2.1.49 Problem 49
Internal
problem
ID
[9120]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
49
Date
solved
:
Monday, January 27, 2025 at 05:43:51 PM
CAS
classification
:
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
Solve
\begin{align*} y^{3} {y^{\prime \prime }}^{2}+y y^{\prime }&=0 \end{align*}
Factoring the ode gives these factors
\begin{align*}
\tag{1} y &= 0 \\
\tag{2} {y^{\prime \prime }}^{2} y^{2}+y^{\prime } &= 0 \\
\end{align*}
Now each of the above equations is solved in
turn.
Solving equation (1)
Solving for \(y\) from
\begin{align*} y = 0 \end{align*}
Solving gives \(y = 0\)
Solving equation (2)
Solved as second order missing x ode
Time used: 5.480 (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*} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2} y^{2}+p \left (y \right ) = 0 \end{align*}
Which is now solved as first order ode for \(p(y)\).
Factoring the ode gives these factors
\begin{align*}
\tag{1} p &= 0 \\
\tag{2} {p^{\prime }}^{2} p y^{2}+1 &= 0 \\
\end{align*}
Now each of the above equations is solved in
turn.
Solving equation (1)
Solving for \(p\) from
\begin{align*} p = 0 \end{align*}
Solving gives \(p = 0\)
Solving equation (2)
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} p^{\prime }&=-\frac {1}{\sqrt {-p}\, y} \\
\tag{2} p^{\prime }&=\frac {1}{\sqrt {-p}\, y} \\
\end{align*}
Now each of the above is solved
separately.
Solving Eq. (1)
The ode
\begin{equation}
p^{\prime } = -\frac {1}{\sqrt {-p}\, y}
\end{equation}
is separable as it can be written as
\begin{align*} p^{\prime }&= -\frac {1}{\sqrt {-p}\, y}\\ &= f(y) g(p) \end{align*}
Where
\begin{align*} f(y) &= -\frac {1}{y}\\ g(p) &= \frac {1}{\sqrt {-p}} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\
\int { \sqrt {-p}\,dp} &= \int { -\frac {1}{y} \,dy} \\
\end{align*}
\[
-\frac {2 \left (-p\right )^{{3}/{2}}}{3}=\ln \left (\frac {1}{y}\right )+c_1
\]
Solving for \(p\) gives
\begin{align*}
p &= -\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4} \\
\end{align*}
Solving Eq. (2)
The ode
\begin{equation}
p^{\prime } = \frac {1}{\sqrt {-p}\, y}
\end{equation}
is separable as it can be written as
\begin{align*} p^{\prime }&= \frac {1}{\sqrt {-p}\, y}\\ &= f(y) g(p) \end{align*}
Where
\begin{align*} f(y) &= \frac {1}{y}\\ g(p) &= \frac {1}{\sqrt {-p}} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\
\int { \sqrt {-p}\,dp} &= \int { \frac {1}{y} \,dy} \\
\end{align*}
\[
-\frac {2 \left (-p\right )^{{3}/{2}}}{3}=\ln \left (y \right )+c_2
\]
Solving for \(p\) gives
\begin{align*}
p &= -\frac {\left (-12 \ln \left (y \right )-12 c_2 \right )^{{2}/{3}}}{4} \\
\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 } = 0 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {0\, dx} + c_3 \\ y &= c_3 \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 {\left (-12 \ln \left (y\right )-12 c_2 \right )^{{2}/{3}}}{4} \end{align*}
Unable to integrate (or intergal too complicated), and since no initial conditions are
given, then the result can be written as
\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}}}d \tau = x +c_4 \]
Singular solutions are found by solving
\begin{align*} -\frac {\left (-12 \ln \left (y \right )-12 c_2 \right )^{{2}/{3}}}{4}&= 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 = {\mathrm e}^{-c_2} \end{align*}
For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = -\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4} \end{align*}
Unable to integrate (or intergal too complicated), and since no initial conditions are
given, then the result can be written as
\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}}}d \tau = x +c_5 \]
Singular solutions are found by solving
\begin{align*} -\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4}&= 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 = {\mathrm e}^{c_1} \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
\int _{}^{y}-\frac {4}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}}}d \tau &= x +c_4 \\
\int _{}^{y}-\frac {4}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}}}d \tau &= x +c_5 \\
y &= c_3 \\
y &= {\mathrm e}^{-c_2} \\
y &= {\mathrm e}^{c_1} \\
\end{align*}
Maple step by step solution
Maple trace
`Methods for second order ODEs:
*** Sublevel 2 ***
Methods for second order ODEs:
Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE.
*** Sublevel 3 ***
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
<- differential order: 2; canonical coordinates successful
<- differential order 2; missing variables successful
-------------------
* Tackling next ODE.
*** Sublevel 3 ***
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
<- differential order: 2; canonical coordinates successful
<- differential order 2; missing variables successful`
Maple dsolve solution
Solving time : 0.102 (sec)
Leaf size : 233
dsolve(y(x)^3*diff(diff(y(x),x),x)^2+y(x)*diff(y(x),x) = 0,y(x),singsol=all)
\begin{align*}
y &= c_{1} \\
y &= 0 \\
-4 \left (\int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-4 \left (\int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\frac {-16 \left (\int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (-x -c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (-i \sqrt {3}-1\right )^{2}} &= 0 \\
\frac {-16 \left (\int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (x +c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\
\frac {-16 \left (\int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (-x -c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (-i \sqrt {3}-1\right )^{2}} &= 0 \\
\frac {-16 \left (\int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (x +c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\
\end{align*}
Mathematica DSolve solution
Solving time : 2.307 (sec)
Leaf size : 459
DSolve[{y[x]^3*D[y[x],{x,2}]^2+y[x]*D[y[x],x]==0,{}},y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to 0 \\
y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2] \\
y(x)\to 0 \\
y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i (-c_1)} (-\log (\text {$\#$1})-i (-1) c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i (-1) c_1-\log (\text {$\#$1})\right )}{(-i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i (-c_1)} (-\log (\text {$\#$1})+i (-c_1)){}^{2/3} \Gamma \left (\frac {1}{3},i (-c_1)-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2] \\
\end{align*}