Analysis graduate school past entrance exam

[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.

Homework Statement

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<br />

We can write

\frac{d}{dx}(xp) - x = 0<br />

\frac{d}{dx}(xp) = x<br />

xp = \frac{1}{2}x^2 + C<br />

p = \frac{1}{2}x + \frac{C}{x}<br />

y&#039; = \frac{1}{2}x + \frac{C}{x}<br />

y = \frac{1}{4}x^2 + C ln|x| + D<br />

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

 

[lanedance]

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.

Homework Statement

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)

there are differenet meanings to homogenous, I've usually used the follwoing for a 2nd order DE, say you have
a_2(x)y&#039;&#039; + a_1(x)y&#039; + a_0(x)y = f(x)

then the definition 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&#039;&#039; + y&#039; = x

it is not homgoenous due to the x term,

after the variable change you have
xp&#039; + p = x

which is not homogenous for the same reason...

 

[lanedance]

(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?

so starting from
xp&#039; + p = x

then as you say subsititute p = xu, then p' = u + xu', so
x(u + xu&#039;) + xu = x

which gives
x^2 u&#039; + 2xu = x

then
xu&#039; + 2u = 1

so by the definition 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&#039; + 2v = 0

 

[lanedance]

wiki appears to have something about your defn of homogenous
http://en.wikipedia.org/wiki/Homogeneous_differential_equation

I haven't seen this before, so can't really comment on it

 

[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.