Elimination of arbitrary constants

[SilverKing]

Hi everyone,

I'm facing some troubles with eliminating constants to make the differential equation from this ordinary equation y=ax^2 - bx + c, where a, b and c are constants.

I'm familiar with eliminating two constants at most like the following example:

Determine the differential equation which has the general solution y=c1 e^2x + c2 e^3x, where c1 and c2 are arbitrary constants.

Solution:

Since we have two arbitrary constants, we differentiate y twice:

y'=2 c1 e^2x + 3 c2 e^3x
y''=4 c1 e^4x + 9 c2 e^3x

Using Wronskian determinant to eliminate the constant:

y 1 1
y' 2 3
y'' 4 9

y(18-12)-y'(9-4)+y''(3-2)=0

y''-5y'+6y=0

So, depending on the preceding example, and since we have three arbitrary constants, we differentiate three time:
y=ax^2 - bx + c
y'=2ax -b
y''=2a
y'''=0

So, the Wrosnkian would be:

y x^2 -x 1
y' 2x -1 0
y'' 2 0 0
y''' 0 0 0

Which leads to: 2y'''=0

Is that correct?

 

[Deleted member 4993]

Hi everyone,

I'm facing some troubles with eliminating constants to make the differential equation from this ordinary equation y=ax^2 - bx + c, where a, b and c are constants.

I'm familiar with eliminating two constants at most like the following example:



So, depending on the preceding example, and since we have three arbitrary constants, we differentiate three time:
y=ax^2 - bx + c
y'=2ax -b
y''=2a
y'''=0

So, the Wrosnkian would be:

y x^2 -x 1
y' 2x -1 0
y'' 2 0 0
y''' 0 0 0

Which leads to: 2y'''=0

Is that correct?

yes

y"' = 0 → y" = 2a → y'= 2ax + b → y = ax² + bx +c

 

[SilverKing]

Thank you for your response.

A little help is needed if you may:

What if I have ln y = ax^2 - bx + c?

Should I take the exponent of both sides? [ e^(ln y) = e^(ax^2 - bx + c --> y = e^(ax^2 - bx + c) ]
But how can I deal with then? Is there another way?

 

[HallsofIvy]

You could but I wouldn't. Instead differentiate both sides as it stands:
\(\displaystyle \frac{1}{y}y'= 2ax- b\)

Differentiate again, using the product rule on the left:
\(\displaystyle \left(\frac{-1}{y^2}\right)y'^2+ \frac{1}{y}y''= 2a\)

Finish by differentiating one more time.