2.5.15 Problem 15

Internal problem ID [8975]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 15
Date solved : Friday, April 25, 2025 at 05:32:28 PM
CAS classification : [_quadrature]

Solve

$$\begin{array}{rl}{h}^2+\frac{2ah}{\sqrt{1+{h^{\prime }}^2}}& =b^2\end{array}$$

Solving for the derivative gives these ODE’s to solve

$$\begin{array}{}\text{(1)}& h^{\prime }& =-\frac{\sqrt{-h^4+4a^2{h}^2+2h^2{b}^2-b^4}}{(h+b)(h-b)}\text{(2)}& h^{\prime }& =\frac{\sqrt{-h^4+4a^2{h}^2+2h^2{b}^2-b^4}}{(h+b)(h-b)}\end{array}$$

Now each of the above is solved separately.

Solving Eq. (1)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^h-\frac{(\tau +b)(\tau -b)}{\sqrt{4a^2{\tau }^2-b^4+2b^2{\tau }^2-{\tau }^4}}d\tau =u+c_1$$

Singular solutions are found by solving

$$\begin{array}{rl}-\frac{\sqrt{4a^2{h}^2-b^4+2h^2{b}^2-h^4}}{(h+b)(h-b)}& =0\end{array}$$

for $h$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}h=-a-\sqrt{a^2+b^2}\\ h=-a+\sqrt{a^2+b^2}\\ h=a-\sqrt{a^2+b^2}\\ h=a+\sqrt{a^2+b^2}\end{array}$$

Solving Eq. (2)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^h\frac{(\tau +b)(\tau -b)}{\sqrt{4a^2{\tau }^2-b^4+2b^2{\tau }^2-{\tau }^4}}d\tau =u+c_2$$

Singular solutions are found by solving

$$\begin{array}{rl}\frac{\sqrt{4a^2{h}^2-b^4+2h^2{b}^2-h^4}}{(h+b)(h-b)}& =0\end{array}$$

for $h$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}h=-a-\sqrt{a^2+b^2}\\ h=-a+\sqrt{a^2+b^2}\\ h=a-\sqrt{a^2+b^2}\\ h=a+\sqrt{a^2+b^2}\end{array}$$

Which simplifies to

$$\begin{array}{rl}-{\int }_{}^h\frac{b^2-{\tau }^2}{\sqrt{4a^2{\tau }^2-b^4+2b^2{\tau }^2-{\tau }^4}}d\tau & =u+c_2\\ h& =-a-\sqrt{a^2+b^2}\\ h& =-a+\sqrt{a^2+b^2}\\ h& =a-\sqrt{a^2+b^2}\\ h& =a+\sqrt{a^2+b^2}\end{array}$$

✓ Maple. Time used: 1.033 (sec). Leaf size: 103

ode:=h(u)^2+2*a*h(u)/(1+diff(h(u),u)^2)^(1/2) = b^2; 
dsolve(ode,h(u), singsol=all);
 

$$\begin{array}{rl}u-{\int }_{}^h\frac{{\mathit{\text{\_a}}}^2-b^2}{\sqrt{-{\mathit{\text{\_a}}}^4+(4a^2+2b^2){\mathit{\text{\_a}}}^2-b^4}}d\mathit{\text{\_a}}-c_1& =0\\ u+{\int }_{}^h\frac{{\mathit{\text{\_a}}}^2-b^2}{\sqrt{-{\mathit{\text{\_a}}}^4+(4a^2+2b^2){\mathit{\text{\_a}}}^2-b^4}}d\mathit{\text{\_a}}-c_1& =0\end{array}$$

