2.1.54 Problem 54 Internal
problem
ID
[10040] Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
54 Date
solved
:
Monday, December 08, 2025 at 07:11:19 PMCAS
classification
:
[[_2nd_order, _quadrature]]
2.1.54.1 second order ode quadrature 0.013 (sec)
\begin{align*}
y^{\prime \prime }&=0 \\
\end{align*}
Entering second order ode quadrature solver Integrating twice gives the solution \[ y= c_1 t + c_2 \]
Summary of solutions found
\begin{align*}
y &= c_1 t +c_2 \\
\end{align*}
Figure 2.120: Slope field \(y^{\prime \prime } = 0\) 2.1.54.2 second order linear constant coeff 0.034 (sec)
\begin{align*}
y^{\prime \prime }&=0 \\
\end{align*}
Entering second order linear constant coefficient ode solver This is second order with constant coefficients homogeneous ODE. In standard form the
ODE is
\[ A y''(t) + B y'(t) + C y(t) = 0 \]
Where in the above \(A=1, B=0, C=0\) . Let the solution be \(y=e^{\lambda t}\) . Substituting this into the ODE
gives \[ \lambda ^{2} {\mathrm e}^{t \lambda } = 0 \tag {1} \]
Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda t}\)
gives \[ \lambda ^{2} = 0 \tag {2} \]
Equation (2) is the characteristic equation of the ODE. Its roots determine the
general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]
Substituting \(A=1, B=0, C=0\) into the above gives
\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {\left (0\right )^2 - (4) \left (1\right )\left (0\right )}\\ &= 0 \end{align*}
Hence this is the case of a double root \(\lambda _{1,2} = 0\) . Therefore the solution is
\[ y= c_1 1 + c_2 t \tag {1} \]
Summary of solutions found
\begin{align*}
y &= c_2 t +c_1 \\
\end{align*}
Figure 2.121: Slope field \(y^{\prime \prime } = 0\) 2.1.54.3 second order linear exact ode 0.169 (sec)
\begin{align*}
y^{\prime \prime }&=0 \\
\end{align*}
Entering second order linear exact ode solver An ode of the form \begin{align*} p \left (t \right ) y^{\prime \prime }+q \left (t \right ) y^{\prime }+r \left (t \right ) y&=s \left (t \right ) \end{align*}
is exact if
\begin{align*} p''(t) - q'(t) + r(t) &= 0 \tag {1} \end{align*}
For the given ode we have
\begin{align*} p(x) &= 1\\ q(x) &= 0\\ r(x) &= 0\\ s(x) &= 0 \end{align*}
Hence
\begin{align*} p''(x) &= 0\\ q'(x) &= 0 \end{align*}
Therefore (1) becomes
\begin{align*} 0- \left (0\right ) + \left (0\right )&=0 \end{align*}
Hence the ode is exact. Since we now know the ode is exact, it can be written as
\begin{align*} \left (p \left (t \right ) y^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) y\right )' &= s(x) \end{align*}
Integrating gives
\begin{align*} p \left (t \right ) y^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) y&=\int {s \left (t \right )\, dt} \end{align*}
Substituting the above values for \(p,q,r,s\) gives
\begin{align*} y^{\prime }&=c_1 \end{align*}
We now have a first order ode to solve which is
\begin{align*} y^{\prime } = c_1 \end{align*}
Entering first order ode quadrature solver Since the ode has the form \(y^{\prime }=f(t)\) , then we only need to
integrate \(f(t)\) .
\begin{align*} \int {dy} &= \int {c_1\, dt}\\ y &= c_1 t + c_2 \end{align*}
Summary of solutions found
\begin{align*}
y &= c_1 t +c_2 \\
\end{align*}
Figure 2.122: Slope field \(y^{\prime \prime } = 0\) 2.1.54.4 second order ode missing y 0.148 (sec)
\begin{align*}
y^{\prime \prime }&=0 \\
\end{align*}
Entering second order ode missing \(y\) solver This is second order ode with missing dependent
variable \(y\) . Let \begin{align*} u(t) &= y^{\prime } \end{align*}
Then
\begin{align*} u'(t) &= y^{\prime \prime } \end{align*}
Hence the ode becomes
\begin{align*} u^{\prime }\left (t \right ) = 0 \end{align*}
Which is now solved for \(u(t)\) as first order ode.
