Is it possible solve equation?`

Question

Is it possible and how to solve equation: $$p\cdot\dfrac{-32}{1+64(1-z)}+\sqrt{1+64\cdot(1-z)}\dfrac{dp}{dz}=\dfrac{-32\cdot\pi\cdot8}{\sqrt{1+64(1-z)}}$$ I need $p$ as results and $p=p(z)$. I am confused because I cannot extract variables on one side of equations? How else I can solve it?

Answer 1

https://www.wolframalpha.com/ with

$\displaystyle p*\frac{-32}{1+64(1-z)} + p’*\sqrt{1+64(1-z)} + \frac{32*\pi*8}{ \sqrt{1+64(1-z)}} = 0 $

gives:

$$p(z)=(C-8\pi Ei(-\frac{1}{\sqrt{65-64z}})) e^{\frac{1}{\sqrt{65-64z}}}$$

with $\enspace\displaystyle Ei(x):=\int_{-\infty}^x \frac{e^t}{t}dt=\gamma +\ln |x| + \sum\limits_{n=1}^\infty\frac{x^n}{n\cdot n!}\enspace$ ($\gamma\,$ is the Euler-Mascheroni constant)

and $\enspace\displaystyle C=p(0)e^{-\frac{1}{\sqrt{65}}}+8\pi Ei(-\frac{1}{\sqrt{65}}) $

or a bit nicer $\enspace\displaystyle C=p(1)e^{-1}+8\pi Ei(-1) $

Note: $\enspace$ Test by derivation.

Look for literature about linear differential equation of the first order (with the variation of the parameters). E.g.: https://www.math.ucdavis.edu/~thomases/W11_16C1_lec_1_7_11.pdf

Answer 2

$p\cdot\dfrac{-32}{1+64(1-z)}+\sqrt{1+64\cdot(1-z)}\dfrac{dp}{dz}=\dfrac{-32\cdot\pi\cdot8}{\sqrt{1+64(1-z)}}$

Divide throughout by $ \sqrt{1+63(1-z)}$

$p\cdot \frac{-32}{(1+64(1-z))^\frac32}+\dfrac{dp}{dz} = \dfrac{-32\cdot 8\pi}{(1+64(1-z))}$

This is a first order linear differential equation ;

The integrating factor is ;

$I = \displaystyle e^{\displaystyle \int\dfrac{-32}{(1+64(1-z))^\frac32}\,dz}$

Let $w = 1+64(1-z) \implies dw = -64dz$

$I = e^{\displaystyle -32\cdot\frac1{-64}\int\frac{1}{w^\frac32}\,dw}$

$I = e^{\frac12\cdot (-2)w^{-\frac12}}$

$I = e^{-(1+64(1-z))^{-\frac12}}$

Multiply the original equation by $I$ ;giving

$I\cdot p\cdot \frac{-32}{(1+64(1-z))^\frac32}+I\dfrac{dp}{dz} = I\cdot\dfrac{-32\cdot 8\pi}{(1+64(1-z))}$

$(I.p)' =I\cdot\dfrac{-32\cdot 8\pi}{(1+64(1-z))}$

$I\cdot p = \displaystyle\int I\cdot\dfrac{-32\cdot 8\pi}{(1+64(1-z))}\,dz $

$I\cdot p = \displaystyle\int e^{-(1+64(1-z))^{-\frac12}}\cdot\dfrac{-32\cdot 8\pi}{(1+64(1-z))}\,dz $

To solve the integral on the RHS ;

let $u = 1+64(1-z)\implies du =-64\,dz$

$J = \displaystyle\int e^{-(u)^{-\frac12}}\cdot\dfrac{-32\cdot 8\pi}{(u)}\,\frac{du}{-64}$

$J = 4\pi\displaystyle\int\frac{e^{-\frac1{\sqrt u}}}{u}\,du$

unfortunately the above (To the best of my knowledge ) cannot be expressed in terms of elementary integrals

Plugging it into Wolfram Alpha gives ;

$J = -8\pi$ Ei$\big[\frac{-1}{\sqrt u}\big]+C$

where Ei is the Exponential integral

So finally we have ;

$I\cdot p = -8\pi$ Ei$\big[\frac{-1}{\sqrt{1+64(1-z)}}\big]+C$

Answer 3

Introduce the following substitution:

$$1+64(1-z)=u^2 \implies\frac{du}{dz}=-\frac{32}{u}$$ $$\frac{dp}{dz}=\frac{dp}{du}\frac{du}{dz}=-\frac{32}{u}\frac{dp}{du}$$

Your differential equation becomes:

$$p \frac{-32}{u^2} + u \frac{-32}{u}\frac{dp}{du}=-\frac{8 \cdot32\pi}{u}$$

or after simplification:

$$ p + u^2\frac{dp}{du}=8\pi u$$

Another substitution:

$$p=e^{\frac1u}q$$

$$\frac{dp}{du}=-\frac{1}{u^2}e^{\frac1u}q+e^{\frac1u}\frac{dq}{du}$$

Now the equation becomes:

$$e^{\frac1u}q+u^2\left(e^{\frac1u}\frac{dq}{du}-\frac{1}{u^2}e^{\frac1u}q\right) =8\pi u$$

$$\frac{dq}{du}=\frac{8 \pi}{u}e^{-\frac1u}$$

$$q(u)=8\pi\int\frac1u e^{-\frac1u}du=C_1 -8\pi Ei(-\frac1u)$$

$$p(u)=e^\frac1u (C_1 - 8\pi Ei(-\frac1u))$$

Now just replace $u$ with $\sqrt{1+64(1-z)}$:

$$p(z)=e^\frac{1}{\sqrt{65-64z}} (C_1 - 8\pi Ei(-\frac{1}{\sqrt{65-64z}}))$$

Answer 4

Calling $x = 1+64(1-z) \Rightarrow dx = - 64 dz$ we have

$$ -\frac{32}{x}p+\sqrt{x}\frac{dp}{dx}\frac{dx}{dz}=-\frac{32\pi 8}{\sqrt x} $$

or

$$ -\frac{32}{\sqrt{x}}p-64x p'=-32\pi 8 $$

This is a non homogeneous linear differential equation with solution

$$ p = p_h+p_p\\ \frac{c_1}{\sqrt{x}}p_h+c_2x p_h'=0\\ \frac{c_1}{\sqrt{x}}p_p+c_2x p_p'=c_3 $$

The solution for $p_h = C_0 e^{\frac{2c_1}{c_2\sqrt x}}$. Now proposing

$p_p = C_0(x) e^{\frac{2c_1}{c_2\sqrt x}}$ we obtain

$$ C'_0(x) = \frac{c_3 e^{-\frac{2c_1}{c_2\sqrt x}}}{c_2 x} $$

and then

$$ p_p = \left(\int\frac{c_3 e^{-\frac{2c_1}{c_2\sqrt x}}}{c_2 x} dx\right) e^{-\frac{2c_1}{c_2\sqrt x}} $$