need to find yP and for this we use variation of parameters; note that undetermined coefficients will not work here, because the coefficients of the left hand side are not constants. So we look for a solution in the form y = ut + vt^3 where u, v are functions of t to be determined, and set u′t + v′t^3 = 0, that is u′^ + v′t^2 = 0. Differentiating, we obtain

y′^ = u′t + u + v′t^3 + 3 t^2 v = u + 3 t^2 v y′′^ = u′^ + 6 tv + 3 t^2 v′.

Plugging back into t^2 y′′^ − 3 ty′^ + 3 y = 2 t^3 , we obtain

t^2 (u′^ + 6 tv + 3 t^2 v′) − 3 t(u + 3 t^2 v) + 3 (ut + t^3 v) = 2 t^3.

As always happens at this point, the terms in u, v cancel and we are left only with terms in u′, v′, that is t^2 u′^ + 3 t^4 v′^ = 2 t^3 , which simplifies to u′^ + 3 t^2 v′^ = 2 t. If multiply the equation u′^ + v′t^2 = 0 from above by 3 and subtract, we obtain − 2 u′^ = 2 t. Therefore

u′^ = −t and we deduce that u = −t^2 /2. Also we get v′^ = 1 /t and we deduce that v = lnt (we don’t need a “C” here because we only need a particular solution). Therefore yP = ut + vt^3 = −t^3 / 2 + t^3 lnt and we conclude that the general solution is y = yC + yP = C 1 t +C 2 t^3 −t^3 / 2 +t^3 lnt. We can tidy this up a little, because the C 2 t^3 and −t^3 / 2 can be combined (replace C 2 with C 2 + 1 /2) to obtain y = C 1 t +C 2 t^3 + t^3 lnt.

  1. Let e m be the extension of the spring when the ball is at rest with no external force. Then at time t the force due to gravity on the ball is 2g N downwards, where g m/sec^2 is the acceleration due to gravity. The force due to air resistance is y′^ in the upward direction when the ball is going down, and the force of the spring is 3(e + y) in the upward direction. Thus F = ma in the downwards direction yields

2 g + 3 cos 3t − 2 sin 3t − y′^ − 3 (e + y) = 2 y′′.

When the system is at rest and there is no force, y′′^ = y′^ = y = 0 and we obtain the equation 2g − 3 e = 0. The equation then becomes 2y′′^ + y′^ + 3 y = 3 cos 3t − 2 sin 3t.

Test on Friday, October 17. Material section 3.1 to section 3.10 on the assignment sheet. Review session on Thursday October 16 at 5 p.m. in McBryde 226.