Internal
problem
ID
[7747] Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
55 Date
solved
:
Monday, October 21, 2024 at 04:01:39 PM CAS
classification
:
[[_2nd_order, _quadrature]]
Solve
\begin{align*} y^{\prime \prime }&=1 \end{align*}
1.55.1 Solved as second order ode quadrature
Time used: 0.021 (sec)
The ODE can be written as
\[ y^{\prime \prime } = 1 \]
Integrating once gives
\[ y^{\prime }= t + c_1 \]
Integrating again gives
\[ y= \frac {t^{2}}{2} + c_1 x + c_2 \]
Will add steps
showing solving for IC soon.
1.55.2 Solved as second order linear constant coeff ode
Time used: 0.086 (sec)
This is second order non-homogeneous ODE. In standard form the ODE is
\[ A y''(t) + B y'(t) + C y(t) = f(t) \]
Where \(A=1, B=0, C=0, f(t)=1\).
Let the solution be
\[ y = y_h + y_p \]
Where \(y_h\) is the solution to the homogeneous ODE \( A y''(t) + B y'(t) + C y(t) = 0\), and \(y_p\) is a
particular solution to the non-homogeneous ODE \(A y''(t) + B y'(t) + C y(t) = f(t)\). \(y_h\) is the solution to
\[ y^{\prime \prime } = 0 \]
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
Hence this is the case of a double root \(\lambda _{1,2} = 0\). Therefore the solution is
\[ y= c_1 1 + c_2 t \tag {1} \]
Therefore the
homogeneous solution \(y_h\) is
\[
y_h = c_2 t +c_1
\]
The particular solution is now found using the method of
undetermined coefficients. Looking at the RHS of the ode, which is
\[ 1 \]
Shows that the
corresponding undetermined set of the basis functions (UC_set) for the trial solution is
\[ [\{1\}] \]
While the set of the basis functions for the homogeneous solution found earlier is
\[ \{1, t\} \]
Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The
UC_set becomes
\[ [\{t\}] \]
Since \(t\) is duplicated in the UC_set, then this basis is multiplied
by extra \(t\). The UC_set becomes
\[ [\{t^{2}\}] \]
Since there was duplication between the basis
functions in the UC_set and the basis functions of the homogeneous solution, the trial
solution is a linear combination of all the basis function in the above updated
UC_set.
\[
y_p = A_{1} t^{2}
\]
The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the
ODE and comparing coefficients. Substituting the trial solution into the ODE
and simplifying gives
\[
2 A_{1} = 1
\]
Solving for the unknowns by comparing coefficients results
in
\[ \left [A_{1} = {\frac {1}{2}}\right ] \]
Substituting the above back in the above trial solution \(y_p\), gives the particular
solution
\[
y_p = \frac {t^{2}}{2}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (c_2 t +c_1\right ) + \left (\frac {t^{2}}{2}\right ) \\
\end{align*}
Which is now solve for \(p(t)\) as first order ode. Since the ode has the form \(p^{\prime }\left (t \right )=f(t)\), then we only need to
integrate \(f(t)\).
\begin{align*} \int {dp} &= \int {1\, dt}\\ p \left (t \right ) &= t + c_1 \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 } = t +c_1 \end{align*}
Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).
\begin{align*} \int {dy} &= \int {t +c_1\, dt}\\ y &= \frac {1}{2} t^{2}+c_1 t + c_2 \end{align*}
Will add steps showing solving for IC soon.
1.55.5 Solved as second order integrable as is ode
Time used: 0.026 (sec)
Integrating both sides of the ODE w.r.t \(t\) gives
\begin{align*} \int y^{\prime \prime }d t &= \int 1d t\\ y^{\prime } = t + c_1 \end{align*}
Which is now solved for \(y\). Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).
\begin{align*} \int {dy} &= \int {t +c_1\, dt}\\ y &= \frac {1}{2} t^{2}+c_1 t + c_2 \end{align*}
Will add steps showing solving for IC soon.
1.55.6 Solved as second order integrable as is ode (ABC method)
Time used: 0.026 (sec)
Writing the ode as
\[
y^{\prime \prime } = 1
\]
Integrating both sides of the ODE w.r.t \(t\) gives
\begin{align*} \int y^{\prime \prime }d t &= \int 1d t\\ y^{\prime } = t +c_1 \end{align*}
Which is now solved for \(y\). Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).
\begin{align*} \int {dy} &= \int {t +c_1\, dt}\\ y &= \frac {1}{2} t^{2}+c_1 t + c_2 \end{align*}
Will add steps showing solving for IC soon.
1.55.7 Solved as second order can be made integrable
\begin{align*} \int \frac {1}{\sqrt {2 y +2 c_1}}d y &= dt\\ \sqrt {2 y +2 c_1}&= t +c_2 \end{align*}
Singular solutions are found by solving
\begin{align*} \sqrt {2 y +2 c_1}&= 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 = -c_1 \end{align*}
Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} y&=-c_1\\ y&=\frac {1}{2} c_2^{2}+c_2 t +\frac {1}{2} t^{2}-c_1 \end{align*}
Solving Eq. (2)
Integrating gives
\begin{align*} \int -\frac {1}{\sqrt {2 y +2 c_1}}d y &= dt\\ -\sqrt {2 y +2 c_1}&= t +c_3 \end{align*}
Singular solutions are found by solving
\begin{align*} -\sqrt {2 y +2 c_1}&= 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 = -c_1 \end{align*}
Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
was found not to satisfy the ode or the IC. Hence it is removed.
1.55.8 Solved as second order ode using Kovacic algorithm
Time used: 0.042 (sec)
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.
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 12: Necessary conditions for each Kovacic case
The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore
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
This is second order nonhomogeneous ODE. Let the solution be
\[
y = y_h + y_p
\]
Where \(y_h\) is the solution
to the homogeneous ODE \( A y''(t) + B y'(t) + C y(t) = 0\), and \(y_p\) is a particular solution to the nonhomogeneous
ODE \(A y''(t) + B y'(t) + C y(t) = f(t)\). \(y_h\) is the solution to
\[
y^{\prime \prime } = 0
\]
The homogeneous solution is found using the Kovacic
algorithm which results in
\[
y_h = c_2 t +c_1
\]
The particular solution is now found using the method of
undetermined coefficients. Looking at the RHS of the ode, which is
\[ 1 \]
Shows that the
corresponding undetermined set of the basis functions (UC_set) for the trial solution is
\[ [\{1\}] \]
While the set of the basis functions for the homogeneous solution found earlier is
\[ \{1, t\} \]
Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The
UC_set becomes
\[ [\{t\}] \]
Since \(t\) is duplicated in the UC_set, then this basis is multiplied
by extra \(t\). The UC_set becomes
\[ [\{t^{2}\}] \]
Since there was duplication between the basis
functions in the UC_set and the basis functions of the homogeneous solution, the trial
solution is a linear combination of all the basis function in the above updated
UC_set.
\[
y_p = A_{1} t^{2}
\]
The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the
ODE and comparing coefficients. Substituting the trial solution into the ODE
and simplifying gives
\[
2 A_{1} = 1
\]
Solving for the unknowns by comparing coefficients results
in
\[ \left [A_{1} = {\frac {1}{2}}\right ] \]
Substituting the above back in the above trial solution \(y_p\), gives the particular
solution
\[
y_p = \frac {t^{2}}{2}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (c_2 t +c_1\right ) + \left (\frac {t^{2}}{2}\right ) \\
\end{align*}