difficult non-linear differential equation

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cgiudice

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In the following equation ∂ = d/dt. Find the solutions (or eigenvalues) for:
(∂4 +2 ∂3 -40 ∂2 +64 ∂) Γ + 3 ∂2 Γ ∂2 Γ + 6 ∂2 Γ ∂ Γ - 80 ∂ Γ ∂ Γ + 40 Γ ∂2 Γ + 192 Γ ∂ Γ = 0
Solutions are of the form Γ~eλt.
 
From Read Before Posting: Show your beginning work, or ask a specific question about the exercise, or explain why you're stuck.

You might also consider a better typesetting of the problem.
 
Never learned how to do this. Thought it might be super easy for someone. I do not know how to start this problem. It is one part of a puzzle. How does one even begin to solve this problem?
 
cgiudice said:
Never learned how to do this. Thought it might be super easy for someone. I do not know how to start this problem. It is one part of a puzzle. How does one even begin to solve this problem?
Click to expand...
We're not sure what's the question is due to your typesetting. Maybe post a picture of the question instead?
 
I wonder if this can be solved by expressing them through a combination of derivatives of [imath]\Gamma^2[/imath] and [imath]\Gamma[/imath] ?
To make this more readable for myself and, partially, as a public service, I've rewritten the equation replacing [imath]\Gamma[/imath] with pedestrian [imath]y[/imath]:
[math]y^{\prime\prime\prime\prime} + 2y^{\prime\prime\prime} - 40y^{\prime\prime}+64y^\prime +3{y^{\prime\prime}}^2 + 6y^{\prime\prime}y^\prime -80y^{\prime\prime} + 40 y^{\prime\prime} y + 192 y^\prime y = 0[/math]I've also written down -- out of sheer altruism -- expressions for the first four derivatives of [imath]y^2[/imath]:
[math](y^2)^\prime = 2y^\prime y[/math][math]\frac{1}{2}(y^2)^{\prime\prime} = y^{\prime\prime} y + {y^\prime}^2[/math][math]\frac{1}{2}(y^2)^{\prime\prime\prime} = y^{\prime\prime\prime}y + 3y^{\prime\prime}y^{\prime}[/math][math]\frac{1}{2}(y^2)^{\prime\prime\prime\prime} = y^{\prime\prime\prime\prime}y + 4y^{\prime\prime\prime}y^\prime + 3{y^{\prime\prime}}^2[/math]Disclaimer: I'll be surprised if the above formulas don't have any typos, so be forewarned.
 
blamocur said:
I wonder if this can be solved by expressing them through a combination of derivatives of [imath]\Gamma^2[/imath] and [imath]\Gamma[/imath] ?
To make this more readable for myself and, partially, as a public service, I've rewritten the equation replacing [imath]\Gamma[/imath] with pedestrian [imath]y[/imath]:
[math]y^{\prime\prime\prime\prime} + 2y^{\prime\prime\prime} - 40y^{\prime\prime}+64y^\prime +3{y^{\prime\prime}}^2 + 6y^{\prime\prime}y^\prime -80y^{\prime\prime} + 40 y^{\prime\prime} y + 192 y^\prime y = 0[/math]I've also written down -- out of sheer altruism -- expressions for the first four derivatives of [imath]y^2[/imath]:
[math](y^2)^\prime = 2y^\prime y[/math][math]\frac{1}{2}(y^2)^{\prime\prime} = y^{\prime\prime} y + {y^\prime}^2[/math][math]\frac{1}{2}(y^2)^{\prime\prime\prime} = y^{\prime\prime\prime}y + 3y^{\prime\prime}y^{\prime}[/math][math]\frac{1}{2}(y^2)^{\prime\prime\prime\prime} = y^{\prime\prime\prime\prime}y + 4y^{\prime\prime\prime}y^\prime + 3{y^{\prime\prime}}^2[/math]Disclaimer: I'll be surprised if the above formulas don't have any typos, so be forewarned.
Click to expand...
This seemed like a promising approach at the time, but I still cannot get to the solution. Time to think again.
 
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