2.1.69 Problem 69

Internal problem ID [8781]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 69
Date solved : Sunday, March 30, 2025 at 01:35:34 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solved as second order missing x ode

Time used: 6.331 (sec)

Solve

$$\begin{array}{rl}a{y}^2{y}^{\prime \prime }+b{y}^2& =c\end{array}$$

This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable $y$ an independent variable. Using

$$\begin{array}{rl}{y}^{\prime }& =p\end{array}$$

Then

$$\begin{array}{rl}{y}^{″}& =\frac{dp}{dx}\\ & =\frac{dp}{dy}\frac{dy}{dx}\\ & =p\frac{dp}{dy}\end{array}$$

Hence the ode becomes

$$\begin{array}{r}a{y}^{2}p\left(y\right)\left(\frac{d}{dy}p\left(y\right)\right)+b{y}^2=c\end{array}$$

Which is now solved as first order ode for $p(y)$.

The ode

$$\begin{array}{}\text{(1)}& p^{\prime }=-\frac{b{y}^2-c}{pa{y}^2}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{p}^{\prime }& =-\frac{b{y}^2-c}{pa{y}^2}\\ & =f(y)g(p)\end{array}$$

Where

$$\begin{array}{rl}f(y)& =-\frac{b{y}^2-c}{a{y}^2}\\ g(p)& =\frac{1}{p}\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(p)}dp& =\int f(y)dy\\ \int pdp& =\int -\frac{b{y}^2-c}{a{y}^2}dy\end{array}$$

$$\frac{p^2}{2}=\frac{-b{y}^2-c}{ay}+c_1$$

For solution (1) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}\frac{{y^{\prime }}^2}{2}=\frac{-b{y}^2-c}{ay}+c_1\end{array}$$

Solving for the derivative gives these ODE’s to solve

$$\begin{array}{}\text{(1)}& y^{\prime }& =\frac{\sqrt{-2ay(b{y}^2-c_{1}ay+c)}}{ay}\text{(2)}& y^{\prime }& =-\frac{\sqrt{-2ay(b{y}^2-c_{1}ay+c)}}{ay}\end{array}$$

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

$${\int }_{}^y\frac{a\tau }{\sqrt{-2a\tau (-c_{1}a\tau +b{\tau }^2+c)}}d\tau =x+c_2$$

Singular solutions are found by solving

$$\begin{array}{rl}\frac{\sqrt{-2ay(-c_{1}ay+b{y}^2+c)}}{ay}& =0\end{array}$$

for $y$. 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.

$$\begin{array}{r}y=\frac{c_{1}a+\sqrt{c_1^2{a}^2-4bc}}{2b}\\ y=-\frac{-c_{1}a+\sqrt{c_1^2{a}^2-4bc}}{2b}\end{array}$$

Solving Eq. (2)

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

$${\int }_{}^y-\frac{a\tau }{\sqrt{-2a\tau (-c_{1}a\tau +b{\tau }^2+c)}}d\tau =x+c_3$$

Singular solutions are found by solving

$$\begin{array}{rl}-\frac{\sqrt{-2ay(-c_{1}ay+b{y}^2+c)}}{ay}& =0\end{array}$$

for $y$. 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.

$$\begin{array}{r}y=\frac{c_{1}a+\sqrt{c_1^2{a}^2-4bc}}{2b}\\ y=-\frac{-c_{1}a+\sqrt{c_1^2{a}^2-4bc}}{2b}\end{array}$$

Will add steps showing solving for IC soon.

The solution

$$y=\frac{c_{1}a+\sqrt{c_1^2{a}^2-4bc}}{2b}$$

was found not to satisfy the ode or the IC. Hence it is removed. The solution

$$y=-\frac{-c_{1}a+\sqrt{c_1^2{a}^2-4bc}}{2b}$$

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

$$\begin{array}{rl}{\int }_{}^y\frac{a\tau }{\sqrt{-2a\tau (-c_{1}a\tau +b{\tau }^2+c)}}d\tau & =x+c_2\end{array}$$

✓ Maple. Time used: 0.027 (sec). Leaf size: 76

ode:=a*y(x)^2*diff(diff(y(x),x),x)+b*y(x)^2 = c; 
dsolve(ode,y(x), singsol=all);
 

$$\begin{array}{rl}a{\int }_{}^y\frac{\mathit{\text{\_a}}}{\sqrt{\mathit{\text{\_a}}a(c_1\mathit{\text{\_a}}a-2b{\mathit{\text{\_a}}}^2-2c)}}d\mathit{\text{\_a}}-x-c_2& =0\\ -a{\int }_{}^y\frac{\mathit{\text{\_a}}}{\sqrt{\mathit{\text{\_a}}a(c_1\mathit{\text{\_a}}a-2b{\mathit{\text{\_a}}}^2-2c)}}d\mathit{\text{\_a}}-x-c_2& =0\end{array}$$

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 
   -> Computing symmetries using: way = exp_sym 
-> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)+(_a^2*b-c)/_a^2/a = 0, 
_b(_a) 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful
 

✓ Mathematica. Time used: 0.85 (sec). Leaf size: 346

ode=a*y[x]^2*D[y[x],{x,2}]+b*y[x]^2==c; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\text{Solve}[-\frac{(\sqrt{-16bc+a^2{c}_1{}^2}-a{c}_1)(\sqrt{-16bc+a^2{c}_1{}^2}+a{c}_1){}^2(1+\frac{4by(x)}{\sqrt{-16bc+a^2{c}_1{}^2}-a{c}_1})(1-\frac{4by(x)}{\sqrt{-16bc+a^2{c}_1{}^2}+a{c}_1})(E\left(i\text{arcsinh}\left(2\sqrt{\frac{b}{\sqrt{a^2{c}_1{}^2-16bc}-a{c}_1}}\sqrt{y(x)}\right)|\frac{a{c}_1-\sqrt{a^2{c}_1{}^2-16bc}}{a{c}_1+\sqrt{a^2{c}_1{}^2-16bc}}\right)-\mathrm{EllipticF}(i\text{arcsinh}\left(2\sqrt{\frac{b}{\sqrt{a^2{c}_1{}^2-16bc}-a{c}_1}}\sqrt{y(x)}\right),\frac{a{c}_1-\sqrt{a^2{c}_1{}^2-16bc}}{a{c}_1+\sqrt{a^2{c}_1{}^2-16bc}})){}^2}{16b^{3}y(x)(-\frac{2(by(x{)}^2+c)}{ay(x)}+c_1)}=(x+c_2){}^2,y(x)]$$

✓ Sympy. Time used: 95.329 (sec). Leaf size: 48

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
y = Function("y") 
ode = Eq(a*y(x)**2*Derivative(y(x), (x, 2)) + b*y(x)**2 - c,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$[\stackrel{y\left(x\right)}{\int }\frac{1}{\sqrt{C_1-\frac{2(ub+\frac{c}{u})}{a}}}du=C_2+x,\stackrel{y\left(x\right)}{\int }\frac{1}{\sqrt{C_1-\frac{2(ub+\frac{c}{u})}{a}}}du=C_2-x]$$