Chapter 4, Section 4

Answers to selected problems

2. The DE is x' = x*sin t or x' -( sint )*x = 0

Homogenenous DE: The function x(t) = 0 is a solution which means the DE is homogeneous.

Linear DE: The operator is L(x) = x' - (sin t)* x and this is linear by Theorem 4.7 ( In operator form, the DE is L(x) = 0 )

If x(t) is a solution then L(x(t)) = 0; For any such x(t) by linearity L(M*x(t)) = M*L(x(t))= M*0 = 0, so that M*x(t) is also a solution.

3. Any linear, homogeneous DE must have x(t) = 0 as a solution (an equilibrium solution at x = 0)

Direction field (a) does not have an equilibrium solution at x = 0, therefore cannot be that from a linear, homogeneous DE.

Direction field (b) has two solution curves, one starting at and the other starting at approximately . If these were from a linear DE, then their sum would also have to be a solution. Note that and . The sum function would begin at x(0) = 1 - .02 = 0.8 (positive) and end at x(2) = 0.5 - 2 = -1.5 (negative). It is clear from the slope field that no solution can drop below 0; a solution can not cross an equilibrium solution; therefore this slope field is not from a linear (homogeneous ) DE

Direction Field (c) and (d) seem to be fields generated from linear, homogeneous DE's.

6. The DE is or and by Theorem 4.7 we see this in not linear.

The solution to the IVP x(0)= 1 is

say (t) =

The solution to the IVP x(0)= 2 is

The solution to the IVP x(0)= 3 is

The solution to the IVP x(0) = c is

The solution (above) to x(0) = c is not

7. If x' = f(t,x) is a linear , homogeneous DE,

(a) If the DE is linear, then for any function x(t) that is a solution, one can conclude that -1*x(t) is also a solution since the DE must be of the form x' + a(t) * x = 0 or L(x) = 0 ,

where L(x) = x' + a(t) *x.

Now if L(x(t)) = 0 by linearity we have L(-1*x(t)) = -1*L(x(t)) = -1 * 0 = 0

Since evey function which is a solution must also have its negative as a solution, the direction field must reflect this symmetry (symmetry about the t-axis)


(b) By definition of a homogeneous DE, the zero function , x(t) = 0, must be a solution. Also the DE must be of the form x' + a(t) * x = 0 in order to be both linear and homogeneous, or the DE is x' = -a(t) * x and so clearly the only equilibrium solution is x(t) = 0

8. (A) The DE is linear by Theorem 4.7 ( a(t) = 0 and ) and the operator is

(b) If , then , so (The computations are similar for )

(c) IN operator for the DE is L(x) = 0 (with L(x) define in part (a) above), so the superposition principle says that any linear combination of solutions produces another solution, so we have that

is a solution for any a and b.

9. L(x) = x''' - 6x'' + 11x' - 6x

(a) L(x + y) = (x + y)''' - 6(x + y) + 11(x + y) - 6(x + y)

= x''' + y''' -6x'' - 6y'' + 11x' + 11y' - 6x - 6y

= x''' - 6x'' + 11x' - 6x + y''' - 6y'' + 11y' - 6y

= L(x) + L(y)

and L(k*x) = (k*x)''' - 6(k*x)'' + 11(k*x)' - 6(k*x)

= k*x''' - k*6x'' + k*11x' - k*6x

= k*(x''' - 6x'' + 11x' - 6x)

= k*L(x) therefore L(x) is linear

(b) L(0) = 0''' - 6*(0)'' + 11*(0)' - 6*(0) = 0 and so L(x) is homogeneous