2.5.4 Internal
problem
ID
[19733]Book
:
Elementary
Differential
Equations.
By
Thornton
C.
Fry.
D
Van
Nostrand.
NY.
First
Edition
(1929) Section
:
Chapter
IV.
Methods
of
solution:
First
order
equations.
section
32.
Problems
at
page
89 Problem
number
:
5Date
solved
:
Friday, November 28, 2025 at 06:41:18 PMCAS
classification
:
[[_homogeneous, `class G`]]
Time used: 12.251 (sec)
Solve
\begin{align*}
\sqrt {t^{2}+T}&=T^{\prime } \\
\end{align*}
Writing the ode as
\begin{align*} T^{\prime }&=\sqrt {t^{2}+T}\\ T^{\prime }&= \omega \left ( t,T\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
\begin{align*} \eta _{t}+\omega \left ( \eta _{T}-\xi _{t}\right ) -\omega ^{2}\xi _{T}-\omega _{t}\xi -\omega _{T}\eta =0\tag {A} \end{align*}
To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as
anstaz gives
\begin{align*}
\tag{1E} \xi &= T a_{3}+t a_{2}+a_{1} \\
\tag{2E} \eta &= T b_{3}+t b_{2}+b_{1} \\
\end{align*}
Where the unknown coefficients are
\[
\{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\}
\]
Substituting equations (1E,2E) and
\(\omega \) into (A)
gives
\begin{equation}
\tag{5E} b_{2}+\sqrt {t^{2}+T}\, \left (b_{3}-a_{2}\right )-\left (t^{2}+T \right ) a_{3}-\frac {t \left (T a_{3}+t a_{2}+a_{1}\right )}{\sqrt {t^{2}+T}}-\frac {T b_{3}+t b_{2}+b_{1}}{2 \sqrt {t^{2}+T}} = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {2 \sqrt {t^{2}+T}\, t^{2} a_{3}+2 \sqrt {t^{2}+T}\, T a_{3}+2 T t a_{3}+4 t^{2} a_{2}-2 t^{2} b_{3}-2 b_{2} \sqrt {t^{2}+T}+2 T a_{2}-T b_{3}+2 t a_{1}+t b_{2}+b_{1}}{2 \sqrt {t^{2}+T}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} -2 \sqrt {t^{2}+T}\, t^{2} a_{3}-2 \sqrt {t^{2}+T}\, T a_{3}-2 T t a_{3}-4 t^{2} a_{2}+2 t^{2} b_{3}+2 b_{2} \sqrt {t^{2}+T}-2 T a_{2}+T b_{3}-2 t a_{1}-t b_{2}-b_{1} = 0
\end{equation}
Simplifying
the above gives
\begin{equation}
\tag{6E} -2 \sqrt {t^{2}+T}\, t^{2} a_{3}-2 \left (t^{2}+T \right ) a_{2}+2 \left (t^{2}+T \right ) b_{3}-2 \sqrt {t^{2}+T}\, T a_{3}-2 T t a_{3}-2 t^{2} a_{2}+2 b_{2} \sqrt {t^{2}+T}-T b_{3}-2 t a_{1}-t b_{2}-b_{1} = 0
\end{equation}
Since the PDE has radicals, simplifying gives
\[
-2 \sqrt {t^{2}+T}\, t^{2} a_{3}-2 \sqrt {t^{2}+T}\, T a_{3}-2 T t a_{3}-4 t^{2} a_{2}+2 t^{2} b_{3}+2 b_{2} \sqrt {t^{2}+T}-2 T a_{2}+T b_{3}-2 t a_{1}-t b_{2}-b_{1} = 0
\]
Looking at the above
PDE shows the following are all the terms with
\(\{T, t\}\) in them.
