Variable coefficient 2nd order DE

[chipotleaway]

Homework Statement

This an example from Boyce+DiPrima's text on ODEs. Original problem is to solve 2t²y''+3ty'-y=0, given that y₁=1/t is a solution.

But I'm stuck at the part where I'm to solve
2tv''-v'=0,
v being the second unknown solution.

The Attempt at a Solution

In the text, they let w=v' which reduces the equation to
2tw'-w=0
and solve by separation of variables.

I 'm trying to solve it by factoring differential operators (not sure what the method is called), so
v''-\frac{1}{2t}v'=0
D(D-\frac{1}{2t})v=0

Then I let w=(D-\frac{1}{2t})v,
\therefore Dw=0
\therefore w=c_1

\therefore v'-\frac{1}{2t}v=c_1

But this isn't right - what am I doing wrong?

 

[Mark44]

Homework Statement

This an example from Boyce+DiPrima's text on ODEs. Original problem is to solve 2t²y''+3ty'-y=0, given that y₁=1/t is a solution.

But I'm stuck at the part where I'm to solve
2tv''-v'=0,
v being the second unknown solution.

The Attempt at a Solution

In the text, they let w=v' which reduces the equation to
2tw'-w=0
and solve by separation of variables.

I 'm trying to solve it by factoring differential operators (not sure what the method is called), so
v''-\frac{1}{2t}v'=0
D(D-\frac{1}{2t})v=0

This looks like the method of annihilators. I'm not sure, but I don't think it will work if you don't have constant coefficients in your DE.

Then I let w=(D-\frac{1}{2t})v,
\therefore Dw=0
\therefore w=c_1

\therefore v'-\frac{1}{2t}v=c_1

But this isn't right - what am I doing wrong?

It looks like you skipped some steps.
If (D - 1/(2t)) v = 0, then dv/dt - 1/(2t)v = 0, or dv/dt = v/(2t). This is separable, and gives you a solution for (D - 1/(2t)) v = 0, but it doesn't also work for the original DE, y''+3ty'-y=0, at the start of your post.

You didn't show how you got from y''+3ty'-y=0 to 2tv''-v'=0, so I'm wondering if you made a mistake in that work.

 

[chipotleaway]

The original equation is 2t²y''+3ty'-y=0
Let v be the unknown solution and y=vt^-1. Differentiate y and y' and substitute it into the DE, work through the algebra and out pops 2tv''-v'=0.

Here're all the steps I took
2tv''-v'=0
v''-\frac{v'}{2t}=0
D(D-\frac{1}{2t})v=0

Let w=(D-\frac{1}{2t})v, therefore the 3rd line becomes
Dw=0
\therefore w=c_1

Since w=(D-\frac{1}{2t})v, then
v'-\frac{1}{2t}v=c_1

Which I solved using integrating factors (\mu(t)=\frac{1}{\sqrt(t)})

v=2c_1t+c_2\sqrt{t}

 

[Mark44]

Please excuse my copy mistake - I neglected to write the 2t² coefficient of y''.

 

[chipotleaway]

Unfortunately the solution I got doesn't satisfy the original DE

Here's how they did it in the text:
2tv''-v'=-
Letting w=v'
2tw'-w=0

Separating variables and solving for w(t)

\therefore w(t)=v'(t)=ct^{\frac{1}{2}}
\therefore v(t)=\frac{2}{3}ct^{\frac{3}{2}}+k

So the annihilator method misses out on an integration...looks like it's the change of variables makes the difference. hmmm

 

[Mark44]

Yeah, I don't know if annihilators are applicable when you have nonconstant coefficients. What you started out with is called variation of parameters or, in some books, variation of constants.

I followed that through and came up with the solutions.

Given that y = t^-1 is a solution, you're looking for a solution of the form y = vt^-1. Differentiating that twice and substituting into the original equation gets you to this equation:
-v' + 2v''t = 0, or equivalently, 2tv'' - v' = 0 (a lot of terms drop out)

Let u = v', so u' = v''
The last equation a couple of lines up becomes 2tu' - u = 0, or
du/dt = u/(2t)

Separating, we get du/u = (1/2) dt/t ==> ln u = (1/2) ln t = ln (t^1/2)
So u = e^ln(t1/2) = t^1/2

Backing out, since u = v', then we integrate u to get v: v = (2/3)t^3/2

I'm being very cavalier about the constants of integration. In the end, we'll work it back in.

One more step. We have y = (1/t)v, so y = (1/t)(2/3)t^3/2 = (2/3)t^1/2.

The 2/3 factor can be discarded. All I needed was to find another linearly independent function to go with y₁ = 1/t. The other is y₂ = t^1/2.

The general solution to the original DE is y = C₁t^-1 + C₂t^1/2
. As you see, the constants I ignored earlier are back in.

An alternate approach is to assume that solutions are of the form y = t^r. Then y' = rt^{r -1} and r'' = r(r - 1)t^{r - 2}

Substitute y, y', y'' into the original DE to get 2t²r(r - 1)t^{r -2} + 3trt^{r - 1} - t^r = 0

Multiplying things out and collecting terms by powers of r, get get
(2r² - r - 1)t^r = 0
The quadratic can be factored to give r = -1 and r = 1/2, which is in agreement with the previous work.

The type of DE in this problem where you have decreasing powers of the independent variable as the orders of the derivatives decrease, has a name. I think this might be called an Euler equation or maybe an Airy equation. Without checking, I don't know for sure.