Question

Diff. Eq.

Looking to see if someone can answer this question or just a part of it. :)

Please see attached picture.

Answer 1

I think I can do a.

Answer 2

The solution is of the form: \[y=c_1 e^{\lambda_1}+c_2 e^{\lambda_2}\]

With c1 and c2 constants and lambda1 and lambda2 the roots.

Answer 3

\[y=c_1 e^{\lambda_1x}+c_2 e^{\lambda_2x}\]

Is better i suppose

Answer 4

if c1 and c2 are zero, then the solution is zero everywhere.

Answer 5

if c1 and c2 are not zero, then the solution can be zero at most in one point, but that needs an argument that I don't see right now.

Answer 6

That's clear. Suppose without loss of generality that c1 < 0 < c2.

Then y = 0 if and only if
\[ e^{(\lambda_1-\lambda_2)x} = - \frac{c_2}{c_1} > 0 \]

As the LHS (left hand side) is strictly monotonically increasing or decreasing, this equation must only have one solution.

Note if c1 and c2 have the same sign, then this equation has no solutions. Hence y = 0 has at most one solution in the case where the roots of the characteristic equation of the ODE are distinct and real.

Answer 7

Yeah, sounds good.

Answer 8

There's one other possibility for part (a) of this question: the roots of the characteristic equation are real and not distinct. Suppose \( \lambda \) is a repeated root. In that case, the solution is
\[ y = c_1 e^{\lambda x} + c_2 xe^{\lambda x} \]
You can also show (although it's slightly less obvious) that this equation also has at most one root.

Answer 9

Makes sense. :).
i'm quite having trouble seeing if the equation is monotonically increasing or decreasing.

Answer 10

e^ax is monotonically increasing if a > 0
and is mono. decreasing if a < 0

As lambda_1 and lambda_2 are not equal by hypothesis, their difference must be positive or negative.

Answer 11

oh i see. Now it's all much clearer.

Answer 12

lol
i total understand (i was lying)