2.1.69 problem 69

Internal problem ID [8457]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 69
Date solved : Thursday, December 12, 2024 at 09:18:15 AM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solve

\begin{align*} a y^{2} y^{\prime \prime }+b y^{2}&=c \end{align*}

Solved as second order missing x ode

Time used: 1.433 (sec)

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{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} a \,y^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+b \,y^{2} = c \end{align*}

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

The ode \(p^{\prime } = -\frac {b \,y^{2}-c}{p a \,y^{2}}\) is separable as it can be written as

\begin{align*} p^{\prime }&= -\frac {b \,y^{2}-c}{p a \,y^{2}}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= -\frac {b \,y^{2}-c}{a \,y^{2}}\\ g(p) &= \frac {1}{p} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy}\\ \int { p\,dp} &= \int { -\frac {b \,y^{2}-c}{a \,y^{2}} \,dy}\\ \frac {p^{2}}{2}&=\frac {-b \,y^{2}-c}{a y}+c_1 \end{align*}

Solving for \(p\) gives

\begin{align*} p &= \frac {\sqrt {2}\, \sqrt {a y \left (c_1 a y -b \,y^{2}-c \right )}}{a y} \\ p &= -\frac {\sqrt {2}\, \sqrt {a y \left (c_1 a y -b \,y^{2}-c \right )}}{a y} \\ \end{align*}

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

\begin{align*} y^{\prime } = \frac {\sqrt {2}\, \sqrt {a y \left (c_1 a y-b y^{2}-c \right )}}{a y} \end{align*}

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

\[ \int _{}^{y}\frac {a \tau \sqrt {2}}{2 \sqrt {a \tau \left (c_1 a \tau -b \,\tau ^{2}-c \right )}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \frac {\sqrt {2}\, \sqrt {a y \left (c_1 a y -b \,y^{2}-c \right )}}{a y}&= 0 \end{align*}

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{align*} y = \frac {a c_1 +\sqrt {c_1^{2} a^{2}-4 b c}}{2 b}\\ y = -\frac {-a c_1 +\sqrt {c_1^{2} a^{2}-4 b c}}{2 b} \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\frac {\sqrt {2}\, \sqrt {a y \left (c_1 a y-b y^{2}-c \right )}}{a y} \end{align*}

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

\[ \int _{}^{y}-\frac {a \tau \sqrt {2}}{2 \sqrt {a \tau \left (c_1 a \tau -b \,\tau ^{2}-c \right )}}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} -\frac {\sqrt {2}\, \sqrt {a y \left (c_1 a y -b \,y^{2}-c \right )}}{a y}&= 0 \end{align*}

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{align*} y = \frac {a c_1 +\sqrt {c_1^{2} a^{2}-4 b c}}{2 b}\\ y = -\frac {-a c_1 +\sqrt {c_1^{2} a^{2}-4 b c}}{2 b} \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = \frac {a c_1 +\sqrt {c_1^{2} a^{2}-4 b c}}{2 b} \]

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

\[ y = -\frac {-a c_1 +\sqrt {c_1^{2} a^{2}-4 b c}}{2 b} \]

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

Summary of solutions found

\begin{align*} \int _{}^{y}\frac {a \tau \sqrt {2}}{2 \sqrt {a \tau \left (c_1 a \tau -b \,\tau ^{2}-c \right )}}d \tau &= x +c_2 \\ \int _{}^{y}-\frac {a \tau \sqrt {2}}{2 \sqrt {a \tau \left (c_1 a \tau -b \,\tau ^{2}-c \right )}}d \tau &= x +c_3 \\ \end{align*}

Maple step by step solution

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`
 

Maple dsolve solution

Solving time : 0.054 (sec)
Leaf size : 76

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

\begin{align*} a \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} a \left (c_{1} \textit {\_a} a -2 b \,\textit {\_a}^{2}-2 c \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ -a \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} a \left (c_{1} \textit {\_a} a -2 b \,\textit {\_a}^{2}-2 c \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}

Mathematica DSolve solution

Solving time : 0.835 (sec)
Leaf size : 346

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

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