graph of solution u intersects graph of solution v

[evinda]

Hi !
Could you give me a hint how to do this exercise:
If the graph of the solution u of the differential equation $\displaystyle y''-4y'+29y=0$ intersects the graph of the solution v of the differential equation $\displaystyle y''+4y'+13y=0$ at the point (0,0),find u,v so that:

$\displaystyle \lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}$

 

[evinda]

My first shot at this makes it look as if a solution doesn't exist.

I get that the solution of u is of the form

y[x] = Exp[-2x](a Cos[3x] + b Sin[3x])

v is of the form

y[x] = Exp[2x](c Cos[5x] + d Sin[5x])

and when you enforce the condition on each that y[0] = 0 you get

u -> y[x] = b Exp[-2x] Sin[3x]

v -> y[x] = d Exp[2x] Sin[5x]

v⁴/u = d⁴/b Exp[10x] Sin⁴[5x]/Sin[3x]

There's no way this is going to a constant as x -> infinity

What progress have you made on the problem. (you ask some pretty hard questions)

I found that the solution of u is $\displaystyle c_{1}e^{2x}sin(5x)+c_{2}e^{2x}cos(5x)$ and that the solution of v is $\displaystyle d_{1}e^{-2x}sin(3x)+d_{2}e^{-2x}cos(3x)$ .

I could't find something else.Would there be a solution,if $\displaystyle x->0$ instead of $\displaystyle \infty$ ??

 

[HallsofIvy]

Hi !
Could you give me a hint how to do this exercise:
If the graph of the solution u of the differential equation $\displaystyle y''-4y'+29y=0$ intersects the graph of the solution v of the differential equation $\displaystyle y''+4y'+13y=0$ at the point (0,0),find u,v so that:

$\displaystyle \lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}$

I can't make heads or tails of this. What do "u" and "v" have to do with y?

 

[evinda]

u and v are solutions to two different differential equations whose trajectories happen to intersect at 0 at t=0

it would have been better notation to say

u''(t) - 4u'(t) + 29u(t) = 0

and

v''(t) + 4v'(t) + 13v(t) = 0

I've seen this sort of verbiage before but usually when referring to different solutions to the same diff eq.

Yes,you are right! Do you have an idea how I could continue?

 

[evinda]

well like I noted above (do people read any of these posts?)

if you enforce the condition that the trajectory of each of these is 0 at t=0 you restrict the solutions to

$\displaystyle u[x] = c_1e^{2x}sin[5x]$
$\displaystyle v[x] = d_1e^{-2x}sin[3x]$

substituting these into your limit of $\displaystyle \displaystyle\frac{v^4}{u}$ we get

$\displaystyle \displaystyle limit_{x\rightarrow\infty}\frac{(d_1e^{-2x}sin[3x])^4}{c_1e^{2x}sin[5x]}$

$\displaystyle \displaystyle limit_{x\rightarrow\infty}\frac{(d_1^4e^{-8x}sin^4[3x])}{c_1e^{2x}sin[5x]}$

$\displaystyle \displaystyle limit_{x\rightarrow\infty}\frac{d_1^4}{c_1}e^{-10x}\frac{sin^4[3x]}{sin[5x]}$

ok, I see in my last post I transposed u and v and ended up with a positive exponential. That's incorrect.

Looking at this limit the exponential term is going to dominate the infinities in the trig term as $\displaystyle x\rightarrow\infty$

but that's just going to force the limit to 0.

I don't see any way to make this converge to a non-zero constant.

Ok,thank you!