Analysis graduate school past entrance exam

In summary: I think the substitution u = v + 1/2 is the best way to solve it.Another thing, I just realized that the original question asks for the general solution, meaning it should include a constant. So the final answer would be y = C1x + C2x ln|x| + x^2/4. In summary, The speaker is studying for a graduate school entrance exam, specifically in analysis. They are trying to learn differential equations by themselves but are having difficulty with some problems. They are asking for help with one particular problem, which involves finding the general solution for a given differential equation using a specific procedure. The speaker is unsure if the equation is already homogeneous and is struggling to solve it using the given procedure
  • #1
agro
46
0
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:

[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!
 
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  • #2
agro said:
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
[tex] a_2(x)y'' + a_1(x)y' + a_0(x)y = f(x) [/tex]

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
[tex] \hat{L} = a_2(x)\frac{\partial^2}{\partial y^2} +a_1(x)\frac{\partial}{\partial y} + a_0(x)y [/tex]

it is linear as
[tex] \hat{L} (y_1 + y_2) = \hat{L} y_1 + \hat{L} y_2 [/tex]

which is not the case with the x term

if you write the original DE as
[tex] xy'' + y' = x[/tex]

it is not homgoenous due to the x term,

after the variable change you have
[tex] xp' + p = x[/tex]

which is not homogenous for the same reason...
 
  • #3
agro said:
(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
[tex] xp' + p = x[/tex]

then as you say subsititute p = xu, then p' = u + xu', so
[tex] x(u + xu') + xu = x[/tex]

which gives
[tex] x^2 u' + 2xu = x[/tex]

then
[tex] xu' + 2u = 1[/tex]

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
[tex] xv' + 2v = 0[/tex]
 
  • #5
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))
[tex]\frac{1}{x}dx + \frac{1}{2u-1}du = 0[/tex]

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

What is the format of the entrance exam for analysis graduate school?

The format of the entrance exam for analysis graduate school varies depending on the institution. However, it typically includes a combination of multiple-choice questions, short answer questions, and essay questions that test your knowledge and understanding of fundamental analysis concepts.

What topics are covered on the analysis graduate school entrance exam?

The topics covered on the analysis graduate school entrance exam also vary by institution, but they generally include calculus, linear algebra, differential equations, and real analysis. Some exams may also include topics such as complex analysis, functional analysis, and topology.

How should I prepare for the analysis graduate school entrance exam?

To prepare for the analysis graduate school entrance exam, it is important to review fundamental concepts in calculus, linear algebra, and differential equations. It is also helpful to practice solving problems and answering questions under timed conditions to improve your speed and accuracy.

Is it necessary to have a strong background in mathematics to pass the analysis graduate school entrance exam?

Yes, a strong background in mathematics is essential to pass the analysis graduate school entrance exam. The exam is designed to test your understanding of advanced mathematical concepts, so a solid foundation in calculus, linear algebra, and other related subjects is crucial.

What is a passing score on the analysis graduate school entrance exam?

The passing score on the analysis graduate school entrance exam varies by institution. Some schools may have a set minimum score required for admission, while others may consider a combination of your exam score, academic record, and other factors in the admissions process.

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