Entering first order ode quadrature solver Since the ode has the form \(u^{\prime }\left (t \right )=f(t)\) , then we only need to
integrate \(f(t)\) .
\begin{align*} \int {du} &= \int {0\, dt} + c_1 \\ u \left (t \right ) &= c_1 \end{align*}
In summary, these are the solution found for \(y\)
\begin{align*}
u \left (t \right ) &= c_1 \\
\end{align*}
For solution \(u \left (t \right ) = c_1\) , since \(u=y^{\prime }\) then we now have a new first
order ode to solve which is \begin{align*} y^{\prime } = c_1 \end{align*}
Entering first order ode quadrature solver Since the ode has the form \(y^{\prime }=f(t)\) , then we only need to
integrate \(f(t)\) .
\begin{align*} \int {dy} &= \int {c_1\, dt}\\ y &= c_1 t + c_2 \end{align*}
In summary, these are the solution found for \((y)\)
\begin{align*}
y &= c_1 t +c_2 \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= c_1 t +c_2 \\
\end{align*}
Figure 2.123: Slope field \(y^{\prime \prime } = 0\) 2.1.54.5 second order integrable as is 0.075 (sec)
\begin{align*}
y^{\prime \prime }&=0 \\
\end{align*}
Entering second order integrable as is solver Integrating both sides of the ODE w.r.t \(t\) gives
\begin{align*} \int y^{\prime \prime }d t &= 0 \\ y^{\prime } = c_1 \end{align*}
Which is now solved for \(y\) . Entering first order ode quadrature solver Since the ode has the form \(y^{\prime }=f(t)\) ,
then we only need to integrate \(f(t)\) .
\begin{align*} \int {dy} &= \int {c_1\, dt}\\ y &= c_1 t + c_2 \end{align*}
Summary of solutions found
\begin{align*}
y &= c_1 t +c_2 \\
\end{align*}
Figure 2.124: Slope field \(y^{\prime \prime } = 0\) 2.1.54.6 second order kovacic 0.048 (sec)
\begin{align*}
y^{\prime \prime }&=0 \\
\end{align*}
Entering kovacic solver Writing the ode as \begin{align*} y^{\prime \prime } &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}
Comparing (1) and (2) shows that
\begin{align*} A &= 1 \\ B &= 0\tag {3} \\ C &= 0 \end{align*}
Applying the Liouville transformation on the dependent variable gives
\begin{align*} z(t) &= y e^{\int \frac {B}{2 A} \,dt} \end{align*}
Then (2) becomes
\begin{align*} z''(t) = r z(t)\tag {4} \end{align*}
Where \(r\) is given by
\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}
Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives
\begin{align*} r &= \frac {0}{1}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= 0\\ t &= 1 \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(t) &= 0 \tag {7} \end{align*}
Equation (7) is now solved. After finding \(z(t)\) then \(y\) is found using the inverse transformation
\begin{align*} y &= z \left (t \right ) e^{-\int \frac {B}{2 A} \,dt} \end{align*}
The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases
depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \) . The following table summarizes these
cases.
Case
Allowed pole order for \(r\)
Allowed value for \(\mathcal {O}(\infty )\)
1
\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)
\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
2
Need to have at least one pole
that is either order \(2\) or odd order
greater than \(2\) . Any other pole order
is allowed as long as the above
condition is satisfied. Hence the
following set of pole orders are all
allowed. \(\{1,2\}\) ,\(\{1,3\}\) ,\(\{2\}\) ,\(\{3\}\) ,\(\{3,4\}\) ,\(\{1,2,5\}\) .
no condition
3
\(\left \{ 1,2\right \} \)
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)
Table 2.20: Necessary conditions for each Kovacic case
The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\) . Therefore
\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - -\infty \\ &= \infty \end{align*}
There are no poles in \(r\) . Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole
larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met. Therefore
\begin{align*} L &= [1] \end{align*}
Since \(r = 0\) is not a function of \(t\) , then there is no need run Kovacic algorithm to obtain a
solution for transformed ode \(z''=r z\) as one solution is
\[ z_1(t) = 1 \]
Using the above, the solution for the
original ode can now be found. The first solution to the original ode in \(y\) is found from
\[
y_1 = z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt}
\]
Since \(B=0\) then the above reduces to
\begin{align*}
y_1 &= z_1 \\
&= 1 \\
\end{align*}
Which simplifies to \[
y_1 = 1
\]
The second solution \(y_2\) to the original ode is
found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{y_1^2} \,dt \]
Since \(B=0\) then the above becomes \begin{align*}
y_2 &= y_1 \int \frac {1}{y_1^2} \,dt \\
&= 1\int \frac {1}{1} \,dt \\
&= 1\left (t\right ) \\
\end{align*}
Therefore the solution
is
\begin{align*}
y &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left (1\right ) + c_2 \left (1\left (t\right )\right ) \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= c_2 t +c_1 \\
\end{align*}
Figure 2.125: Slope field \(y^{\prime \prime } = 0\) 1.198 (sec)
\begin{align*}
y^{\prime \prime }&=0 \\
\end{align*}
Applying change of variable \(t = \arccos \left (\tau \right )\) to the above ode results in the following new ode
\[
\left (-\tau ^{2}+1\right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )-\left (\frac {d}{d \tau }y \left (\tau \right )\right ) \tau = 0
\]
Which is now
solved for \(y \left (\tau \right )\) . Entering second order ode missing \(y\) solver This is second order ode with missing
dependent variable \(y \left (\tau \right )\) . Let \begin{align*} u(\tau ) &= \frac {d}{d \tau }y \left (\tau \right ) \end{align*}
Then
\begin{align*} u'(\tau ) &= \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right ) \end{align*}
Hence the ode becomes
\begin{align*} \left (-\tau ^{2}+1\right ) \left (\frac {d}{d \tau }u \left (\tau \right )\right )-u \left (\tau \right ) \tau = 0 \end{align*}
Which is now solved for \(u(\tau )\) as first order ode.
