# Thread: second-order and no independed variable

1. ## second-order differential equations without independent variable

Dear friends,

I have the following question.

Question. What is the general solution of the differential equation $yy^{\prime\prime}-2(y^{\prime})^{2}=y^{2}\ln(y)$?

I have a solution but it involves an indefinite integral, I wonder if one will be able to give an explicit result.

Thanks.

bkarpuz

2. Originally Posted by bkarpuz
Dear friends,

I have the following question.

Question. What is the general solution of the differential equation $yy^{\prime\prime}-2(y^{\prime})^{2}=y^{2}\ln(y)$?

I have a solution but it involves an indefinite integral, I wonder if one will be able to give an explicit result.

Thanks.

bkarpuz
I remember in my very first homework assignment of my first differential equations course, getting the solution to an integral and then struggling mightely to integrate. When I finally gave up and looked in the back of the book, I found that the answer was given in terms of that integral!

Are you sure you are required to give the answer in "closed form" and not in terms of an integral?

In any case, what is the integral you are having trouble with?

3. ## My solution.

Originally Posted by bkarpuz
Dear friends,

I have the following question.

Question. What is the general solution of the differential equation $yy^{\prime\prime}-2(y^{\prime})^{2}=y^{2}\ln(y)$?

I have a solution but it involves an indefinite integral, I wonder if one will be able to give an explicit result.

Thanks.

bkarpuz
Here is my solution.

Solution. Let $p:=\frac{\mathrm{d}y}{\mathrm{d}x
}$
, then by the chain rule we have $\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}
}=p\frac{\mathrm{d}p}{\mathrm{d}y}$
.
Hence the given differential equation reduces to
$yp\mathrm{d}p-\big(2p^{2}+y^{2}\ln(y)\big)\mathrm{d}y=0.$........(1)
Clearly, this equation is not exact since $\frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}\neq0$, where $U(p,y):=yp$ and $V(p,y):=-\big(2p^{2}+y^{2}\ln(y)\big)$.
The integrating factor for this equation is
$\mu(y):=\exp\bigg\{-\int\frac{\frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}}{U}\mathrm{d}y\bigg\}=\exp\bigg\{-5\int\frac{1}{y}\mathrm{d}y\bigg\}=\frac{1}{y^{5}} .$
Grouping the terms after multiplying through (1) by $\mu$, we get
$\frac{1}{2}\bigg(\frac{2py^{4}\mathrm{d}p-4p^{2}y^{3}\mathrm{d}y}{\big(y^{4}\big)^{2}}\bigg) +\frac{\ln(y)}{y^{3}}\mathrm{d}y=0$
or equivalently
$\frac{1}{2}\mathrm{d}\bigg(\frac{p^{2}}{y^{4}}\big g)+\frac{\ln(y)}{y^{3}}\mathrm{d}y=0.$
From this, we are lead to
$p=\pm\sqrt{c_{1}+\frac{y^{2}}{2}\big(1+2\ln(y)\big )},$
which gives us
$x=\pm\int\frac{1}{\sqrt{c_{1}+\frac{y^{2}}{2}\big( 1+2\ln(y)\big)}}\mathrm{d}y+c_{2}$
by using the relation $\mathrm{d}x=p\mathrm{d}y$.
This is the desired solution to (1)..... $\rule{0.3cm}{0.3cm}$

It seems not possible to solve the last integral.
I think there may be other solutions for this equation which may give explicit result.

4. Originally Posted by bkarpuz
Here is my solution.

Solution. Let $p:=\frac{\mathrm{d}y}{\mathrm{d}x
}$
, then by the chain rule we have $\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}
}=p\frac{\mathrm{d}p}{\mathrm{d}y}$
.
Hence the given differential equation reduces to
$yp\mathrm{d}p-\big(2p^{2}+y^{2}\ln(y)\big)\mathrm{d}y=0.$........(1)
Clearly, this equation is not exact since $\frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}\neq0$, where $U(p,y):=yp$ and $V(p,y):=-\big(2p^{2}+y^{2}\ln(y)\big)$.
The integrating factor for this equation is
$\mu(y):=\exp\bigg\{-\int\frac{\frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}}{U}\mathrm{d}y\bigg\}=\exp\bigg\{-5\int\frac{1}{y}\mathrm{d}y\bigg\}=\frac{1}{y^{5}} .$
Grouping the terms after multiplying through (1) by $\mu$, we get
$\frac{1}{2}\bigg(\frac{2py^{4}\mathrm{d}p-4p^{2}y^{3}\mathrm{d}y}{\big(y^{4}\big)^{2}}\bigg) +\frac{\ln(y)}{y^{3}}\mathrm{d}y=0$
or equivalently
$\frac{1}{2}\mathrm{d}\bigg(\frac{p^{2}}{y^{4}}\big g)+\frac{\ln(y)}{y^{3}}\mathrm{d}y=0.$
From this, we are lead to
$p=\pm\sqrt{c_{1}+\frac{y^{2}}{2}\big(1+2\ln(y)\big )},$
which gives us
$x=\pm\int\frac{1}{\sqrt{c_{1}+\frac{y^{2}}{2}\big( 1+2\ln(y)\big)}}\mathrm{d}y+c_{2}$
by using the relation $\mathrm{d}x=p\mathrm{d}y$.
This is the desired solution to (1)..... $\rule{0.3cm}{0.3cm}$

It seems not possible to solve the last integral.
I think there may be other solutions for this equation which may give explicit result.
I think you should get this

$x = \pm \int\!\frac{dy}{\sqrt{C_1 y^4 - \dfrac{y^2}{2}(1+2\ln y)}}.$

5. Originally Posted by DeMath
I think you should get this

$x = \pm \int\!\frac{dy}{\sqrt{C_1 y^4 - \dfrac{y^2}{2}(1+2\ln y)}}.$
Yes, it seems I had mistakes.
But this integral looks so bad and not possible to integrate up to me...