Maple trace

Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
<- differential order: 1; missing  x  successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & h{\left(u\right)}^2+\frac{2ah\left(u\right)}{\sqrt{1+{\left(\frac{d}{du}h\left(u\right)\right)}^2}}=b^2\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{du}h\left(u\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & [\frac{d}{du}h\left(u\right)=\frac{\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}{(h\left(u\right)+b)(h\left(u\right)-b)},\frac{d}{du}h\left(u\right)=-\frac{\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}{(h\left(u\right)+b)(h\left(u\right)-b)}]\\ ◻& & \text{Solve the equation}\frac{d}{du}h\left(u\right)=\frac{\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}{(h\left(u\right)+b)(h\left(u\right)-b)}\\ & \circ & \text{Separate variables}\\ & & \frac{\left(\frac{d}{du}h\left(u\right)\right)(h\left(u\right)+b)(h\left(u\right)-b)}{\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}=1\\ & \circ & \text{Integrate both sides with respect to}u\\ & & \int \frac{\left(\frac{d}{du}h\left(u\right)\right)(h\left(u\right)+b)(h\left(u\right)-b)}{\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}du=\int 1du+\mathit{\text{\_C1}}\\ & \circ & \text{Evaluate integral}\\ & & \frac{2b^4\sqrt{1+\frac{(2a\sqrt{a^2+b^2}-2a^2-b^2)h{\left(u\right)}^2}{b^4}}\sqrt{1-\frac{(2a\sqrt{a^2+b^2}+2a^2+b^2)h{\left(u\right)}^2}{b^4}}(\mathit{EllipticF}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}})-\mathit{EllipticE}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}}))}{\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}}\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}(4a^2+2b^2+4a\sqrt{a^2+b^2})}-\frac{b^2\sqrt{1+\frac{(2a\sqrt{a^2+b^2}-2a^2-b^2)h{\left(u\right)}^2}{b^4}}\sqrt{1-\frac{(2a\sqrt{a^2+b^2}+2a^2+b^2)h{\left(u\right)}^2}{b^4}}\mathit{EllipticF}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}})}{\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}}\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}=u+\mathit{\text{\_C1}}\\ ◻& & \text{Solve the equation}\frac{d}{du}h\left(u\right)=-\frac{\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}{(h\left(u\right)+b)(h\left(u\right)-b)}\\ & \circ & \text{Separate variables}\\ & & \frac{\left(\frac{d}{du}h\left(u\right)\right)(h\left(u\right)+b)(h\left(u\right)-b)}{\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}=-1\\ & \circ & \text{Integrate both sides with respect to}u\\ & & \int \frac{\left(\frac{d}{du}h\left(u\right)\right)(h\left(u\right)+b)(h\left(u\right)-b)}{\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}du=\int (-1)du+\mathit{\text{\_C1}}\\ & \circ & \text{Evaluate integral}\\ & & \frac{2b^4\sqrt{1+\frac{(2a\sqrt{a^2+b^2}-2a^2-b^2)h{\left(u\right)}^2}{b^4}}\sqrt{1-\frac{(2a\sqrt{a^2+b^2}+2a^2+b^2)h{\left(u\right)}^2}{b^4}}(\mathit{EllipticF}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}})-\mathit{EllipticE}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}}))}{\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}}\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}(4a^2+2b^2+4a\sqrt{a^2+b^2})}-\frac{b^2\sqrt{1+\frac{(2a\sqrt{a^2+b^2}-2a^2-b^2)h{\left(u\right)}^2}{b^4}}\sqrt{1-\frac{(2a\sqrt{a^2+b^2}+2a^2+b^2)h{\left(u\right)}^2}{b^4}}\mathit{EllipticF}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}})}{\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}}\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}=-u+\mathit{\text{\_C1}}\\ \bullet & & \text{Set of solutions}\\ & & \{\frac{2b^4\sqrt{1+\frac{(2a\sqrt{a^2+b^2}-2a^2-b^2)h{\left(u\right)}^2}{b^4}}\sqrt{1-\frac{(2a\sqrt{a^2+b^2}+2a^2+b^2)h{\left(u\right)}^2}{b^4}}(\mathit{EllipticF}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}})-\mathit{EllipticE}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}}))}{\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}}\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}(4a^2+2b^2+4a\sqrt{a^2+b^2})}-\frac{b^2\sqrt{1+\frac{(2a\sqrt{a^2+b^2}-2a^2-b^2)h{\left(u\right)}^2}{b^4}}\sqrt{1-\frac{(2a\sqrt{a^2+b^2}+2a^2+b^2)h{\left(u\right)}^2}{b^4}}\mathit{EllipticF}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}})}{\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}}\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}=-u+\mathit{C}\mathit{1},\frac{2b^4\sqrt{1+\frac{(2a\sqrt{a^2+b^2}-2a^2-b^2)h{\left(u\right)}^2}{b^4}}\sqrt{1-\frac{(2a\sqrt{a^2+b^2}+2a^2+b^2)h{\left(u\right)}^2}{b^4}}(\mathit{EllipticF}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}})-\mathit{EllipticE}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}}))}{\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}}\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}(4a^2+2b^2+4a\sqrt{a^2+b^2})}-\frac{b^2\sqrt{1+\frac{(2a\sqrt{a^2+b^2}-2a^2-b^2)h{\left(u\right)}^2}{b^4}}\sqrt{1-\frac{(2a\sqrt{a^2+b^2}+2a^2+b^2)h{\left(u\right)}^2}{b^4}}\mathit{EllipticF}(h\left(u\right)\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}},\sqrt{-1+\frac{(4a^2+2b^2)(2a\sqrt{a^2+b^2}+2a^2+b^2)}{b^4}})}{\sqrt{-\frac{2a\sqrt{a^2+b^2}-2a^2-b^2}{b^4}}\sqrt{-h{\left(u\right)}^4+4a^{2}h{\left(u\right)}^2+2h{\left(u\right)}^2{b}^2-b^4}}=u+\mathit{C}\mathit{1}\}\end{array}$$

✓ Mathematica. Time used: 35.557 (sec). Leaf size: 913

ode=h[u]^2 + 2*a*h[u]/Sqrt[1 + (D[ h[u],u])^2] == b^2; 
ic={}; 
DSolve[{ode,ic},h[u],u,IncludeSingularSolutions->True]
 

$$\begin{array}{r}\text{Solution too large to show}\end{array}$$

✗ Sympy

from sympy import * 
u = symbols("u") 
a = symbols("a") 
b = symbols("b") 
h = Function("h") 
ode = Eq(2*a*h(u)/sqrt(Derivative(h(u), u)**2 + 1) - b**2 + h(u)**2,0) 
ics = {} 
dsolve(ode,func=h(u),ics=ics)
 

Timed Out