What Method Solves the ODE x^3y'+4x^2y=1/x?

[pat666]

Homework Statement

x^3y'+4x^2y=1/x

Homework Equations

NA

The Attempt at a Solution

I've tried separation of variable but I can't get the ys on 1 side and the xs on the other.

Please help the exam is soon and I don't know what method to use?

 

[vela]

It's a first-order differential equation, so you can always use an integration factor.

 

[pat666]

How do I do that for the non homogenous ODE?

 

[SteamKing]

As always, find solution to homogeneous equation. Then find particular solution. Add the two solutions together.

 

[pat666]

Ok, ill try to figure it out. I can't find the y_p value for r(x) though. Do you know what it is for 1/x?

 

[tiny-tim]

hi pat666!

hint: the LHS is almost an exact integral, isn't it?

ok, multiply it by something to make it an exact integral (that's vela's integrating factor)

 

[pat666]

Question. what is an exact integral?

 

[tiny-tim]

oops!

i meant an exact derivative ​

 

[hunt_mat]

Suppose you had a general differential equation:
<br /> p(x)\frac{dy}{dx}+r(x)y=r(x)<br />
The first thing to do is divide through by p(x) to obtain:
<br /> \frac{dy}{dx}+\frac{q(x)}{p(x)}y=\frac{r(x)}{p(x)}<br />
Then the trick is to multiply through by:
<br /> e^{\int\frac{q(x)}{p(x)}dx}<br />
and note that:
<br /> \frac{d}{dx}e^{\int\frac{q(x)}{p(x)}dx}=e^{\int \frac{q(x)}{p(x)}dx}\frac{q(x)}{p(x)}<br />
and so the LHS of the equation can be written as:
<br /> \frac{d}{dx}\left( e^{\int\frac{q(x)}{p(x)}dx}y\right) =\frac{r(x)}{p(x)}e^{\int\frac{q(x)}{p(x)}dx}<br />
From this I think you can solve your equation.

 

[HallsofIvy]

hunt_mat gave the general formula for the integrating factor. I perfer to think like this:
We are looking for a function u(x) so that multiplying by u(x) makes the left side of the equation,
x^3u(x)y&#039;+ 4x^2u(x)y
an "exact derivative". That is, it is of the form
\frac{d(x^3u(x)y)}{dx}= x^3u(x)y&#039;+ 4x^2u
by the product rule, that is the same as
x^3u(x)y&#039;+ 3x^2uy+ x^3u&#039;y= x^3u(x)y&#039;+ 4x^2u
so we want
3x^2u+ x^3u&#039;= 4x^2u

Then
x^3u&#039;= x^2u
which is a separable equation

\frac{du}{u}= \frac{dx}{x}
Which is easily integrable.

 

[pat666]

I still can't do this, when I try hund_mat's method I get x^4*y'+4x^3*y=1 I don't know how to solve this either?

 

[pat666]

I need help urgently/ ODE

Homework Statement

Solve this ODE

x^3y&#039;+4x^2y=1/x

Separation of variables won't work and I can't find an integrating factor

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

hi pat666!

(try using the X² icon just above the Reply box )

hint: what is d/dx (x⁴y) ?

 

[pat666]

That is 4x^3*y but where did that come from and what do I do with it?

Thanks

 

[tiny-tim]

ahhh!

nooo … use the product rule ​

 

[pat666]

I'm not sure what you mean? how can the answer change by using a different rule?

 

[tiny-tim]

you didn't differentiate the "y"

 

[ehild]

This is a linear first order differential equation. Substitute y by the product f˙g.

ehild

 

[pat666]

What is f and g? Does that mean by the product of d/dx(x^3)*?

Thanks

 

[ehild]

f(x) and g(x) are two arbitrary functions.
You have a linear differential equation y'+a(x)y=b(x) (The comma ' means d/dx)
Replace y=f(x)g(x), y'=f'g+fg':
f'g+fg'+a(x)fg=b(x), and choose f in such way that f'g+a(x)fg=0. Eliminate g. Solve for f (you need a particular solution only) and plug it into the rest of the original equation:
fg'=b(x) and find the general solution for g(x).
Try. :)

ehild

 

[TylerH]

Is it an exact differential equation? I looks like, if you multiple by x, you get x⁴y'+4x³y=1, let f(x, y)=x⁴y, and you can see df/dx=x⁴y'+4x³y=1.

 

[hunt_mat]

what he's saying is:
<br /> \frac{d}{dx}(x^{4}y(x))=4x^{3}y(x)+x^{4}\frac{dy}{dx}<br />

 

[hunt_mat]

This question was asked before, multiply through by x and notice that you can write the LHS as a total derivative.

 

[berkeman]

Homework Statement

Solve this ODE

x^3y&#039;+4x^2y=1/x

Separation of variables won't work and I can't find an integrating factor

Homework Equations

The Attempt at a Solution

Two threads merged. Please do not multiple post the same question.