2.1.69 problem 69
Internal
problem
ID
[8207]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
69
Date
solved
:
Sunday, November 10, 2024 at 03:22:00 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: 1.408 (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_3 \\
\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 \\
\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]
\[
\text {Solve}\left [-\frac {\left (\sqrt {-16 b c+a^2 c_1{}^2}-a c_1\right ) \left (\sqrt {-16 b c+a^2 c_1{}^2}+a c_1\right ){}^2 \left (1+\frac {4 b y(x)}{\sqrt {-16 b c+a^2 c_1{}^2}-a c_1}\right ) \left (1-\frac {4 b y(x)}{\sqrt {-16 b c+a^2 c_1{}^2}+a c_1}\right ) \left (E\left (i \text {arcsinh}\left (2 \sqrt {\frac {b}{\sqrt {a^2 c_1{}^2-16 b c}-a c_1}} \sqrt {y(x)}\right )|\frac {a c_1-\sqrt {a^2 c_1{}^2-16 b c}}{a c_1+\sqrt {a^2 c_1{}^2-16 b c}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (2 \sqrt {\frac {b}{\sqrt {a^2 c_1{}^2-16 b c}-a c_1}} \sqrt {y(x)}\right ),\frac {a c_1-\sqrt {a^2 c_1{}^2-16 b c}}{a c_1+\sqrt {a^2 c_1{}^2-16 b c}}\right )\right ){}^2}{16 b^3 y(x) \left (-\frac {2 \left (b y(x)^2+c\right )}{a y(x)}+c_1\right )}=(x+c_2){}^2,y(x)\right ]
\]