# Homework Help: A relatively simple ODE problem

1. Jan 28, 2010

### bennyska

So many years ago, i took elementary differential equations. unfortunately, i've forgotten too much of it, and now i'm in a not so elementary class. i'm confident i'll catch up to where i was, but at the moment i'm stuck on a relatively simple problem.

i'm having trouble separating them (not sure if this is separable). maybe if someone could give me a pointer to get started, i would be very appreciative. thanks!
1. The problem statement, all variables and given/known data

x3 + 3y -xy' = 0.

2. Relevant equations

3. The attempt at a solution
stuck
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 28, 2010

### Dick

It doesn't look separable to me. Try finding an integrating factor. Does that ring a bell?

3. Jan 29, 2010

### bennyska

okay, so i'm not getting the correct answer, but here's the work i've done so far:
x3 + 3y -xy' = 0
3y - xy' = -x3
x(3y/x) - y' = -x3
-y' + 3/x*y = -x2

let mu(x) = eintegral(-3/x) = e-3 ln x = 1/x3

d/dx[1/x3*y] = 1/x3* x2
1/x3*y = x-1
y=x2

is this the right path?

editing:
i saw i didn't integrate the left hand side at x-1. afterwards, it seemed to work.
y = x3ln(x)

Last edited: Jan 29, 2010
4. Jan 29, 2010

### vela

Staff Emeritus
Almost. You forgot the constant of integration, which will result in another term, cx3, which is the solution to the homogeneous equation.

5. Jan 29, 2010

### bennyska

cool, thanks so much.

6. Jan 29, 2010

### bennyska

it's slowly coming back to me. okay, linear equations, put them in the proper form, use integrating factor, got it.
now i'm kind of stuck again with a problem from the same set.

xy + y2 - x2y'=0.

so this isn't separable, and it's not linear, and i'm kind of stuck. any help, or advice where to start looking? gracias.

7. Jan 29, 2010

### Dick

If you divide both sides by x^2, you've got y/x+(y/x)^2-y'=0. Suggests that a change of variable v=y/x might help.

8. Jan 29, 2010

### bennyska

how's this:

$$\frac{y}{x}$$ + ($$\frac{y}{x}$$)2 - y' = 0
let v= $$\frac{y}{x}$$, with $$\frac{dy}{dx}$$ = v + x$$\frac{dv}{dx}$$

v + v2 - v - x$$\frac{dv}{dx}$$ = 0
v2 - x$$\frac{dv}{dx}$$ = 0

$$\frac{1}{v2}$$ dv = $$\frac{1}{x}$$dx

integrate, get

v = $$\frac{-1}{ln x}$$

substitute and solve for y

y = x ($$\frac{-1}{ln x}$$

and do i just throw a constant on at the end?

9. Jan 29, 2010

### Dick

It's pretty good. But you don't just "throw a constant on at the end". You put in a constant at the point you integrate then carry it through.

10. Jan 30, 2010

### bennyska

yeah, i kind of figured that, it just seems that that's how it usually works out, and i can't remember to put them in when they come up.
so i'm having another hard time working out a different problem.

(for future reference, should i start a new thread if i have a different problem?)

so it's a bernoulli equation
4xy2 + y' = 5x4y2
y' = 5x4y2 - 4xy2
y' = y2( 5x4 - 4x)

let n = 2, p(x) = 0, g(x) = 5x4 - 4x, v = y-1
divide both sides by y2
y-2y' + 0y = 5x4 - 4x
substitute v= y-1

-dv/dx + 0v = 5x4 - 4x
let mu(x) = eint. 0 dx = ec0 = c1
-c1 dv/dx + 1*0v = c1(5x4 - 4x)
d/dx c1*v = c1(5x4 - 4x)
integrate both sides
c1*v=-c1*x2(x3 + 2)
v = -x2(x3 + 2)
replace v
y-1 = -x2(x3 + 2)
y = 1/ (-x2(x3 + 2))
however, this doesn't seem to work. i'm guessing maybe letting p(x) = 0 might be a problem? i don't see why, it's continuous, but that's the only thing i can think of.

btw, thanks for your help. it's coming back to me. i was able to solve 2 other bernoulli equations and got stuck on this one.

11. Jan 30, 2010

### vela

Staff Emeritus
You're making things hard for yourself. You should be able to see at this point that the equation is separable.

12. Jan 30, 2010

### vela

Staff Emeritus
You dropped a sign or two, and again, you forgot the constant of integration.