2.1.56 problem 56
Internal
problem
ID
[8194]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
56
Date
solved
:
Sunday, November 10, 2024 at 03:08:31 AM
CAS
classification
:
[[_2nd_order, _quadrature]]
Solve
\begin{align*} y^{\prime \prime }&=f \left (t \right ) \end{align*}
Solved as second order ode quadrature
Time used: 0.039 (sec)
Integrating once gives
\[ y^{\prime }= \int f \left (t \right )d t + c_1 \]
Integrating again gives
\[ y= \int \left (\int f \left (t \right )d t\right ) \,dt + c_1 x + c_2 \]
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= \int \left (\int f \left (t \right )d t \right )d t +c_1 t +c_2 \\
\end{align*}
Solved as second order linear constant coeff ode
Time used: 0.356 (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 \(A=1, B=0, C=0, f(t)=f \left (t \right )\).
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
\[ \lambda ^{2} {\mathrm e}^{t \lambda } = 0 \tag {1} \]
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
\[ y= c_1 1 + c_2 t \tag {1} \]
Therefore the
homogeneous solution \(y_h\) is
\[
y_h = c_2 t +c_1
\]
The particular solution \(y_p\) can be found using either the method of
undetermined coefficients, or the method of variation of parameters. The method of variation
of parameters will be used as it is more general and can be used when the coefficients of the
ODE depend on \(t\) as well. Let
\begin{equation}
\tag{1} y_p(t) = u_1 y_1 + u_2 y_2
\end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the
two linearly independent solutions of the homogeneous ODE) found earlier when
solving the homogeneous ODE as
\begin{align*}
y_1 &= 1 \\
y_2 &= t \\
\end{align*}
In the Variation of parameters \(u_1,u_2\) are found using
\begin{align*}
\tag{2} u_1 &= -\int \frac {y_2 f(t)}{a W(t)} \\
\tag{3} u_2 &= \int \frac {y_1 f(t)}{a W(t)} \\
\end{align*}
Where \(W(t)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The
Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence
\[ W = \begin {vmatrix} 1 & t \\ \frac {d}{dt}\left (1\right ) & \frac {d}{dt}\left (t\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} 1 & t \\ 0 & 1 \end {vmatrix} \]
Therefore
\[
W = \left (1\right )\left (1\right ) - \left (t\right )\left (0\right )
\]
Which simplifies to
\[
W = 1
\]
Which
simplifies to
\[
W = 1
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {t f \left (t \right )}{1}\,dt
\]
Which simplifies to
\[
u_1 = - \int t f \left (t \right )d t
\]
Hence
\[
u_1 = -\left (\int _{0}^{t}\alpha f \left (\alpha \right )d \alpha \right )
\]
And Eq. (3)
becomes
\[
u_2 = \int \frac {f \left (t \right )}{1}\,dt
\]
Which simplifies to
\[
u_2 = \int f \left (t \right )d t
\]
Hence
\[
u_2 = \int _{0}^{t}f \left (\alpha \right )d \alpha
\]
Therefore the particular solution, from equation
(1) is
\[
y_p(t) = -\left (\int _{0}^{t}\alpha f \left (\alpha \right )d \alpha \right )+t \left (\int _{0}^{t}f \left (\alpha \right )d \alpha \right )
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (c_2 t +c_1\right ) + \left (-\left (\int _{0}^{t}\alpha f \left (\alpha \right )d \alpha \right )+t \left (\int _{0}^{t}f \left (\alpha \right )d \alpha \right )\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= -\left (\int _{0}^{t}\alpha f \left (\alpha \right )d \alpha \right )+t \left (\int _{0}^{t}f \left (\alpha \right )d \alpha \right )+c_2 t +c_1 \\
\end{align*}
Solved as second order linear exact ode
Time used: 0.049 (sec)
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) &= f \left (t \right ) \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 }&=\int {f \left (t \right )\, dt} \end{align*}
We now have a first order ode to solve which is
\begin{align*} y^{\prime } = \int f \left (t \right )d 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 {\int f \left (t \right )d t +c_1\, dt}\\ y &= \int \left (\int f \left (t \right )d t +c_1 \right )d t + c_2 \end{align*}
\begin{align*} y&= \int \left (\int f \left (t \right )d t +c_1 \right )d t +c_2 \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \int \left (\int f \left (t \right )d t +c_1 \right )d t +c_2 \\
\end{align*}
Solved as second order missing y ode
Time used: 0.047 (sec)
This is second order ode with missing dependent variable \(y\). Let
