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.(adsbygoogle = window.adsbygoogle || []).push({});

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:

[tex]\frac{d}{dx}(xp) = x\frac{d}{dx}p + p

[/tex]

We can write

[tex]\frac{d}{dx}(xp) - x = 0

[/tex]

[tex]\frac{d}{dx}(xp) = x

[/tex]

[tex]xp = \frac{1}{2}x^2 + C

[/tex]

[tex]p = \frac{1}{2}x + \frac{C}{x}

[/tex]

[tex]y' = \frac{1}{2}x + \frac{C}{x}

[/tex]

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

[/tex]

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Analysis graduate school past entrance exam

**Physics Forums | Science Articles, Homework Help, Discussion**