Download Practice Explanations: Solutions 1. Suppose y1 and y2 are both

Practice Explanations: Solutions
1. Suppose y1 and y2 are both solutions to the same homogenous, second order, linear differential
equation. Explain why, as long as y1 and y2 are not multiples of each other, that y = c1 y1 +c2 y2
can satisfy any initial condition by choosing c1 and c2 correctly.
Answer: Let’s look at the differential equation y 00 = y. Two solutions to this equation are
y1 = et and y2 = e−t . What the question claims is that, since et and e−t are clearly not
multiples of each other, that y = c1 et + c2 e−t should be able to satisfy any initial condition,
say, at t0 = 0. So, that means, we want to be able to find c1 and c2 so that y(0) = y0 and
y 0 (0) = y00 . These equations, for this example, would reduce to the equations c1 + c2 = y0 and
c1 − c2 = y00 . This kind of system of equations is solvable as long as the rows of the matrix
1 1
1 −1
are not multiples of each other, since we are trying to solve
1 1
c1
y
= 00
1 −1 c2
y0
In this case, it is obvious, and so we can always find c1 and c2 so that y will satisfy the initial
conditions.
This is more thorough than you would have to be, but: In general, the columns are (y1 (0), y10 (0))
and (y2 (0), y20 (0)). If these were multiples of each other, then there would be some number c so
that cy1 would satisfy the same differential equation and initial conditions as y2 . That means
that cy1 = y2 , so they are multiples of each other. So, unless the functions are multiples of
each other, we will always be able to find c1 and c2 so that y satisfies the initial conditions.
2. Suppose the characteristic equation of the differential equation ay 00 + by 0 + cy = 0 has two
(distinct) real roots. When does the solution y(t) go to zero at infinity (i.e., when is lim y(t) =
t→∞
0)? Why do we know this?
Answer: Since this is a constant coefficient differential equation, if r1 and r2 are the roots of
the characteristic equation ar2 + br + c = 0, then y1 = er1 t and y2 = er2 t are solutions of the
differential equation. Since r1 6= r2 , we know that y = c1 er1 t + c2 er2 t is the general solution of
the differential equation. From this equation, if r1 and r2 are real, it is clear that this goes to
zero at infinity as long as r1 and r2 are both negative, since then we have an exponential with
a negative power. As a more unusual circumstance, it would decay to zero as long as one of
the roots, say r1 , was negative, and the other constant, c2 in this case, was zero for the initial
conditions given.
3. Suppose B(x, y) is a function such that Bx (x, y) = M (x, y) and By (x, y) = N (x, y). Explain
dy
why B(x, y) = c is the solution of the differential equation M (x, y) + N (x, y) dx
= 0. Also,
∂
∂
explain why we know that ∂y M = ∂x N .
Answer: If we look at B(x, y) = c, and take the derivative of both sides with respect to x, we
1
get, using the chain rule
d
d
B(x, y) =
c
dx
dx
∂
dx
∂
dy
B(x, y) +
B(x, y)
=0
∂x
dx ∂y
dx
dy
M (x, y) + N (x, y)
= 0,
dx
which is the differential equation. This means that B(x, y) = c implicitly defines the solutions
to the differential equation.
For the second question, remember that derivatives commute, i.e., that fxy = fyx . Another
way to say this is that you can change the order of differentiation. In this case, this means we
can calculate
My = Bxy = Byx = Nx ,
and so My = Nx .
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