2.1.13 Problem 13
Internal
problem
ID
[10372]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
13
Date
solved
:
Monday, January 26, 2026 at 09:47:10 PM
CAS
classification
:
[[_2nd_order, _missing_x]]
2.1.13.1 second order ode missing x
190.280 (sec)
\begin{align*}
{y^{\prime \prime }}^{2}+y^{\prime }&=0 \\
\end{align*}
Entering second order ode missing \(x\) solverThis 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}+p \left (y \right ) = 0 \end{align*}
Which is now solved as first order ode for \(p(y)\).
2.1.13.2 Solved by factoring the differential equation
Time used: 0.990 (sec)
\begin{align*}
p^{2} {p^{\prime }}^{2}+p&=0 \\
\end{align*}
Writing the ode as \begin{align*} \left (p\right )\left ({p^{\prime }}^{2} p+1\right )&=0 \end{align*}
Therefore we need to solve the following equations
\begin{align*}
\tag{1} p &= 0 \\
\tag{2} {p^{\prime }}^{2} p+1 &= 0 \\
\end{align*}
Now each of the above equations is solved in
turn.
Solving equation (1)
Entering zero order ode solverSolving for \(p\) from
\begin{align*} p = 0 \end{align*}
Solving gives
\begin{align*}
p &= 0 \\
\end{align*}
Solving equation (2)
Entering first order ode homog type G solverMultiplying the right side of the ode, which is \(-\frac {1}{\sqrt {-p}}\) by \(\frac {y}{p}\)
gives
\begin{align*} p^{\prime } &= \left (\frac {y}{p}\right ) -\frac {1}{\sqrt {-p}}\\ &= -\frac {y}{p \sqrt {-p}}\\ &= F(y,p) \end{align*}
Since \(F \left (y , p\right )\) has \(p\), then let
\begin{align*} f_y&= y \left (\frac {\partial }{\partial y}F \left (y , p\right )\right )\\ &= -\frac {y}{p \sqrt {-p}}\\ f_p&= p \left (\frac {\partial }{\partial p}F \left (y , p\right )\right )\\ &= -\frac {3 y}{2 \left (-p \right )^{{3}/{2}}}\\ \alpha &= \frac {f_y}{f_p} \\ &=-{\frac {2}{3}} \end{align*}
Since \(\alpha \) is independent of \(y,p\) then this is Homogeneous type G.
Let
\begin{align*} p&=\frac {z}{y^ \alpha }\\ &=\frac {z}{\frac {1}{y^{{2}/{3}}}} \end{align*}
Substituting the above back into \(F(y,p)\) gives
\begin{align*} F \left (z \right ) &=\frac {1}{\left (-z \right )^{{3}/{2}}} \end{align*}
We see that \(F \left (z \right )\) does not depend on \(y\) nor on \(p\). If this was not the case, then this method will not
work.
Therefore, the implicit solution is given by
\begin{align*} \ln \left (y \right )- c_1 - \int ^{p y^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}
Which gives
\[
\ln \left (y \right )-c_1 +\int _{}^{\frac {p}{y^{{2}/{3}}}}\frac {1}{z \left (\frac {2}{3}-\frac {1}{\left (-z \right )^{{3}/{2}}}\right )}d z = 0
\]
The value of the above is \[
\ln \left (y \right )-c_1 +\ln \left (2 \left (-\frac {p}{y^{{2}/{3}}}\right )^{{3}/{2}}-3\right ) = 0
\]
Multiplying the right side of the ode, which is \(\frac {1}{\sqrt {-p}}\) by \(\frac {y}{p}\) gives
\begin{align*} p^{\prime } &= \left (\frac {y}{p}\right ) \frac {1}{\sqrt {-p}}\\ &= \frac {y}{p \sqrt {-p}}\\ &= F(y,p) \end{align*}
Since \(F \left (y , p\right )\) has \(p\), then let
\begin{align*} f_y&= y \left (\frac {\partial }{\partial y}F \left (y , p\right )\right )\\ &= \frac {y}{p \sqrt {-p}}\\ f_p&= p \left (\frac {\partial }{\partial p}F \left (y , p\right )\right )\\ &= \frac {3 y}{2 \left (-p \right )^{{3}/{2}}}\\ \alpha &= \frac {f_y}{f_p} \\ &=-{\frac {2}{3}} \end{align*}
Since \(\alpha \) is independent of \(y,p\) then this is Homogeneous type G.
Let
\begin{align*} p&=\frac {z}{y^ \alpha }\\ &=\frac {z}{\frac {1}{y^{{2}/{3}}}} \end{align*}
Substituting the above back into \(F(y,p)\) gives
\begin{align*} F \left (z \right ) &=-\frac {1}{\left (-z \right )^{{3}/{2}}} \end{align*}
We see that \(F \left (z \right )\) does not depend on \(y\) nor on \(p\). If this was not the case, then this method will not
work.
Therefore, the implicit solution is given by
\begin{align*} \ln \left (y \right )- c_1 - \int ^{p y^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}
Which gives
\[
\ln \left (y \right )-c_2 +\int _{}^{\frac {p}{y^{{2}/{3}}}}\frac {1}{z \left (\frac {2}{3}+\frac {1}{\left (-z \right )^{{3}/{2}}}\right )}d z = 0
\]
The value of the above is \[
\ln \left (y \right )-c_2 +\ln \left (2 \left (-\frac {p}{y^{{2}/{3}}}\right )^{{3}/{2}}+3\right ) = 0
\]
Summary of solutions found
\begin{align*}
\ln \left (y \right )-c_1 +\ln \left (2 \left (-\frac {p}{y^{{2}/{3}}}\right )^{{3}/{2}}-3\right ) &= 0 \\
\ln \left (y \right )-c_2 +\ln \left (2 \left (-\frac {p}{y^{{2}/{3}}}\right )^{{3}/{2}}+3\right ) &= 0 \\
p &= 0 \\
\end{align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to
solve is \begin{align*} \ln \left (y\right )-c_1 +\ln \left (2 \left (-\frac {y^{\prime }}{y^{{2}/{3}}}\right )^{{3}/{2}}-3\right ) = 0 \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=-{\left (-\frac {{\left (\left (4 \,{\mathrm e}^{c_1}+12 y\right ) y^{2}\right )}^{{1}/{3}}}{4 y}-\frac {i \sqrt {3}\, {\left (\left (4 \,{\mathrm e}^{c_1}+12 y\right ) y^{2}\right )}^{{1}/{3}}}{4 y}\right )}^{2} y^{{2}/{3}} \\
\tag{2} y^{\prime }&=-{\left (-\frac {{\left (\left (4 \,{\mathrm e}^{c_1}+12 y\right ) y^{2}\right )}^{{1}/{3}}}{4 y}+\frac {i \sqrt {3}\, {\left (\left (4 \,{\mathrm e}^{c_1}+12 y\right ) y^{2}\right )}^{{1}/{3}}}{4 y}\right )}^{2} y^{{2}/{3}} \\
\tag{3} y^{\prime }&=-\frac {{\left (\left (4 \,{\mathrm e}^{c_1}+12 y\right ) y^{2}\right )}^{{2}/{3}}}{4 y^{{4}/{3}}} \\
\end{align*}
Now each of the above is solved
separately.
Solving Eq. (1)
Entering first order ode autonomous solverIntegrating gives
\begin{align*} \int -\frac {16 y^{{4}/{3}}}{{\left (\left (4 \,{\mathrm e}^{c_1}+12 y \right ) y^{2}\right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d y &= dx\\ -\frac {4 \left ({\mathrm e}^{c_1}+3 y \right ) 4^{{1}/{3}} y^{{4}/{3}}}{{\left (\left ({\mathrm e}^{c_1}+3 y \right ) y^{2}\right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}&= x +c_9 \end{align*}
Singular solutions are found by solving
\begin{align*} -\frac {{\left (\left (4 \,{\mathrm e}^{c_1}+12 y \right ) y^{2}\right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}{16 y^{{4}/{3}}}&= 0 \end{align*}
for \(y\). This is because of dividing by the above earlier. This gives the following singular solution(s),
which also has to satisfy the given ODE.
\begin{align*} y = -\frac {{\mathrm e}^{c_1}}{3} \end{align*}
Simplifying the above gives
\begin{align*}
\frac {\left (4 \,{\mathrm e}^{c_1}+12 y\right ) 2^{{2}/{3}} y^{{4}/{3}}}{{\left (\left ({\mathrm e}^{c_1}+3 y\right ) y^{2}\right )}^{{2}/{3}} \left (i-\sqrt {3}\right )^{2}} &= x +c_9 \\
y &= -\frac {{\mathrm e}^{c_1}}{3} \\
\end{align*}
Solving Eq. (2)
Entering first order ode autonomous solverIntegrating gives
\begin{align*} \int -\frac {16 y^{{4}/{3}}}{{\left (\left (4 \,{\mathrm e}^{c_1}+12 y \right ) y^{2}\right )}^{{2}/{3}} \left (-1+i \sqrt {3}\right )^{2}}d y &= dx\\ -\frac {4 \left ({\mathrm e}^{c_1}+3 y \right ) 4^{{1}/{3}} y^{{4}/{3}}}{{\left (\left ({\mathrm e}^{c_1}+3 y \right ) y^{2}\right )}^{{2}/{3}} \left (-1+i \sqrt {3}\right )^{2}}&= x +\textit {\_C10} \end{align*}
Singular solutions are found by solving
\begin{align*} -\frac {{\left (\left (4 \,{\mathrm e}^{c_1}+12 y \right ) y^{2}\right )}^{{2}/{3}} \left (-1+i \sqrt {3}\right )^{2}}{16 y^{{4}/{3}}}&= 0 \end{align*}
for \(y\). This is because of dividing by the above earlier. This gives the following singular solution(s),
which also has to satisfy the given ODE.
\begin{align*} y = -\frac {{\mathrm e}^{c_1}}{3} \end{align*}
Simplifying the above gives
\begin{align*}
\frac {\left (4 \,{\mathrm e}^{c_1}+12 y\right ) 2^{{2}/{3}} y^{{4}/{3}}}{{\left (\left ({\mathrm e}^{c_1}+3 y\right ) y^{2}\right )}^{{2}/{3}} \left (\sqrt {3}+i\right )^{2}} &= x +\textit {\_C10} \\
y &= -\frac {{\mathrm e}^{c_1}}{3} \\
\end{align*}
Solving Eq. (3)
Entering first order ode autonomous solverIntegrating gives
\begin{align*} \int -\frac {4 y^{{4}/{3}}}{{\left (\left (4 \,{\mathrm e}^{c_1}+12 y \right ) y^{2}\right )}^{{2}/{3}}}d y &= dx\\ -\frac {\left ({\mathrm e}^{c_1}+3 y \right ) y^{{4}/{3}} 4^{{1}/{3}}}{{\left (\left ({\mathrm e}^{c_1}+3 y \right ) y^{2}\right )}^{{2}/{3}}}&= x +\textit {\_C11} \end{align*}
Singular solutions are found by solving
\begin{align*} -\frac {{\left (\left (4 \,{\mathrm e}^{c_1}+12 y \right ) y^{2}\right )}^{{2}/{3}}}{4 y^{{4}/{3}}}&= 0 \end{align*}
for \(y\). This is because of dividing by the above earlier. This gives the following singular solution(s),
which also has to satisfy the given ODE.
\begin{align*} y = -\frac {{\mathrm e}^{c_1}}{3} \end{align*}
Simplifying the above gives
\begin{align*}
\frac {\left (-{\mathrm e}^{c_1}-3 y\right ) 2^{{2}/{3}} y^{{4}/{3}}}{{\left (\left ({\mathrm e}^{c_1}+3 y\right ) y^{2}\right )}^{{2}/{3}}} &= x +\textit {\_C11} \\
y &= -\frac {{\mathrm e}^{c_1}}{3} \\
\end{align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then the new first order ode to
solve is \begin{align*} \ln \left (y\right )-c_2 +\ln \left (2 \left (-\frac {y^{\prime }}{y^{{2}/{3}}}\right )^{{3}/{2}}+3\right ) = 0 \end{align*}
Entering first order ode dAlembert solverLet \(p=y^{\prime }\) the ode becomes
\begin{align*} \ln \left (y \right )-c_2 +\ln \left (2 \left (-\frac {p}{y^{{2}/{3}}}\right )^{{3}/{2}}+3\right ) = 0 \end{align*}
Solving for \(y\) from the above results in
\begin{align*}
\tag{1} y &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{6 p \sqrt {-p}+3 \,{\mathrm e}^{c_2}} \\
\tag{2} y &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{3 \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{3} \left (2 p \sqrt {-p}+{\mathrm e}^{c_2}\right )} \\
\tag{3} y &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{3 \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{3} \left (2 p \sqrt {-p}+{\mathrm e}^{c_2}\right )} \\
\tag{4} y &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{-6 p \sqrt {-p}+3 \,{\mathrm e}^{c_2}} \\
\tag{5} y &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{3 \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{3} \left (-2 p \sqrt {-p}+{\mathrm e}^{c_2}\right )} \\
\tag{6} y &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{3 \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{3} \left (-2 p \sqrt {-p}+{\mathrm e}^{c_2}\right )} \\
\end{align*}
This has the form \begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}
Where \(f,g\) are functions of \(p=y'(x)\). Each of the above ode’s is dAlembert ode which is now solved.
Solving ode 1A
Taking derivative of (*) w.r.t. \(x\) gives
\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}
Comparing the form \(y=x f + g\) to (1A) shows that
\begin{align*} f &= 0\\ g &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p = \left (\frac {12 p^{2}}{-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}-\frac {36 \sqrt {-p}\, p^{3}}{\left (-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \sqrt {-p}\, {\mathrm e}^{2 c_2}}{\left (-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right )
\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=0 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = \frac {{\mathrm e}^{c_2}}{3} \end{align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
\begin{equation}
\tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {12 p \left (x \right )^{2}}{-6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}-\frac {36 \sqrt {-p \left (x \right )}\, p \left (x \right )^{3}}{\left (-6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \sqrt {-p \left (x \right )}\, {\mathrm e}^{2 c_2}}{\left (-6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}}
\end{equation}
This ODE is now solved for \(p \left (x \right )\).
No inversion is needed.
Integrating gives
\begin{align*} \int \frac {-4 \left (-p \right )^{{7}/{2}}+4 \,{\mathrm e}^{c_2} p^{2}-\sqrt {-p}\, {\mathrm e}^{2 c_2}}{p \left (-2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )^{2}}d p &= dx\\ -2 \sqrt {-p}-\frac {4 \,{\mathrm e}^{2 c_2} {\mathrm e}^{-c_2} \sqrt {-p}}{3 \left (-2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )}+\frac {4 \,{\mathrm e}^{c_2} \sqrt {-p}}{3 \left (-2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )}&= x +\textit {\_C12} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*}
y &= \frac {-\frac {\left (x +\textit {\_C12} \right )^{6}}{16}+{\mathrm e}^{2 c_2}}{-\frac {3 \sqrt {4}\, \left (\left (x +\textit {\_C12} \right )^{2}\right )^{{3}/{2}}}{8}+3 \,{\mathrm e}^{c_2}} \\
\end{align*}
Solving ode 2A
Taking derivative of (*) w.r.t. \(x\) gives
\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}
Comparing the form \(y=x f + g\) to (1A) shows that
\begin{align*} f &= 0\\ g &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p = \left (\frac {12 p^{2}}{-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}-\frac {36 \sqrt {-p}\, p^{3}}{\left (-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \sqrt {-p}\, {\mathrm e}^{2 c_2}}{\left (-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right )
\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=0 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = \frac {{\mathrm e}^{c_2}}{3} \end{align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
\begin{equation}
\tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {12 p \left (x \right )^{2}}{-6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}-\frac {36 \sqrt {-p \left (x \right )}\, p \left (x \right )^{3}}{\left (-6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \sqrt {-p \left (x \right )}\, {\mathrm e}^{2 c_2}}{\left (-6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}}
\end{equation}
This ODE is now solved for \(p \left (x \right )\).
No inversion is needed.
Integrating gives
\begin{align*} \int \frac {-4 \left (-p \right )^{{7}/{2}}+4 \,{\mathrm e}^{c_2} p^{2}-\sqrt {-p}\, {\mathrm e}^{2 c_2}}{p \left (-2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )^{2}}d p &= dx\\ -2 \sqrt {-p}-\frac {4 \,{\mathrm e}^{2 c_2} {\mathrm e}^{-c_2} \sqrt {-p}}{3 \left (-2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )}+\frac {4 \,{\mathrm e}^{c_2} \sqrt {-p}}{3 \left (-2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )}&= x +\textit {\_C13} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*}
y &= \frac {-\frac {\left (x +\textit {\_C13} \right )^{6}}{16}+{\mathrm e}^{2 c_2}}{-\frac {3 \sqrt {4}\, \left (\left (x +\textit {\_C13} \right )^{2}\right )^{{3}/{2}}}{8}+3 \,{\mathrm e}^{c_2}} \\
\end{align*}
Solving ode 3A
Taking derivative of (*) w.r.t. \(x\) gives
\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}
Comparing the form \(y=x f + g\) to (1A) shows that
\begin{align*} f &= 0\\ g &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p = \left (\frac {12 p^{2}}{-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}-\frac {36 \sqrt {-p}\, p^{3}}{\left (-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \sqrt {-p}\, {\mathrm e}^{2 c_2}}{\left (-6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right )
\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=0 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = \frac {{\mathrm e}^{c_2}}{3} \end{align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
\begin{equation}
\tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {12 p \left (x \right )^{2}}{-6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}-\frac {36 \sqrt {-p \left (x \right )}\, p \left (x \right )^{3}}{\left (-6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \sqrt {-p \left (x \right )}\, {\mathrm e}^{2 c_2}}{\left (-6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}}
\end{equation}
This ODE is now solved for \(p \left (x \right )\).
No inversion is needed.
Integrating gives
\begin{align*} \int \frac {-4 \left (-p \right )^{{7}/{2}}+4 \,{\mathrm e}^{c_2} p^{2}-\sqrt {-p}\, {\mathrm e}^{2 c_2}}{p \left (-2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )^{2}}d p &= dx\\ -2 \sqrt {-p}-\frac {4 \,{\mathrm e}^{2 c_2} {\mathrm e}^{-c_2} \sqrt {-p}}{3 \left (-2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )}+\frac {4 \,{\mathrm e}^{c_2} \sqrt {-p}}{3 \left (-2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )}&= x +\textit {\_C14} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*}
y &= \frac {-\frac {\left (x +\textit {\_C14} \right )^{6}}{16}+{\mathrm e}^{2 c_2}}{-\frac {3 \sqrt {4}\, \left (\left (x +\textit {\_C14} \right )^{2}\right )^{{3}/{2}}}{8}+3 \,{\mathrm e}^{c_2}} \\
\end{align*}
Solving ode 4A
Taking derivative of (*) w.r.t. \(x\) gives
\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}
Comparing the form \(y=x f + g\) to (1A) shows that
\begin{align*} f &= 0\\ g &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p = \left (\frac {12 p^{2}}{6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}+\frac {36 \sqrt {-p}\, p^{3}}{\left (6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \sqrt {-p}\, {\mathrm e}^{2 c_2}}{\left (6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right )
\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=0 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = \frac {{\mathrm e}^{c_2}}{3} \end{align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
\begin{equation}
\tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {12 p \left (x \right )^{2}}{6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}+\frac {36 \sqrt {-p \left (x \right )}\, p \left (x \right )^{3}}{\left (6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \sqrt {-p \left (x \right )}\, {\mathrm e}^{2 c_2}}{\left (6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}}
\end{equation}
This ODE is now solved for \(p \left (x \right )\).
No inversion is needed.
Integrating gives
\begin{align*} \int \frac {4 \left (-p \right )^{{7}/{2}}+4 \,{\mathrm e}^{c_2} p^{2}+\sqrt {-p}\, {\mathrm e}^{2 c_2}}{p \left (2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )^{2}}d p &= dx\\ 2 \sqrt {-p}&= x +\textit {\_C15} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*}
y &= \frac {4 \left (-\frac {1}{4} x^{2}-\frac {1}{2} \textit {\_C15} x -\frac {1}{4} \textit {\_C15}^{2}\right )^{3}+{\mathrm e}^{2 c_2}}{6 \left (\frac {1}{4} x^{2}+\frac {1}{2} \textit {\_C15} x +\frac {1}{4} \textit {\_C15}^{2}\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}} \\
\end{align*}
Solving ode 5A
Taking derivative of (*) w.r.t. \(x\) gives
\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}
Comparing the form \(y=x f + g\) to (1A) shows that
\begin{align*} f &= 0\\ g &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p = \left (\frac {12 p^{2}}{6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}+\frac {36 \sqrt {-p}\, p^{3}}{\left (6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \sqrt {-p}\, {\mathrm e}^{2 c_2}}{\left (6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right )
\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=0 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = \frac {{\mathrm e}^{c_2}}{3} \end{align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
\begin{equation}
\tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {12 p \left (x \right )^{2}}{6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}+\frac {36 \sqrt {-p \left (x \right )}\, p \left (x \right )^{3}}{\left (6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \sqrt {-p \left (x \right )}\, {\mathrm e}^{2 c_2}}{\left (6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}}
\end{equation}
This ODE is now solved for \(p \left (x \right )\).
No inversion is needed.
Integrating gives
\begin{align*} \int \frac {4 \left (-p \right )^{{7}/{2}}+4 \,{\mathrm e}^{c_2} p^{2}+\sqrt {-p}\, {\mathrm e}^{2 c_2}}{p \left (2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )^{2}}d p &= dx\\ 2 \sqrt {-p}&= x +\textit {\_C16} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*}
y &= \frac {4 \left (-\frac {1}{4} x^{2}-\frac {1}{2} \textit {\_C16} x -\frac {1}{4} \textit {\_C16}^{2}\right )^{3}+{\mathrm e}^{2 c_2}}{6 \left (\frac {1}{4} x^{2}+\frac {1}{2} \textit {\_C16} x +\frac {1}{4} \textit {\_C16}^{2}\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}} \\
\end{align*}
Solving ode 6A
Taking derivative of (*) w.r.t. \(x\) gives
\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}
Comparing the form \(y=x f + g\) to (1A) shows that
\begin{align*} f &= 0\\ g &= \frac {4 p^{3}+{\mathrm e}^{2 c_2}}{6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p = \left (\frac {12 p^{2}}{6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}+\frac {36 \sqrt {-p}\, p^{3}}{\left (6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \sqrt {-p}\, {\mathrm e}^{2 c_2}}{\left (6 \left (-p \right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right )
\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=0 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = \frac {{\mathrm e}^{c_2}}{3} \end{align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
\begin{equation}
\tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {12 p \left (x \right )^{2}}{6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}}+\frac {36 \sqrt {-p \left (x \right )}\, p \left (x \right )^{3}}{\left (6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \sqrt {-p \left (x \right )}\, {\mathrm e}^{2 c_2}}{\left (6 \left (-p \left (x \right )\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}\right )^{2}}}
\end{equation}
This ODE is now solved for \(p \left (x \right )\).
No inversion is needed.
Integrating gives
\begin{align*} \int \frac {4 \left (-p \right )^{{7}/{2}}+4 \,{\mathrm e}^{c_2} p^{2}+\sqrt {-p}\, {\mathrm e}^{2 c_2}}{p \left (2 \left (-p \right )^{{3}/{2}}+{\mathrm e}^{c_2}\right )^{2}}d p &= dx\\ 2 \sqrt {-p}&= x +\textit {\_C17} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*}
y &= \frac {4 \left (-\frac {1}{4} x^{2}-\frac {1}{2} \textit {\_C17} x -\frac {1}{4} \textit {\_C17}^{2}\right )^{3}+{\mathrm e}^{2 c_2}}{6 \left (\frac {1}{4} x^{2}+\frac {1}{2} \textit {\_C17} x +\frac {1}{4} \textit {\_C17}^{2}\right )^{{3}/{2}}+3 \,{\mathrm e}^{c_2}} \\
\end{align*}
The above solution was found not to satisfy the
ode or the IC. Hence it is removed.
The above solution was found not to satisfy the ode or the IC. Hence it is removed.
The above solution was found not to satisfy the ode or the IC. Hence it is removed.
The above solution was found not to satisfy the ode or the IC. Hence it is removed.
The above solution was found not to satisfy the ode or the IC. Hence it is removed.
The above solution was found not to satisfy the ode or the IC. Hence it is removed.
For solution (3) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is
\begin{align*} y^{\prime } = 0 \end{align*}
Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to
integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {0\, dx} + \textit {\_C18} \\ y &= \textit {\_C18} \end{align*}
Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} y&=\textit {\_C18}\\ y&=-\frac {{\mathrm e}^{c_1}}{3}\\ y&=\frac {{\mathrm e}^{c_2}}{3}\\ y&=-\frac {c_9^{3}}{12}-\frac {c_9^{2} x}{4}-\frac {c_9 \,x^{2}}{4}-\frac {x^{3}}{12}-\frac {{\mathrm e}^{c_1}}{3} \end{align*}
Summary of solutions found
\begin{align*}
y &= \textit {\_C18} \\
y &= -\frac {{\mathrm e}^{c_1}}{3} \\
y &= \frac {{\mathrm e}^{c_2}}{3} \\
y &= -\frac {c_9^{3}}{12}-\frac {c_9^{2} x}{4}-\frac {c_9 \,x^{2}}{4}-\frac {x^{3}}{12}-\frac {{\mathrm e}^{c_1}}{3} \\
\end{align*}
2.1.13.3 second order ode missing y
0.465 (sec)
\begin{align*}
{y^{\prime \prime }}^{2}+y^{\prime }&=0 \\
\end{align*}
Entering second order ode missing \(y\) solverThis is second order ode with missing dependent
variable \(y\). Let \begin{align*} u(x) &= y^{\prime } \end{align*}
Then
\begin{align*} u'(x) &= y^{\prime \prime } \end{align*}
Hence the ode becomes
\begin{align*} {u^{\prime }\left (x \right )}^{2}+u \left (x \right ) = 0 \end{align*}
Which is now solved for \(u(x)\) as first order ode.
Entering first order ode dAlembert solverLet \(p=u^{\prime }\left (x \right )\) the ode becomes
\begin{align*} p^{2}+u = 0 \end{align*}
Solving for \(u \left (x \right )\) from the above results in
\begin{align*}
\tag{1} u \left (x \right ) &= -p^{2} \\
\end{align*}
This has the form \begin{align*} u=x f(p)+g(p)\tag {*} \end{align*}
Where \(f,g\) are functions of \(p=u'(x)\). The above ode is dAlembert ode which is now solved.
Taking derivative of (*) w.r.t. \(x\) gives
\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}
Comparing the form \(u \left (x \right )=x f + g\) to (1A) shows that
\begin{align*} f &= 0\\ g &= -p^{2} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p = -2 p p^{\prime }\left (x \right )
\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=0 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} u \left (x \right ) = 0 \end{align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
\begin{equation}
\tag{3} p^{\prime }\left (x \right ) = -{\frac {1}{2}}
\end{equation}
This ODE is now solved for \(p \left (x \right )\).
No inversion is needed.
Since the ode has the form \(p^{\prime }\left (x \right )=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dp} &= \int {-{\frac {1}{2}}\, dx}\\ p \left (x \right ) &= -\frac {x}{2} + c_1 \end{align*}
\begin{align*} p \left (x \right )&= -\frac {x}{2}+c_1 \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*}
u \left (x \right ) &= -\frac {\left (2 c_1 -x \right )^{2}}{4} \\
\end{align*}
Simplifying the above gives \begin{align*}
u \left (x \right ) &= 0 \\
u \left (x \right ) &= -\frac {\left (x -2 c_1 \right )^{2}}{4} \\
\end{align*}
In summary, these
are the solution found for \(y\) \begin{align*}
u \left (x \right ) &= 0 \\
u \left (x \right ) &= -\frac {\left (x -2 c_1 \right )^{2}}{4} \\
\end{align*}
For solution \(u \left (x \right ) = 0\), since \(u=y^{\prime }\) then the new first order ode to solve is
\begin{align*} y^{\prime } = 0 \end{align*}
Entering first order ode quadrature solverSince 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_2 \\ y &= c_2 \end{align*}
In summary, these are the solution found for \((y)\)
\begin{align*}
y &= c_2 \\
\end{align*}
For solution \(u \left (x \right ) = -\frac {\left (x -2 c_1 \right )^{2}}{4}\), since \(u=y^{\prime }\) then the new first order ode to
solve is \begin{align*} y^{\prime } = -\frac {\left (x -2 c_1 \right )^{2}}{4} \end{align*}
Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to
integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {-\frac {\left (2 c_1 -x \right )^{2}}{4}\, dx}\\ y &= \frac {\left (2 c_1 -x \right )^{3}}{12} + c_3 \end{align*}
\begin{align*} y&= \frac {\left (2 c_1 -x \right )^{3}}{12}+c_3 \end{align*}
In summary, these are the solution found for \((y)\)
\begin{align*}
y &= \frac {\left (2 c_1 -x \right )^{3}}{12}+c_3 \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= c_2 \\
y &= \frac {\left (2 c_1 -x \right )^{3}}{12}+c_3 \\
\end{align*}
2.1.13.4 ✓ Maple. Time used: 0.054 (sec). Leaf size: 27
ode:=diff(diff(y(x),x),x)^2+diff(y(x),x) = 0;
dsolve(ode,y(x), singsol=all);
\begin{align*}
y &= c_{1} \\
y &= -\frac {1}{12} x^{3}+\frac {1}{2} c_{1} x^{2}-c_{1}^{2} x +c_{2} \\
\end{align*}
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 e\
ach 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
-> Calling odsolve with the ODE, diff(diff(diff(y(x),x),x),x)+1/2, y(x)
*** Sublevel 4 ***
Methods for third order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful
<- 2nd order ODE linearizable_by_differentiation 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
<- 2nd order ODE linearizable_by_differentiation successful
-> Calling odsolve with the ODE, diff(y(x),x) = 0, y(x), singsol = none
*** Sublevel 2 ***
Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )^{2}+\frac {d}{d x}y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}u \left (x \right )\right )^{2}+u \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}u \left (x \right )=\sqrt {-u \left (x \right )}, \frac {d}{d x}u \left (x \right )=-\sqrt {-u \left (x \right )}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}u \left (x \right )=\sqrt {-u \left (x \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {-u \left (x \right )}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {-u \left (x \right )}}d x =\int 1d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -2 \sqrt {-u \left (x \right )}=x +\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{4} x^{2}-\frac {1}{2} \textit {\_C1} x -\frac {1}{4} \textit {\_C1}^{2} \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & u \left (x \right )=-\frac {\left (x +\textit {\_C1} \right )^{2}}{4} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}u \left (x \right )=-\sqrt {-u \left (x \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {-u \left (x \right )}}=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {-u \left (x \right )}}d x =\int \left (-1\right )d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -2 \sqrt {-u \left (x \right )}=-x +\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{4} x^{2}+\frac {1}{2} \textit {\_C1} x -\frac {1}{4} \textit {\_C1}^{2} \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & u \left (x \right )=-\frac {\left (x -\textit {\_C1} \right )^{2}}{4} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & u \left (x \right )=-\frac {\left (x +\textit {\_C1} \right )^{2}}{4} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{u \left (x \right )=-\frac {\left (x +\mathit {C1} \right )^{2}}{4}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {\left (x +\mathit {C1} \right )^{2}}{4} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {\left (x +\mathit {C1} \right )^{2}}{4} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int -\frac {\left (x +\mathit {C1} \right )^{2}}{4}d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=-\frac {\left (x +\mathit {C1} \right )^{3}}{12}+\mathit {C2} \end {array} \]
2.1.13.5 ✓ Mathematica. Time used: 0.014 (sec). Leaf size: 69
ode=(D[y[x],{x,2}])^2+D[y[x],x]==0;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to -\frac {x^3}{12}-\frac {1}{4} i c_1 x^2+\frac {c_1{}^2 x}{4}+c_2\\ y(x)&\to -\frac {x^3}{12}+\frac {1}{4} i c_1 x^2+\frac {c_1{}^2 x}{4}+c_2 \end{align*}
2.1.13.6 ✓ Sympy. Time used: 0.548 (sec). Leaf size: 22
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(Derivative(y(x), x) + Derivative(y(x), (x, 2))**2,0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
\[
y{\left (x \right )} = C_{1} - \frac {C_{2}^{2} x}{4} + \frac {C_{2} x^{2}}{4} - \frac {x^{3}}{12}
\]
Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0]
Sympy version 1.14.0
classify_ode(ode,func=y(x))
('nth_order_reducible',)