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!