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AlfredFlaßhaar15-Moonstone
15-Moonstone
April 15, 2025
Solved

Strange initial values...

  • April 15, 2025
  • 2 replies
  • 792 views

For some time now, I've been interested in ordinary differential equations of the following type: x*y´´ + y = 0 for the (incomplete) initial value y(0) = 1. These differential equations in the complex plane are discussed extensively in the literature (e.g., Schlömilch, Smirnov 3/2,...) – closed-form solutions exist only using well-known transcendental functions or in series form. Now, however, this linear equation must be considered in the real realm. Is there a way to construct solutions for a sequence of initial values ​​(eps;1) with eps-->0 and to create plots close to the initial value? The second initial value for y´(0) can be chosen "appropriately." My MC14 knowledge is insufficient for this.

Best answer by LucMeekes

Hi Alfred,

 

The general solution to the DE is:

LucMeekes_0-1744751633818.png

Its derivative is:

LucMeekes_1-1744751669945.png

(And the second derivative is:

LucMeekes_2-1744751782098.png

so that:

LucMeekes_3-1744751800307.png

and that's the negative of y(x). which proves that y(x) shown above is a/the solution to the DE.)

 

Let's guess that C0=1 and C1=0 and plot the function and its derivative:

LucMeekes_4-1744751944523.png

Seems like y(0)=0 and y'(0) can be chosen to be 1.

Simple evaluation of y(0,C) results in a singularity error, but a limit does the job:

LucMeekes_5-1744752002878.png

Hope this helps.

Success!
Luc

 

2 replies

L
LucMeekes23-Emerald IV
23-Emerald IV
April 15, 2025

Do you mean that y is a function of x, or are y and x both functions (of t, e.g.)?

 

Luc

A
AlfredFlaßhaar15-MoonstoneAuthor
15-Moonstone
April 15, 2025

It is assumed that y = y(x). This is also indicated by the incomplete initial value y(0) = 0. Perhaps the problem can be solved by substituting y' = z and the resulting first-order linear system? For eps --> 0, the "appropriate" initial value for y'(0) must be kept constant.

L
LucMeekes23-Emerald IVAnswer
23-Emerald IV
April 15, 2025

Hi Alfred,

 

The general solution to the DE is:

LucMeekes_0-1744751633818.png

Its derivative is:

LucMeekes_1-1744751669945.png

(And the second derivative is:

LucMeekes_2-1744751782098.png

so that:

LucMeekes_3-1744751800307.png

and that's the negative of y(x). which proves that y(x) shown above is a/the solution to the DE.)

 

Let's guess that C0=1 and C1=0 and plot the function and its derivative:

LucMeekes_4-1744751944523.png

Seems like y(0)=0 and y'(0) can be chosen to be 1.

Simple evaluation of y(0,C) results in a singularity error, but a limit does the job:

LucMeekes_5-1744752002878.png

Hope this helps.

Success!
Luc

 

    A
    AlfredFlaßhaar15-MoonstoneAuthor
    15-Moonstone
    April 16, 2025

    Thank you, perfect ;-). The initial value y(0) = 1 is therefore not realizable by the general solution.

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