Analysis graduate school past entrance exam

1. Jul 30, 2010

agro

Hello there. I'm studying for graduate school entrance exam (nagoya university), and analysis is part of it. I've learned calculus on my undergraduate course but since I didn't get differential equations, I'm kinda learning it by myself right now.

The questions are from past problems which are publicly available. I can solve some already (like second order homogenous linear ones with constant coefficients), but I'm stuck on some. I would really appreciate some help, even if only hints or keywords so that I can search more about it myself.

1. The problem statement, all variables and given/known data

We are supposed to find the general solution y for the given differential equation below:

xy'' + y' - x = 0

However we are to follow the following procedure:

a) By substituting p = y', express p' in terms of p and x
b) Make the equation in (a) homogeneous by setting p=xu. Then solve the resulting differential equation and express u in terms of x
c) Using the result in (b) and the fact that p = xu = y', give the general solution for the original differential equation (express y in terms of x)

(the original question is in Japanese. This is my best translation attempt)

2. The attempt at a solution

(a)

if p = y', then clearly p' = y''. So

xy'' + y' - x = 0 implies
xp' + p - x = 0
xp' = x - p
p' = (x - p)/x

My question here is, isn't the equation homogeneous already? if f(x, p) = (x-p)/x, then clearly f(tx, tp) = f(x, p) (my understanding of the definition of homogeneous)

(b)

If we substitute p = xu, then p' = u + xu', so

p' = (x - p)/x implies
u + xu' = (x - xu)/x
u' = (1-2u)/x

Which isn't homogeneous at all! What I'm I doing wrong here?

Note that I can solve the differential equation by deviating from the hinted procedure. From

xp' + p - x = 0

And the fact that:

$$\frac{d}{dx}(xp) = x\frac{d}{dx}p + p$$

We can write

$$\frac{d}{dx}(xp) - x = 0$$

$$\frac{d}{dx}(xp) = x$$

$$xp = \frac{1}{2}x^2 + C$$

$$p = \frac{1}{2}x + \frac{C}{x}$$

$$y' = \frac{1}{2}x + \frac{C}{x}$$

$$y = \frac{1}{4}x^2 + C ln|x| + D$$

How do I solve it by following the procedure? Thanks!

2. Jul 31, 2010

lanedance

there are differenet meanings to homogenous, i've usually used the follwoing for a 2nd order DE, say you have
$$a_2(x)y'' + a_1(x)y' + a_0(x)y = f(x)$$

then the defintion of homogenous is that f(x) = 0

this means if you think of the differential equation as an operator, L, it is linear in y, so let
$$\hat{L} = a_2(x)\frac{\partial^2}{\partial y^2} +a_1(x)\frac{\partial}{\partial y} + a_0(x)y$$

it is linear as
$$\hat{L} (y_1 + y_2) = \hat{L} y_1 + \hat{L} y_2$$

which is not the case with the x term

if you write the original DE as
$$xy'' + y' = x$$

it is not homgoenous due to the x term,

after the variable change you have
$$xp' + p = x$$

which is not homogenous for the same reason...

3. Jul 31, 2010

lanedance

so starting from
$$xp' + p = x$$

then as you say subsititute p = xu, then p' = u + xu', so
$$x(u + xu') + xu = x$$

which gives
$$x^2 u' + 2xu = x$$

then
$$xu' + 2u = 1$$

so by the defintion I gave, this is still not homogenous, but it is very close, if you let u = v+1/2, then u' = v' and you get
$$xv' + 2v = 0$$

4. Jul 31, 2010

lanedance

Last edited: Jul 31, 2010
5. Jul 31, 2010

agro

Thanks for the reply lanedance. Your explanation from the viewpoint of operator is interesting.

Anyway I tried following the substitution as is, and after playing around a bit I can transform it into a separable form...

u' = (1-2u)/x
du/dx = (1/x)/(1/(1-2u))
$$\frac{1}{x}dx + \frac{1}{2u-1}du = 0$$

Integrating both sides and performing back substitutions, we will get the same result but with a more tedious process.

Last edited: Jul 31, 2010