# Math 0250 Homework 9

• 3.3 Number 2
Let u1, u2 be functions of t, and let c1, c2 be constants. Then

• 3.3 Number 4

• 3.3 Number 5

L(c1u+c2v) = c1L(u) + c2L(v) = c1f(t)+c2f(t) = (c1+c2)f(t)=f(t)

• 3.4 Number 2
Let v1 = u1+u2 and v2=u1-u2. By the superposition principle, v1 and v2 are both solutions to the differential equation, so to show that they form a basic set of solutions, it's only necessary to show that they're independent. You can either do this directly, or by using the Wronskian. Let's take the direct approach. Suppose that c1v1+c2v2=0, and try to show that the constants c1 and c2 must both be 0. You have

0 = c1(u1+u2) + c2(u1-u2) = (c1+c2)u1 + (c1-c2)u2

Since u1 and u2 are independent, it follows that

Solving for c1 and c2 gives c1=c2=0, and so the functions are independent.

For the other pair of functions, v1=u1 and v2 = u1+u2, you could follow the same procedure as above, but for the sake of variety, let's use the Wronskian instead. Since u1 and u2 form a basic set of solutions, you know that their Wronskian determinant is non-zero. The Wronskian determinant for v1 and v2 is

Subtracting the first column from the second gives

so v1 and v2 are independent, and therefore form a basic set of solutions.

• 3.4 Number 5
The fact that u1 and u2 are both solutions is a direct calculation. You have

t(t-2)u1'' - 2(t-1)u1' + 2u1 = t(t-2) (2) - 2(t-1) (2t-2) + 2(t2-2t+2) = 0

and

t(t-2)u2'' - 2(t-1)u2' + 2u2 = t(t-2) (0) - 2 (t-1) + 2 (t-1) = 0

To verify that these form a basic set of solutions, you must show that they're independent, which can be verified either directly, or by using the Wronskian. The Wronskian determinant is

which vanishes only at t=0 and t=2, so the solutions u1 and u2 are independent. The general solution is therefore

u = c1 u1 + c2 u2 = c1 (t2-2t+2) + c2 (t-1)

• 3.4 Number 6
Verifying that the given functions are all solutions is straightforward. Any two of the four will be independent, and therefore any two form a basic set of solutions. However, the first two, and satisfy the initial conditions for a fundamental set of solutions. The general solution is

• 3.5 Number 1

u = 1 + c1 (t2-2t+2) + c2 (t-1)

• 3.5 Number 2
a
The given initial conditions give

Solving gives c1 = -1 and c2 = 0, so

u = 1 - (t2-2t+2)

b
The given initial conditions give

Solving gives c1 = 0 and c2 = 0, so

u = 1

c
The given initial conditions give

Solving gives c1 = 0 and c2 = 1, so

u = 1 + (t-1) = t

d
The given initial conditions give

Solving gives c1 = -3 and c2 = 4, so

u = 1 - 3 (t2-2t+2) + 4(t-1)

e
The given initial conditions give

Solving gives c1 = 3 and c2 = -10, so

u = 1 +3 (t2-2t+2) -10(t-1)

• 3.6 Number 1
The characteristic equation is

which has roots 3 and -2, so a basic set of solutions to the differential equation are u1 = e3t and u2 = e-2t, and the general solution is

u = c1 u1 + c2 u2 = c1 e3t + c2 e-2t

• 3.6 Number 8
The characteristic equation is

which has roots 2 and -3, so a basic set of solutions to the differential equation are u1 = e2t and u2 = e-3t, and the general solution is

u = c1 u1 + c2 u2 = c1 e2t + c2 e-3t

Plugging in the initial condition gives

Solving gives c1=-2/5, c2=2/5, so

Frank Beatrous
1998-11-17