193_pdfsam_math 54 differential equation solutions odd

# 193_pdfsam_math 54 differential equation solutions odd - 11...

This preview shows page 1. Sign up to view the full content.

Exercises 4.4 EXERCISES 4.4: Nonhomogeneous Equations: The Method of Undetermined Coeﬃcients, page 186 1. We cannot use the method of undetermined coeﬃcients to find a particular solution because of the t 1 term, which is not a polynomial. 3. Rewriting the right-hand side in the form 3 t = e (ln 3) t = e rt , where r = ln 3, we conclude that the method of undetermined coeﬃcients can be applied. 5. Since sec θ = 1 / cos θ , we cannot use the method of undetermined coeﬃcients. 7. Given equation is not an equation with constant coeﬃcients. Thus the method of undeter- mined coeﬃcients cannot be applied. 9. The roots of the auxiliary equation, r 2 + 3 = 0, are r = ± 3 i . Since they are different from zero, we look for a particular solution of the form y p ( t ) A . Substitution into the original equation yields ( A ) + 3 A = 9 3 A = 9 A = 3 . Thus, y p ( t ) ≡ − 3 is a particular solution to the given nonhomogeneous equation.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 11. The auxiliary equation in this problem, 2 r 2 + 1 = 0, has complex roots. Therefore, e 2 t is not a solution to the corresponding homogeneous equation, and a particular solution to the original nonhomogeneous equation has the form z p ( t ) = Ae 2 t . Substituting this expression into the equation, we Fnd the constant A . 2 ( Ae 2 t ) + Ae 2 t = 2 ( 4 Ae 2 t ) + Ae 2 t = 9 Ae 2 t = 9 e 2 t ⇒ A = 1 . Hence, z p ( t ) = e 2 t . 12. This equation is a linear Frst order di±erential equation with constant coeﬃcients. The corresponding homogeneous equation, 2 x + x = 0, can be solved by the methods of Chapter 2. Alternatively, one can use the result of Problem 21 in Section 4.2. Either approach yields 189...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern