2.6.4 Problem 4 (eq 50)

Internal problem ID [18497]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter IV. Methods of solution: First order equations. section 33. Problems at page 91
Problem number : 4 (eq 50)
Date solved : Monday, March 31, 2025 at 05:37:44 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solved as second order ode by reversing roles of dependent and independent variables

Time used: 1.933 (sec)

Solve

$$\begin{array}{rl}{\varphi }^{\prime \prime }& =\frac{4\pi nc}{\sqrt{v_0^2+\frac{2e(\varphi -V_0)}{m}}}\end{array}$$

reversing the roles of the dependent and independent variables, the ode becomes

$$\begin{array}{r}\frac{d^2}{d{\varphi }^2}x\left(\varphi \right)=-\frac{4\pi nc{\left(\frac{d}{d\varphi }x\left(\varphi \right)\right)}^3}{\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}}\end{array}$$

Which is now solved for $x\left(\varphi \right)$ instead for $\varphi$

Solved as second order missing y ode

Time used: 1.299 (sec)

This is second order ode with missing dependent variable $x$. Let

$$\begin{array}{rl}u(\varphi )& =x^{\prime }\end{array}$$

Then

$$\begin{array}{rl}{u}^{\prime }(\varphi )& =x^{\prime \prime }\end{array}$$

Hence the ode becomes

$$\begin{array}{r}{u}^{\prime }\left(\varphi \right)+\frac{4\pi ncu{\left(\varphi \right)}^3}{\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}}=0\end{array}$$

Which is now solved for $u(\varphi )$ as first order ode.

The ode

$$\begin{array}{}\text{(1)}& u^{\prime }=-\frac{4\pi nc{u}^3}{\sqrt{\frac{v_0^{2}m-2e{V}_0+2e\varphi }{m}}}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{u}^{\prime }& =-\frac{4\pi nc{u}^3}{\sqrt{\frac{v_0^{2}m-2e{V}_0+2e\varphi }{m}}}\\ & =f(\varphi )g(u)\end{array}$$

Where

$$\begin{array}{rl}f(\varphi )& =-\frac{4\pi nc}{\sqrt{\frac{v_0^{2}m-2e{V}_0+2e\varphi }{m}}}\\ g(u)& =u^3\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(u)}du& =\int f(\varphi )d\varphi \\ \int \frac{1}{u^3}du& =\int -\frac{4\pi nc}{\sqrt{\frac{v_0^{2}m-2e{V}_0+2e\varphi }{m}}}d\varphi \end{array}$$

$$\frac{1}{2u^2}=\frac{4\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}\pi ncm}{e}-c_5$$

We now need to find the singular solutions, these are found by finding for what values $g(u)$ is zero, since we had to divide by this above. Solving $g(u)=0$ or

$$u^3=0$$

for $u$ gives

$$\begin{array}{rl}u& =0\end{array}$$

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

$$\begin{array}{rl}\frac{1}{2u^2}& =\frac{4\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}\pi ncm}{e}-c_5\\ u& =0\end{array}$$

In summary, these are the solution found for $u(\varphi )$

$$\begin{array}{rl}\frac{1}{2u^2}& =\frac{4\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}\pi ncm}{e}-c_5\\ u& =0\end{array}$$

For solution $\frac{1}{2u^2}=\frac{4\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}\pi ncm}{e}-c_5$, since $u=x^{\prime }\left(\varphi \right)$ then we now have a new first order ode to solve which is

$$\begin{array}{r}\frac{1}{2{x^{\prime }\left(\varphi \right)}^2}=\frac{4\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}\pi ncm}{e}-c_5\end{array}$$

Solving for the derivative gives these ODE’s to solve

$$\begin{array}{}\text{(1)}& x^{\prime }& =-\frac{\sqrt{2}\sqrt{(4\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}\pi ncm-c_{5}e)e}}{2(4\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}\pi ncm-c_{5}e)}\text{(2)}& x^{\prime }& =\frac{\sqrt{2}\sqrt{(4\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}\pi ncm-c_{5}e)e}}{8\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}\pi ncm-2c_{5}e}\end{array}$$

Now each of the above is solved separately.

Solving Eq. (1)

Since the ode has the form $x^{\prime }=f(\varphi )$, then we only need to integrate $f(\varphi )$.

$$\begin{array}{rl}\int dx& =\int -\frac{\sqrt{2}\sqrt{(4\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}\pi ncm-c_{5}e)e}}{2(4\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}\pi ncm-c_{5}e)}d\varphi \\ x& =-\frac{\sqrt{2}(e^2{c}_5\sqrt{4e\pi mn\sqrt{\frac{2e\varphi }{m}-\frac{-v_0^{2}m+2e{V}_0}{m}}c-e^2{c}_5}+\frac{{(4e\pi mn\sqrt{\frac{2e\varphi }{m}-\frac{-v_0^{2}m+2e{V}_0}{m}}c-e^2{c}_5)}^{3/2}}{3})}{16m{e}^2{\pi }^2{c}^2{n}^2}+c_6\end{array}$$

$$\begin{array}{rl}x& =\frac{-2(\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}\pi ncm+\frac{c_{5}e}{2})\sqrt{2}\sqrt{4cmn\pi e\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}-e^2{c}_5}+24c_{6}me{\pi }^2{c}^2{n}^2}{24em{\pi }^2{c}^2{n}^2}\end{array}$$

Solving Eq. (2)

Since the ode has the form $x^{\prime }=f(\varphi )$, then we only need to integrate $f(\varphi )$.

$$\begin{array}{rl}\int dx& =\int \frac{\sqrt{2}\sqrt{(4\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}\pi ncm-c_{5}e)e}}{8\sqrt{-\frac{-v_0^{2}m+2e{V}_0-2e\varphi }{m}}\pi ncm-2c_{5}e}d\varphi \\ x& =\frac{\sqrt{2}(e^2{c}_5\sqrt{4e\pi mn\sqrt{\frac{2e\varphi }{m}-\frac{-v_0^{2}m+2e{V}_0}{m}}c-e^2{c}_5}+\frac{{(4e\pi mn\sqrt{\frac{2e\varphi }{m}-\frac{-v_0^{2}m+2e{V}_0}{m}}c-e^2{c}_5)}^{3/2}}{3})}{16m{e}^2{\pi }^2{c}^2{n}^2}+c_7\end{array}$$

$$\begin{array}{rl}x& =\frac{2(\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}\pi ncm+\frac{c_{5}e}{2})\sqrt{2}\sqrt{4cmn\pi e\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}-e^2{c}_5}+24c_{7}me{\pi }^2{c}^2{n}^2}{24em{\pi }^2{c}^2{n}^2}\end{array}$$

For solution $u\left(\varphi \right)=0$, since $u=x^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{x}^{\prime }=0\end{array}$$

Since the ode has the form $x^{\prime }=f(\varphi )$, then we only need to integrate $f(\varphi )$.

$$\begin{array}{rl}\int dx& =\int 0d\varphi +c_8\\ x& =c_8\end{array}$$

In summary, these are the solution found for $(x)$

$$\begin{array}{rl}x& =\frac{-2(\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}\pi ncm+\frac{c_{5}e}{2})\sqrt{2}\sqrt{4cmn\pi e\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}-e^2{c}_5}+24c_{6}me{\pi }^2{c}^2{n}^2}{24em{\pi }^2{c}^2{n}^2}\\ x& =c_8\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}x& =c_8\\ x& =\frac{-2(\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}\pi ncm+\frac{c_{5}e}{2})\sqrt{2}\sqrt{4cmn\pi e\sqrt{\frac{(-2V_0+2\varphi )e+v_0^{2}m}{m}}-e^2{c}_5}+24c_{6}me{\pi }^2{c}^2{n}^2}{24em{\pi }^2{c}^2{n}^2}\end{array}$$

Now that the reversed roles ode was solved, we will change back to the original roles. This results in the above solution becoming the following.

$$\begin{array}{rl}x& =\frac{-2(\sqrt{\frac{(-2V_0+2\varphi \left(x\right))e+v_0^{2}m}{m}}\pi ncm+\frac{c_{5}e}{2})\sqrt{2}\sqrt{4cmn\pi e\sqrt{\frac{(-2V_0+2\varphi \left(x\right))e+v_0^{2}m}{m}}-e^2{c}_5}+24c_{6}me{\pi }^2{c}^2{n}^2}{24em{\pi }^2{c}^2{n}^2}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}x& =\frac{-2(\sqrt{\frac{(-2V_0+2\varphi \left(x\right))e+v_0^{2}m}{m}}\pi ncm+\frac{c_{5}e}{2})\sqrt{2}\sqrt{4cmn\pi e\sqrt{\frac{(-2V_0+2\varphi \left(x\right))e+v_0^{2}m}{m}}-e^2{c}_5}+24c_{6}me{\pi }^2{c}^2{n}^2}{24em{\pi }^2{c}^2{n}^2}\end{array}$$

✓ Maple. Time used: 0.086 (sec). Leaf size: 210

ode:=diff(diff(phi(x),x),x) = 4*Pi*n*c/(v__0^2+2*e/m*(phi(x)-V__0))^(1/2); 
dsolve(ode,phi(x), singsol=all);
 

$$\begin{array}{rl}e{\int }_{}^{\varphi }\frac{\sqrt{\frac{(-2V_0+2\mathit{\text{\_a}})e+v_0^{2}m}{m}}}{4\sqrt{e\sqrt{\frac{(-2V_0+2\mathit{\text{\_a}})e+v_0^{2}m}{m}}(\frac{c_1\sqrt{(-2\mathit{\text{\_a}}+2V_0)e-v_0^{2}m}}{16}+n\pi ((\mathit{\text{\_a}}-V_0)e+\frac{v_0^{2}m}{2})c)}}d\mathit{\text{\_a}}-x-c_2& =0\\ -e{\int }_{}^{\varphi }\frac{\sqrt{\frac{(-2V_0+2\mathit{\text{\_a}})e+v_0^{2}m}{m}}}{4\sqrt{e\sqrt{\frac{(-2V_0+2\mathit{\text{\_a}})e+v_0^{2}m}{m}}(\frac{c_1\sqrt{(-2\mathit{\text{\_a}}+2V_0)e-v_0^{2}m}}{16}+n\pi ((\mathit{\text{\_a}}-V_0)e+\frac{v_0^{2}m}{2})c)}}d\mathit{\text{\_a}}-x-c_2& =0\end{array}$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
trying 2nd order JacobiSN 
differential order: 2; trying a linearization to 3rd order 
trying 2nd order ODE linearizable_by_differentiation 
trying 2nd order, 2 integrating factors of the form mu(x,y) 
trying differential order: 2; missing variables 
   -> Computing symmetries using: way = 3 
-> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)-4*Pi*n*c/(-(-m*v__0^2+2 
*V__0*e-2*_a*e)/m)^(1/2) = 0, _b(_a), HINT = [[-2/3*(-m*v__0^2+2*V__0*e-2*_a*e) 
/e, 1/3*_b]] 
   *** Sublevel 2 *** 
   symmetry methods on request 
   1st order, trying reduction of order with given symmetries: 
[-2/3*(-m*v__0^2+2*V__0*e-2*_a*e)/e, 1/3*_b] 
   1st order, trying the canonical coordinates of the invariance group 
      -> Calling odsolve with the ODE, diff(y(x),x) = y(x)/((4*x-4*V__0)*e+2* 
v__0^2*m)*e, y(x) 
         *** Sublevel 3 *** 
         Methods for first order ODEs: 
         --- Trying classification methods --- 
         trying a quadrature 
         trying 1st order linear 
         <- 1st order linear successful 
   <- 1st order, canonical coordinates successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful
 

✓ Mathematica. Time used: 79.952 (sec). Leaf size: 2754

ode=D[phi[x],{x,2}]==4*Pi*n*c/Sqrt[v0^2+2*e/m*(phi[x]-V0)]; 
ic={}; 
DSolve[{ode,ic},phi[x],x,IncludeSingularSolutions->True]
 

Too large to display

✗ Sympy

from sympy import * 
x = symbols("x") 
V__0 = symbols("V__0") 
c = symbols("c") 
e = symbols("e") 
m = symbols("m") 
n = symbols("n") 
v__0 = symbols("v__0") 
phi = Function("phi") 
ode = Eq(-4*pi*c*n/sqrt(2*e*(-V__0 + phi(x))/m + v__0**2) + Derivative(phi(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=phi(x),ics=ics)
 

Timed Out