Entering first order ode linear solver In canonical form a linear first order is
\begin{align*} \frac {d}{d \tau }u \left (\tau \right ) + q(\tau )u \left (\tau \right ) &= p(\tau ) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(\tau ) &=\frac {\tau }{\tau ^{2}-1}\\ p(\tau ) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,d\tau }}\\ &= {\mathrm e}^{\int \frac {\tau }{\tau ^{2}-1}d \tau }\\ &= \sqrt {\tau ^{2}-1} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }} \left (u \sqrt {\tau ^{2}-1}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} u \sqrt {\tau ^{2}-1}&= \int {0 \,d\tau } + c_1 \\ &=c_1 \end{align*}
Dividing throughout by the integrating factor \(\sqrt {\tau ^{2}-1}\) gives the final solution
\[ u \left (\tau \right ) = \frac {c_1}{\sqrt {\tau ^{2}-1}} \]
In summary, these are the
solution found for \(y \left (\tau \right )\) \begin{align*}
u \left (\tau \right ) &= \frac {c_1}{\sqrt {\tau ^{2}-1}} \\
\end{align*}
For solution \(u \left (\tau \right ) = \frac {c_1}{\sqrt {\tau ^{2}-1}}\) , since \(u=\frac {d}{d \tau }y \left (\tau \right )\) then we now have a new first order ode to solve which is
\begin{align*} \frac {d}{d \tau }y \left (\tau \right ) = \frac {c_1}{\sqrt {\tau ^{2}-1}} \end{align*}
Entering first order ode quadrature solver Since the ode has the form \(\frac {d}{d \tau }y \left (\tau \right )=f(\tau )\) , then we only need to
integrate \(f(\tau )\) .
\begin{align*} \int {dy} &= \int {\frac {c_1}{\sqrt {\tau ^{2}-1}}\, d\tau }\\ y \left (\tau \right ) &= c_1 \ln \left (\tau +\sqrt {\tau ^{2}-1}\right ) + c_2 \end{align*}
In summary, these are the solution found for \((y \left (\tau \right ))\)
\begin{align*}
y \left (\tau \right ) &= c_1 \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_2 \\
\end{align*}
Applying change of variable \(\tau = \cos \left (t \right )\) to the solutions
above gives \begin{align*}
y &= c_1 \ln \left (\cos \left (t \right )+\sqrt {\cos \left (t \right )^{2}-1}\right )+c_2 \\
\end{align*}
Figure 2.126: Slope field \(y^{\prime \prime } = 0\) 2.1.54.8 ✓ Maple. Time used: 0.000 (sec). Leaf size: 9 ode := diff ( diff ( y ( t ), t ), t ) = 0;
dsolve ( ode , y ( t ), singsol=all);
\[
y = c_1 t +c_2
\]
Maple trace
Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=t \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} y_{1}\left (t \right )+\mathit {C2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C2} t +\mathit {C1} \end {array} \]
2.1.54.9 ✓ Mathematica. Time used: 0.002 (sec). Leaf size: 12 ode = D [ y [ t ],{ t ,2}]==0; ic ={}; DSolve [{ ode , ic }, y [ t ], t , IncludeSingularSolutions -> True ] \begin{align*} y(t)&\to c_2 t+c_1 \end{align*}
2.1.54.10 ✓ Sympy. Time used: 0.016 (sec). Leaf size: 7 from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(Derivative(y(t), (t, 2)),0)
ics = {}
dsolve ( ode , func = y ( t ), ics = ics ) \[
y{\left (t \right )} = C_{1} + C_{2} t
\]