Constructing a Multiplicative Inverse Differential Equation with Ease

[epkid08]

If I know a n-th order differential equation and a particular solution for it, can I easily construct a separate n-th order differential equation if I want the particular solution to be $$\frac{1}{y_p}$$?

Key word is easily

 

[tiny-tim]

Hi epkid08!

If I know a n-th order differential equation and a particular solution for it, can I easily construct a separate n-th order differential equation if I want the particular solution to be $$\frac{1}{y_p}$$?

Key word is easily

What's y_p?

 

[epkid08]

The differential equation is

$$(x+\alpha)^2y'' + (x + \alpha)y' + a^2y = 0$$

and the particular solution is

$$y_p =c_1\cos(a\ln(x+\alpha)) + c_2\sin(a\ln(x+\alpha))$$

So I want to find a differential equation that has a particular solution $$\frac{1}{y_p}$$.
I've found an 2nd order differential equation that fits what I want, but It's messy and I want to know if there's a faster way to find it than what I did, and also if it simplifies into something similar to the above differential equation.

 

[tiny-tim]

The differential equation is

$$(x^2+\alpha)y'' + (x + \alpha)y' + a^2y = 0$$

and the particular solution is

$$y_p =c_1\cos(a\ln(x+\alpha)) + c_2\sin(a\ln(x+\alpha))$$

Isn't that the same as A(x + α)^a + B(x + α)^-a ?

And are you sure the equation doesn't begin (x + α)²y'' ?

 

[epkid08]

Isn't that the same as A(x + α)^a + B(x + α)^-a ?

And are you sure the equation doesn't begin (x + α)²y'' ?

OOPS! Yes, my mistake, it does begin with $$(x+\alpha)^2y''$$.

 

[epkid08]

Isn't that the same as A(x + α)^a + B(x + α)^-a ?

I'm not sure, are you saying $$y_p = A(x + \alpha)^a + B(x + \alpha)^{-a}$$?
Anyways, I'll post the differential equation that fits $$1/y_p$$, and maybe you can help me simplify it, also this is when alpha equals zero and a equals one:

$$x^2\sin(\frac{\pi}{4}+\ln(x))y'' + 2x\cos(\frac{\pi}{4}+\ln(x))y' + [\cos(\frac{\pi}{4}+\ln(x)) - \sin(\frac{\pi}{4}+\ln(x))]y = 0$$It's very similar to the original differential equation, though it seems like it could be simplified!

 

[matematikawan]

If I know a n-th order differential equation and a particular solution for it, can I easily construct a separate n-th order differential equation if I want the particular solution to be $$\frac{1}{y_p}$$?

Key word is easily

Just let $y_p=\frac{1}{Y}$ and form the DE that is satisfies by Y(x).

Key word is easy!

 

[tiny-tim]

Hi epkid08!

First thing: those are not particular solutions, they are general solutions …

If P(y) = 0 has solutions Af(x) + Bg(x),

then P(y) = q(x) has solutions Af(x) + Bg(x) + h(x), where h(x) is any solution of the whole equation.

In other words, the general solution of the whole equation is the general solution of the "incomplete" equation (with 0 on the RHS), plus any particular solution of the whole equation.

I'm not sure, are you saying $$y_p = A(x + \alpha)^a + B(x + \alpha)^{-a}$$?

oops!

i left out an i … A(x + α)^ia + B(x + α)^-ia.

Apart from that, yes …

any combination of cos and sin is also a combination of e^+i… and e^-i…​

$$x^2\sin(\frac{\pi}{4}+\ln(x))y'' + 2x\cos(\frac{\pi}{4}+\ln(x))y' + [\cos(\frac{\pi}{4}+\ln(x)) - \sin(\frac{\pi}{4}+\ln(x))]y = 0$$

well, cosπ/4 = sinπ/4, so you can divide by that throughout (after expanding the brackets), and get either coslnx and sinlnx or just xⁱ and x^-i.

If I know a n-th order differential equation and a particular solution for it, can I easily construct a separate n-th order differential equation if I want the particular solution to be $$\frac{1}{y_p}$$?

Key word is easily

Just let $y_p=\frac{1}{Y}$ and form the DE that is satisfies by Y(x).

Key word is easy!

nice one, matematikawan!