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## Homework Statement

(1): [tex]\frac{d^{2}y}{dx^{2}} + 4\frac{dy}{dx} + 20y = 10e^{-2x}\cos (rx)[/tex], [tex]y(0) = 1[/tex], [tex]y'(0) = 0[/tex]

Show that there is precisely one value for

*r*for which [tex] e^{2x}y(x)[/tex] can become arbitrarily large where [tex]y(x)[/tex] is a solution of (1)

## The Attempt at a Solution

I used Maple to solve the DE and found a particular solution [tex]y(x)[/tex].

[tex] e^{2x}y(x) = -\frac{10(-\cos 4x + \cos rx)}{-16+r^2}[/tex]

When

*r*tends to 4 or -4 then [tex]e^{2x}y(x)[/tex] can become arbitrarily large, but this doesn't seem to me like it satisfies the question - it's not "precisely one value" and it's not an exact value of

*r*. I graphed [tex] e^{2x}y(x)[/tex] and can't see anywhere else it can become arbitrarily large. I feel like I'm missing something, can anyone point me in the right direction? Cheers.