\[
\left \{T, t, \sqrt {t^{2}+T}\right \}
\]
The following substitution is
now made to be able to collect on all terms with
\(\{T, t\}\) in them
\[
\left \{T = v_{1}, t = v_{2}, \sqrt {t^{2}+T} = v_{3}\right \}
\]
The above PDE (6E) now
becomes
\begin{equation}
\tag{7E} -2 v_{3} v_{2}^{2} a_{3}-4 v_{2}^{2} a_{2}-2 v_{1} v_{2} a_{3}-2 v_{3} v_{1} a_{3}+2 v_{2}^{2} b_{3}-2 v_{2} a_{1}-2 v_{1} a_{2}-v_{2} b_{2}+2 b_{2} v_{3}+v_{1} b_{3}-b_{1} = 0
\end{equation}
Collecting the above on the terms
\(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}, v_{3}\}
\]
Equation (7E) now
becomes
\begin{equation}
\tag{8E} -2 v_{1} v_{2} a_{3}-2 v_{3} v_{1} a_{3}+\left (-2 a_{2}+b_{3}\right ) v_{1}-2 v_{3} v_{2}^{2} a_{3}+\left (-4 a_{2}+2 b_{3}\right ) v_{2}^{2}+\left (-2 a_{1}-b_{2}\right ) v_{2}+2 b_{2} v_{3}-b_{1} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} -2 a_{3}&=0\\ -b_{1}&=0\\ 2 b_{2}&=0\\ -2 a_{1}-b_{2}&=0\\ -4 a_{2}+2 b_{3}&=0\\ -2 a_{2}+b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=0\\ a_{2}&=a_{2}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=2 a_{2} \end{align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown
in the RHS) gives
\begin{align*}
\xi &= t \\
\eta &= 2 T \\
\end{align*}
Shifting is now applied to make
\(\xi =0\) in order to simplify the rest of the computation
\begin{align*} \eta &= \eta - \omega \left (t,T\right ) \xi \\ &= 2 T - \left (\sqrt {t^{2}+T}\right ) \left (t\right ) \\ &= -t \sqrt {t^{2}+T}+2 T\\ \xi &= 0 \end{align*}
The next step is to determine the canonical coordinates \(R,S\) . The canonical coordinates map \(\left ( t,T\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\)
are the canonical coordinates which make the original ode become a quadrature and hence solved
by integration.
The characteristic pde which is used to find the canonical coordinates is
\begin{align*} \frac {d t}{\xi } &= \frac {d T}{\eta } = dS \tag {1} \end{align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial t} + \eta \frac {\partial }{\partial T}\right ) S(t,T) = 1\) . Starting with the first pair of ode’s in (1) gives an
ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\) . Since \(\xi =0\) then in this
special case
\begin{align*} R = t \end{align*}
\(S\) is found from
\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{-t \sqrt {t^{2}+T}+2 T}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\ln \left (-t \sqrt {t^{2}+T}+2 T \right )}{4}-\frac {\sqrt {17}\, \operatorname {arctanh}\left (\frac {\left (-t +4 \sqrt {t^{2}+T}\right ) \sqrt {17}}{17 t}\right )}{34}-\frac {\ln \left (t \sqrt {t^{2}+T}+2 T \right )}{4}-\frac {\sqrt {17}\, \operatorname {arctanh}\left (\frac {\left (4 \sqrt {t^{2}+T}+t \right ) \sqrt {17}}{17 t}\right )}{34}+\frac {\ln \left (-t^{4}-T \,t^{2}+4 T^{2}\right )}{4}-\frac {\sqrt {17}\, \operatorname {arctanh}\left (\frac {\left (-t^{2}+8 T \right ) \sqrt {17}}{17 t^{2}}\right )}{34} \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating
\begin{align*} \frac {dS}{dR} &= \frac { S_{t} + \omega (t,T) S_{T} }{ R_{t} + \omega (t,T) R_{T} }\tag {2} \end{align*}
Where in the above \(R_{t},R_{T},S_{t},S_{T}\) are all partial derivatives and \(\omega (t,T)\) is the right hand side of the original ode given
by
\begin{align*} \omega (t,T) &= \sqrt {t^{2}+T} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{t} &= 1\\ R_{T} &= 0\\ S_{t} &= \frac {t^{6}+2 T \,t^{4}-3 T^{2} t^{2}-4 T^{3}}{\left (t \sqrt {t^{2}+T}+2 T \right ) \sqrt {t^{2}+T}\, \left (t \sqrt {t^{2}+T}-2 T \right )^{2}}\\ S_{T} &= \frac {\left (t +2 \sqrt {t^{2}+T}\right ) T +t^{3}}{\left (-t^{4}-T \,t^{2}+4 T^{2}\right ) \sqrt {t^{2}+T}} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= 0\tag {2A} \end{align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(t,T\) in terms of \(R,S\) from
the result obtained earlier and simplifying. This gives
\begin{align*} \frac {dS}{dR} &= 0 \end{align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an
ode, no matter how complicated it is, to one that can be solved by integration when the ode is in
the canonical coordiates \(R,S\) .
