Exam 3 Answers

1. (Section 4.3, problem 13)  y''' - 2y'' + y' = t3 + 2et

The particular solution has the form

Y(t) = t(A0t3 + A1t2 + A2t + A3) + t2Bet

       = tG1(t) + t2G2(t)et

       = Y1(t) + Y2(t).

Without knowing the answer, we know that Y1 and Y2 must have the forms

Y1(t) = tsG1(t) and Y2(t) = trG2(t), for some s and r.

First, we will determine Y1(t).  A quick inspection of L[Y1] shows that there is no t3 term to match when s = 0.  Hence, s > 0.  Trying s = 1 gives us

Y1'(t) = 4A0t3 + 3A1t2 + 2A2t + 2A3

Y1''(t) = 12A0t2 + 6A1t + 2A2

Y1'''(t) = 24A0t + 6A1

This is promising.  We get

L[Y1](t) = A0t3 + (-24A0 + 3A1)t2 +
(24A0 - 12A1 + 2A2) + (6A1 - 4A2 + 2A3)

We need to match the t3 term (which gives us A0 = 1) and then eliminate the rest of the coefficients.  We do this from left to right, term by term in the polynomial:

A1 = -(-24A0)/3 or A1 = 8,

A2 = -(24A0 - 12A1)/2 or A2 = 36.

A3 = -(6A1 - 4A2)/2 or A3 = 48.

Now consider Y2(t).  When r < 2, we have L[Y2](t)=0 (r = 0 is obvious, but r = 1 is a little less obvious).  For r = 2 we get

L[Y2](t) = 2Bet or B = 1.

Finally, Y(t) = Y1(t) + Y2(t) by Theorem 3.2.2.

Points off: -20 for no explanation, -5 for a wrong explanation for either the t3 or the 2et terms.

2. Solve y''' + y' = 0, y(0) = 1, y'(0) = 1, y''(0) = 2.

The roots of the characteristic r3 + r = r(r2 + 1) = 0 are r = 0 and r = +i.  Thus,

y(t) = c1 + c2cost + c3sint

(the c1 term actually multiplies e0t = 1).  By noting that cos(0) = 1 and that sin(0) = 0, we must solve the 3x3 system of linear equations where the right hand sides are from the initial conditions.

c1 + c2 = 0,

c3 = 1,

-c2 = 2

Hence, c3 = 1, c2 = -2, and c1 = 2.

Points off:  -10 for the wrong general solution and no constants and -5 for either the wrong general solution or the wrong constants for your general solution.

3. y' = et + t, y(0) = 1.

First we need to determine the solution, which is just the integral of et + t (note that there is no y dependence to make this hard).  Hence, y(t) = et + t2/2.  Next we need to compute a table.

t Solution Heun Runge-Kutta Adams-Moulton
0.0 1.0000000 1.0000000 1.0000000 1.0000000
0.1 1.1101709 1.1102585 1.1107092 1.1107092
0.2 1.2414028 1.2415872 1.2414028 1.2414028
0.3 1.3948588 1.3951503 1.3948588 1.3948588
0.4 1.5718247 1.5722344 1.5718247 1.5718250
0.5 1.7737213 1.7742618 1.7737213 1.7732201

Just by inspection, we see that the Runge-Kutta method is the most accurate numerical method for this problem.  We can actually quantify this by looking at the 2-norm of the difference between the exact solution at the mesh points and the computed values.  Specifically, if we define

|| y(t) - Y(t) || = ( S (yi - Yi )2 )1/2, Y a computed solution,

then we can determine the 2-norms:

Heun 7.660x10-4
Runge-Kutta 3.191x10-8
Adams-Moulton 8.236x10-7
Points off:  -5 for missing the exact solution, each missing method, missing the best method, -2 for the wrong best method (you had to have an argument for one), wrong numerical values for the exact solution or each of the methods, and -1 each for not having a table that went far enough on the number line.

 

Cheers,
Craig C. Douglas

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