# Math Help - Solve non-linear differential analytically

1. ## Solve non-linear differential analytically

Hi,

I am trying to solve the following equation analytically. I think the solution shouldnt be that hard but I'm really rusty on these kind of things.

V'' - k*J*V^-1/2 = 0

I have then said u = V' and therefore u' = u(du/dV)

u(du/dV) - k*J*V^-1/2 = 0

now you can use separation of variables but assuming I did it correctly you get a u in the answer and since u = V' you have to start all over.

The goal of this is to prove J proportional to V^3/2

Any help on how to do this is appreciated, got to turn this in tomorrow!

2. ## Re: Solve non-linear differential analytically

Originally Posted by pdizzle0
Hi,

I am trying to solve the following equation analytically. I think the solution shouldnt be that hard but I'm really rusty on these kind of things.

V'' - k*J*V^-1/2 = 0

I have then said u = V' and therefore u' = u(du/dV)

u(du/dV) - k*J*V^-1/2 = 0

now you can use separation of variables but assuming I did it correctly you get a u in the answer and since u = V' you have to start all over.

The goal of this is to prove J proportional to V^3/2

Any help on how to do this is appreciated, got to turn this in tomorrow!
Are you using $\displaystyle V''$ to represent $\displaystyle \frac{d^2V}{dJ^2}$? In other words, is J your independent variable?

3. ## Re: Solve non-linear differential analytically

No, sorry for not making that clear. J is constant. k*J essentially could be considered 1 constant in this context just for the final answer J proportional to V^3/2

The independent variable in this case is x -> differentiating with respect to space

4. ## Re: Solve non-linear differential analytically

Originally Posted by pdizzle0
Hi,

I am trying to solve the following equation analytically. I think the solution shouldnt be that hard but I'm really rusty on these kind of things.

V'' - k*J*V^-1/2 = 0

I have then said u = V' and therefore u' = u(du/dV)

u(du/dV) - k*J*V^-1/2 = 0

now you can use separation of variables but assuming I did it correctly you get a u in the answer and since u = V' you have to start all over.

The goal of this is to prove J proportional to V^3/2

Any help on how to do this is appreciated, got to turn this in tomorrow!
For semplicity we write the DE as...

$y^{''}= \frac{a}{\sqrt{y}}$ (1)

... where a is a constant. Setting $y^{'}=u$ we have...

$y^{''}= \frac{d u}{d x}= \frac{d u}{d y}\ \frac{dy}{dx}= u\ \frac{d u}{dy}$ (2)

... so that the (1) becomes...

$u\ \frac{du}{dy}= \frac{a}{\sqrt{y}}$ (3)

In (3) the variables are separable so that the solution is easily found...

$u^{2}= 4\ a\ \sqrt{y} + c_{1} \implies y^{'}= \pm \sqrt{4\ a\ \sqrt{y} + c_{1}}$ (4)

The (4) is a firdt order DE the solution of which leads us to y... are You able to proceed?...

Kind regards

$\chi$ $\sigma$