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\) , then we only need to integrate \(f(R)\) .
\begin{align*} \int {dS} &= \int {0\, dR} + c_2 \\ S \left (R \right ) &= c_2 \end{align*}
To complete the solution, we just need to transform the above back to \(t,T\) coordinates. This results in
\begin{align*} \frac {\ln \left (-t \sqrt {t^{2}+T}+2 T\right )}{4}+\frac {\sqrt {17}\, \operatorname {arctanh}\left (\frac {\left (t -4 \sqrt {t^{2}+T}\right ) \sqrt {17}}{17 t}\right )}{34}-\frac {\ln \left (t \sqrt {t^{2}+T}+2 T\right )}{4}-\frac {\sqrt {17}\, \operatorname {arctanh}\left (\frac {\left (4 \sqrt {t^{2}+T}+t \right ) \sqrt {17}}{17 t}\right )}{34}+\frac {\ln \left (-t^{4}-t^{2} T+4 T^{2}\right )}{4}-\frac {\sqrt {17}\, \operatorname {arctanh}\left (\frac {\left (-t^{2}+8 T\right ) \sqrt {17}}{17 t^{2}}\right )}{34} = c_2 \end{align*}
The following diagram shows solution curves of the original ode and how they transform in the
canonical coordinates space using the mapping shown.
Original ode in \(t,T\) coordinates
Canonical
coordinates
transformation
ODE in canonical coordinates \((R,S)\)
\( \frac {dT}{dt} = \sqrt {t^{2}+T}\)
\( \frac {d S}{d R} = 0\)
\(\!\begin {aligned} R&= t\\ S&= \frac {\ln \left (-t \sqrt {t^{2}+T}+2 T \right )}{4}+\frac {\sqrt {17}\, \operatorname {arctanh}\left (\frac {\left (t -4 \sqrt {t^{2}+T}\right ) \sqrt {17}}{17 t}\right )}{34}-\frac {\ln \left (t \sqrt {t^{2}+T}+2 T \right )}{4}-\frac {\sqrt {17}\, \operatorname {arctanh}\left (\frac {\left (4 \sqrt {t^{2}+T}+t \right ) \sqrt {17}}{17 t}\right )}{34}+\frac {\ln \left (-t^{4}-T \,t^{2}+4 T^{2}\right )}{4}-\frac {\sqrt {17}\, \operatorname {arctanh}\left (\frac {\left (-t^{2}+8 T \right ) \sqrt {17}}{17 t^{2}}\right )}{34} \end {aligned} \)
Simplifying the above gives
\begin{align*}
\frac {\ln \left (-t^{4}-t^{2} T+4 T^{2}\right )}{4}+\frac {\ln \left (-t \sqrt {t^{2}+T}+2 T\right )}{4}-\frac {\ln \left (t \sqrt {t^{2}+T}+2 T\right )}{4}+\frac {\left (2 \,\operatorname {arctanh}\left (\frac {\left (t -4 \sqrt {t^{2}+T}\right ) \sqrt {17}}{17 t}\right )+2 \,\operatorname {arctanh}\left (\frac {\left (t^{2}-8 T\right ) \sqrt {17}}{17 t^{2}}\right )-2 \,\operatorname {arctanh}\left (\frac {\left (4 \sqrt {t^{2}+T}+t \right ) \sqrt {17}}{17 t}\right )\right ) \sqrt {17}}{68} &= c_2 \\
\end{align*}
Figure 2.65: Slope field \(\sqrt {t^{2}+T} = T^{\prime }\) Summary of solutions found
\begin{align*}
