A relatively simple ODE problem

[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!

Homework Statement

x³ + 3y -xy' = 0.

Homework Equations

The Attempt at a Solution

stuck

 

[Dick]

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

 

[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) = e^{integral(-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)

 

[vela]

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

 

[bennyska]

cool, thanks so much.

 

[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.

 

[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.

 

[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

[tex]\frac{1}{v²}[/tex] 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?

 

[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.

 

[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) = e^{int. 0 dx} = e^c0 = 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.

 

[vela]

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)

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

 

[vela]

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) = e^{int. 0 dx} = e^c0 = 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.

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