- #1
agro
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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.
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!
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!