Question

Match the following nonhomogeneous linear equations with the form of the particular solution yp for the method of undetermined coefficients.
? A B C D 1. y′′+y=t(1+sint)

? A B C D 2. y′′+4y=t2sin(2t)+(5t−7)cos(2t)

? A B C D 3. y′′+2y′+2y=3e−t+2e−tcost+4e−tt2sint

? A B C D 4. y′′−4y′+4y=2t2+4te2t+tsin(2t)


A. yp=t(A0t2+A1t+A2)sin(2t)+t(B0t2+B1t+B2)cos(2t)
B. yp=A0t2+A1t+A2+t2(B0t+B1)e2t+(C0t+C1)sin(2t)+(D0t+D1)cos(2t)
C. yp=Ae−t+t(B0t2+B1t+B2)e−tcost+t(C0t2+C1t+C2)e−tsint
D. yp=A0t+A1+t(B0t+B1)sint+t(C0t+C1)cost

Answer 1

Answer:

1. D

2. A

3. C

4. B

Step-by-step explanation:

1.

The particular function is:

$t(1 + \sin{t}) = t + t\sin{t}$

We have a first degree polynomial and a first degree polynomial multiplying a sine function.

The particular solution of a polynomial of degree n is another polynomial of degree n.

The particular solution of a sin(at) function is a sum of Asin(at) and Bcos(at).

So, the particular solution is

$A_{0}t + A_{1} + (B_{0} + B_{1})\sin{t} + (C_{0}t + C_{1})\cos{t}$

So 1 and D.

2.

The particular function is:

$t^{2}\sin{2t}+(5t−7)\cos{2t}$

$\sin{2t}$ and $\cos{2t}$ have particular solutions in the same format. This means that we can multiply the particular solutions by t. The highest degree of the polynomials here is 2, so we have a sum of sin(2t) and cos(2t) each multiplied by a second order polynomial.

So

$yp=t(A_{0}t^{2}+A_{1}t+A2)\sin{2t}+t(B_{0}t^{2}+B1t+B2)\cos{2t}$

The particular solution of 2 is A.

3.

The particular function is

$3e^{−t}+2e{−t}\cos{t}+4e{−t}t^{2}\sin{t}$

The particular solution of an exponential is another exponential.

So the solution is C.

4.

The particular function is:

$2t^{2} + 4te^{2t} + t\sin{2t}$

Polynomial, polynomial multiplying an exponential and polynomial multiplying a sine functions.

So the answer is B.