MATH 315 FALL 1998 TEST 2 SOLUTION KEY

1.

(15 pts.) Consider the differential equation y'' +2y' + y = 0.

(a)

Find the general solution for the equation.
Answer:
Characteristic polynomial is r²+2r+1 = (r+1)², with roots -1, -1.
The general solution is therefore

c₁e^-t+c₂te^-t.

(b)

Determine the Wronskian for fundamental solution set.
Answer:

$$W = \left\vert \begin{array}{cc}e^{-t}&te^{-t}\\ -e^{-t}&-te^{-t}+e^{-t}\end{array}\right\vert = e^{-2t} .$$

2.

(15 pts.) Find the general solution for the differential equation

y''' + 2y'' - y' - 2y = 0.

Answer:
Characteristic polynomial is r³+2r²-r-2 = (r-1)(r+1)(r+2), with roots 1, -1, -2.
The general solution is therefore

c₁e^t+c₂e^-t+c₃e^-2t.

3.

(15 pts.) Find the general solution for the differential equation

y'''' + 3y'' - 4y = 4e^-t,

assuming that the general solution to the homogeneous equation is given by

$$y_c(t) = c_1e^t+c_2e^{-t}+c_3\cos(2t)+c_4\sin(2t).$$

Answer:
A particular solution should have the form Y(t) = Ate^-t.

$$\begin{eqnarray*}Y(t) & = &\ Ate^{-t}\\ Y'(t) & = &-Ate^{-t}+ \ Ae^{-t}\\ Y'... ...5.\\ \mbox{Therefore\ } y(t) & = & y_c(t) - \frac{2}{5}te^{-t}. \end{eqnarray*}$$

4.

(20 pts.) Solve the differential equation

y'' + 2y' + 2y = e^-t,

with y(0) = 1, y'(0) = 1.
Answer:
Characteristic polynomial is r²+2r+2, with roots $-1\pm i$, so $y_c(t)=e^{-t}(c_1\cos(t)+c_2\sin(t))$. A particular solution should have the form Y(t) = Ae^-t.

$$\begin{eqnarray*}Y(t) & = &\ Ae^{-t}\\ Y'(t) & = &-Ae^{-t}\\ Y''(t) & = &\ A... ...plies\ } A = 1.\\ \mbox{Therefore\ } y(t) & = & y_c(t) + e^{-t} \end{eqnarray*}$$

$$\begin{eqnarray*}y(0) &=& c_1 + 1 = 1, \mbox{\ so\ } c_1 = 0;\\ y'(0) &=& -c_1 ... ...2 = 2;\\ \mbox{Therefore\ } y(t) & = & 2e^{-t}\sin(t) + e^{-t}. \end{eqnarray*}$$

5.

(15 pts.) Suppose the general solution to an n^th order homogeneous equation L[y]=0 is

$$y_c(t) = c_1e^t+c_2t\sin(3t)+c_3t\cos(3t)+c_4\cos(3t)+c_5\sin(3t) +c_6e^t\cos(t)+c_7e^t\sin(t).$$

Determine the general form that would be used with the method of undetermined coefficients for the solution to the nonhomogeneous equation

$$L[y] = 7te^t - \cos(3t) + 14e^t\sin(t).$$

Answer:

$$\begin{eqnarray*}g_1(t) = 7te^t, &\mbox{\ so\ }& Y_1(t) = t(A+Bt)e^t;\\ g_2(t) ... ... \mbox{Therefore\ } y(t) & = & y_c(t) + Y_1(t)+Y_2(t)+Y_3(t).\\ \end{eqnarray*}$$

6.

(20 pts.) Use the variation of parameters method to solve the differential equation

y''+3y'+2y = 4e^t.

Answer:
Characteristic polynomial is r²+3r+2 = (r+1)(r+2), with roots -1, -2; y_c(t) = c₁e^-t+c₂e^-2t.

$$W = \left\vert\begin{array}{cc}e^{-t}&e^{-2t}\\ -e^{-t}&-2e^{-2t}\end{array}\right\vert = -e^{-3t}.$$

Let Y(t) = u₁(t)e^-t + u₂(t)e^-2t; then

$$u'_1(t) = \left\vert\begin{array}{cc}0&e^{-2t}\\ 4e^{t}&-2e^{... ...-4e^{-t}/(-e^{-3t}) = 4e^{2t}, \mbox{\ so\ } u_1(t) = 2e^{2t};$$

$$u'_2(t) = \left\vert\begin{array}{cc}e^{-t}&0\\ -e^{-t}&4e^{t... ...= 4/(-e^{-3t}) = -4e^{3t}, \mbox{\ so\ } u_2(t) = -4e^{3t}/3 .$$

Therefore

y(t) = y_c(t) + 2e^2te^-t - 4e^3te^-2t/3 = y_c(t) + 2e^t/3.