\frac {\ln \left (-t^{4}-t^{2} T+4 T^{2}\right )}{4}+\frac {\ln \left (-t \sqrt {t^{2}+T}+2 T\right )}{4}-\frac {\ln \left (t \sqrt {t^{2}+T}+2 T\right )}{4}+\frac {\left (2 \,\operatorname {arctanh}\left (\frac {\left (t -4 \sqrt {t^{2}+T}\right ) \sqrt {17}}{17 t}\right )+2 \,\operatorname {arctanh}\left (\frac {\left (t^{2}-8 T\right ) \sqrt {17}}{17 t^{2}}\right )-2 \,\operatorname {arctanh}\left (\frac {\left (4 \sqrt {t^{2}+T}+t \right ) \sqrt {17}}{17 t}\right )\right ) \sqrt {17}}{68} &= c_2 \\
\end{align*}
✓ Time used: 0.007 (sec). Leaf size: 136 ode :=( t ^2+ T ( t ))^(1/2) = diff(T(t),t);
dsolve ( ode , T ( t ), singsol=all);
\[
-17 \ln \left (-t^{4}-T t^{2}+4 T^{2}\right )-17 \ln \left (-\sqrt {t^{2}+T}\, t +2 T\right )+17 \ln \left (\sqrt {t^{2}+T}\, t +2 T\right )+\left (2 \,\operatorname {arctanh}\left (\frac {\left (4 \sqrt {t^{2}+T}+t \right ) \sqrt {17}}{17 t}\right )-2 \,\operatorname {arctanh}\left (\frac {\left (t^{2}-8 T\right ) \sqrt {17}}{17 t^{2}}\right )-2 \,\operatorname {arctanh}\left (\frac {\left (t -4 \sqrt {t^{2}+T}\right ) \sqrt {17}}{17 t}\right )\right ) \sqrt {17}-c_1 = 0
\]
Maple trace
Methods for first order ODEs:
--- Trying classification methods ---
trying homogeneous types:
trying homogeneous G
1 st order, trying the canonical coordinates of the invariance group
<- 1st order, canonical coordinates successful
<- homogeneous successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \sqrt {t^{2}+T \left (t \right )}=\frac {d}{d t}T \left (t \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d t}T \left (t \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}T \left (t \right )=\sqrt {t^{2}+T \left (t \right )} \end {array} \]
✓ Time used: 0.171 (sec). Leaf size: 135 ode = Sqrt [ t ^2+ T [ t ]]== D [ T [ t ], t ]; ic ={}; DSolve [{ ode , ic }, T [ t ], t , IncludeSingularSolutions -> True ] \[
\text {Solve}\left [\frac {1}{34} \left (-34 \log \left (\sqrt {t^2+T(t)}-t\right )-\left (\sqrt {17}-17\right ) \log \left (2 \left (\sqrt {17}-4\right ) t \sqrt {t^2+T(t)}-2 \left (\sqrt {17}-4\right ) t^2-\left (\sqrt {17}-3\right ) T(t)\right )+\left (17+\sqrt {17}\right ) \log \left (2 \left (4+\sqrt {17}\right ) t \sqrt {t^2+T(t)}-2 \left (4+\sqrt {17}\right ) t^2-\left (3+\sqrt {17}\right ) T(t)\right )\right )=c_1,T(t)\right ]
\]
✗ from sympy import *
t = symbols("t")
T = Function("T")
ode = Eq(sqrt(t**2 + T(t)) - Derivative(T(t), t),0)
ics = {}
dsolve ( ode , func = T ( t ), ics = ics ) NotImplementedError : The given ODE -sqrt(t**2 + T(t)) + Derivative(T(t), t) cannot be solved by the