Differential equation, what method should I use?

[tarmon.gaidon]

Homework Statement

Solve the D.E.:

6xy - y³ = (3xy²-3x²-4y)y'

Not really looking for much here. I am just not sure what method I should use to solve this.

Also how would you use Laplace Transforms to solve this tx'' + (3t-1)x' + 3x =0?

Thanks,
Rob

 

[Mark44]

Homework Statement

Solve the D.E.:

6xy - y³ = (3xy²-3x²-4y)y'

Not really looking for much here. I am just not sure what method I should use to solve this.

What have you tried? What methods do you have to choose from?

Also how would you use Laplace Transforms to solve this tx'' + (3t-1)x' + 3x =0?

Thanks,
Rob

 

[LCKurtz]

Homework Statement

Solve the D.E.:

6xy - y³ = (3xy²-3x²-4y)y'

Not really looking for much here. I am just not sure what method I should use to solve this.

Try writing it as M(x,y)dx + N(x,y)dy = 0 and see if anything comes to you.

 

[gomunkul51]

as said by LCKurtz the first equation can be written as:

<br /> (6xy-y^3)+(4y+3x^2-3xy^2)\frac{\mathrm{dy} }{\mathrm{d} x}=0<br />

first order exact ODE rings a bell?

As to the second ODE I don't think it could be done in this form by the Laplace Transform, as Laplace Transform is (as far as I studied) applicable only to differential equations with constant coefficient.

The second ODE could be done with a series solution.

 

[LawlQuals]

Homework Statement

Solve the D.E.:

6xy - y³ = (3xy²-3x²-4y)y'

Not really looking for much here. I am just not sure what method I should use to solve this.

Also how would you use Laplace Transforms to solve this tx'' + (3t-1)x' + 3x =0?

Thanks,
Rob

While this was posted over a month ago, I did have some words to say on it.

As stated, please investigate exact differentials, this equation is exact all ready as you may verify on your own.

Regarding the second query, there is certainly a way to Laplace transform differential equations with variable coefficients. I will not spell it entirely out for you, but will do one of the terms for you so you can see the general idea.

Let L be the laplace transform operator.

Then, by definition, L\{ty&#039;&#039;\} = \int_0^{\infty} e^{-st}(ty&#039;&#039;)dt

To facilitate the integration, the observation is made that

e^{-st}(ty&#039;&#039;) = -\frac{d}{ds}e^{-st}y&#039;&#039;

Such that,

L\{ty&#039;&#039;\} = \int_0^{\infty} e^{-st}(ty&#039;&#039;)dt = -\frac{d}{ds} \int_0^{\infty}e^{-st}y&#039;&#039; dt = -\frac{d}{ds}L\{y&#039;&#039;\}

where you may readily obtain the laplace transform of the second derivative, or confer with tables. The end result of Laplace transforming a differential equation with variable coefficients is to obtain a new differential equation in the Laplace transform variable s that is presumably easier to solve (e.g. maybe it reduces to a Cauchy-Euler type ODE). The rest of the terms in your original ODE are transformed thusly.