Differential equations: Elimination of arbitrary constants

[JasonHathaway]

Homework Statement

Find the differential equation of ln y = ax^2 + bx + c by eliminating the arbitrary constants a, b and c.

Homework Equations

Wrosnkian determinant.

The Attempt at a Solution

I've solved a similar problem (y=ax^2+bx+c --> y'''=0), but couldn't do the same with this one.
All what I could is taking the exponent of both sides --> y=e^(ax^2 + bx + c).

 

[dirk_mec1]

I do not see a DE...

 

[Simon Bridge]

I think this is the calc form of Jeopardy ... you are given the solution to a DE, and you have to find the DE.

Taking the exponential of both sides looks promising - you can use you knowledge of how powers combine to simplify it further or investigate what happens as you differentiate it.

note. y=e^x comes from y'=y

 

[pasmith]

Homework Statement

Find the differential equation of ln y = ax^2 + bx + c by eliminating the arbitrary constants a, b and c.

Homework Equations

Wrosnkian determinant.

The Attempt at a Solution

I've solved a similar problem (y=ax^2+bx+c --> y'''=0), but couldn't do the same with this one.
All what I could is taking the exponent of both sides --> y=e^(ax^2 + bx + c).

Does it not follow from \ln y = ax^2 + bx + c and <br /> \frac{d^3}{dx^3}(ax^2 + bx + c) = 0 that <br /> \frac{d^3}{dx^3}(\ln y) = 0? Some would regard that as an adequate ODE for y; others might insist that you expand the left hand side and re-arrange the result into the form <br /> y&#039;&#039;&#039; = F(y, y&#039;, y&#039;&#039;).<br />

 

[JasonHathaway]

I'm using a method in which I differentiate depending on the number of constants. So, in this case I shall differentiate three times.

y=e^{ax^{2}+bx+c}
y&#039;=e^{ax^{2}+bx+c} (2ax+b)
y&#039;&#039;=e^{ax^{2}+bx+c} (2ax^{2}+4axb+b^{2}+2a)
y&#039;&#039;&#039;=e^{ax^{2}+bx+c} (4a^{2}x^{3}+8^{2}x^{2}b+2axb^{2}+4a^{2}x+2ax^{2}b+4axb^{2}+b^{3}+2ab)

And then I shall put the coefficients of a, b and c in Wrosnkian and then find determinant.

\begin{matrix}<br /> y &amp; * &amp; * &amp; * \\<br /> y&#039; &amp; * &amp; * &amp; * \\<br /> y&#039;&#039; &amp; * &amp; * &amp; *\\<br /> y&#039;&#039;&#039; &amp; * &amp; * &amp; *<br /> \end{matrix}

Where the stars (*) are the coefficients. And that my problem right now, in some terms of y'' and y''' there are a and b together. How can I deal with it?

 

[ehild]

First check if the equations are correct.

Note that you do not have a linear system of equations for a,b,c.
The equations can be divided by y so the exponential factors cancel, and there are three equations to solve.
y&#039;/y=2ax+b
y&#039;&#039;/y=4a^2x^{2}+4axb+b^{2}+2a

y'''/y=...

Isolate b from the first one, substitute for b into the second one, and isolate a. Substitute a and b in terms of y'/y and y"/y into the third one.

But it is much simpler to follow pasmith's hint.


ehild