\begin{align*} p(t) &= y^{\prime } \end{align*}
Then
\begin{align*} p'(t) &= y^{\prime \prime } \end{align*}
Hence the ode becomes
\begin{align*} p^{\prime }\left (t \right )-f \left (t \right ) = 0 \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 {f \left (t \right )\, dt}\\ p \left (t \right ) &= \int f \left (t \right )d t + c_1 \end{align*}
\begin{align*} p \left (t \right )&= \int f \left (t \right )d 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 } = \int f \left (t \right )d 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 {\int f \left (t \right )d t +c_1\, dt}\\ y &= \int \left (\int f \left (t \right )d t +c_1 \right )d t + c_2 \end{align*}
\begin{align*} y&= \int \left (\int f \left (t \right )d t +c_1 \right )d t +c_2 \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \int \left (\int f \left (t \right )d t +c_1 \right )d t +c_2 \\
\end{align*}
Solved as second order ode using Kovacic algorithm
Time used: 0.080 (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.
| | |
| 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 2.13: 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*}
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 \(y_p\) can be found using either the method of undetermined coefficients,
or the method of variation of parameters. The method of variation of parameters will be
used as it is more general and can be used when the coefficients of the ODE depend
on \(t\) as well. Let
\begin{equation}
\tag{1} y_p(t) = u_1 y_1 + u_2 y_2
\end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the
two linearly independent solutions of the homogeneous ODE) found earlier when
solving the homogeneous ODE as
\begin{align*}
y_1 &= 1 \\
y_2 &= t \\
\end{align*}
In the Variation of parameters \(u_1,u_2\) are found using
\begin{align*}
\tag{2} u_1 &= -\int \frac {y_2 f(t)}{a W(t)} \\
\tag{3} u_2 &= \int \frac {y_1 f(t)}{a W(t)} \\
\end{align*}
Where \(W(t)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The
Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence
\[ W = \begin {vmatrix} 1 & t \\ \frac {d}{dt}\left (1\right ) & \frac {d}{dt}\left (t\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} 1 & t \\ 0 & 1 \end {vmatrix} \]
Therefore
\[
W = \left (1\right )\left (1\right ) - \left (t\right )\left (0\right )
\]
Which simplifies to
\[
W = 1
\]
Which
simplifies to
\[
W = 1
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {t f \left (t \right )}{1}\,dt
\]
Which simplifies to
\[
u_1 = - \int t f \left (t \right )d t
\]
Hence
\[
u_1 = -\left (\int _{0}^{t}\alpha f \left (\alpha \right )d \alpha \right )
\]
And Eq. (3)
becomes
\[
u_2 = \int \frac {f \left (t \right )}{1}\,dt
\]
Which simplifies to
\[
u_2 = \int f \left (t \right )d t
\]
Hence
\[
u_2 = \int _{0}^{t}f \left (\alpha \right )d \alpha
\]
Therefore the particular solution, from equation
(1) is
\[
y_p(t) = -\left (\int _{0}^{t}\alpha f \left (\alpha \right )d \alpha \right )+t \left (\int _{0}^{t}f \left (\alpha \right )d \alpha \right )
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (c_2 t +c_1\right ) + \left (-\left (\int _{0}^{t}\alpha f \left (\alpha \right )d \alpha \right )+t \left (\int _{0}^{t}f \left (\alpha \right )d \alpha \right )\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= -\left (\int _{0}^{t}\alpha f \left (\alpha \right )d \alpha \right )+t \left (\int _{0}^{t}f \left (\alpha \right )d \alpha \right )+c_2 t +c_1 \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=f \left (t \right ) \\ \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 homogeneous 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 homogeneous 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 )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} +\mathit {C2} t +y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \left (\int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right )+y_{2}\left (t \right ) \left (\int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=f \left (t \right )\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} 1 & t \\ 0 & 1 \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\left (\int t f \left (t \right )d t \right )+t \left (\int f \left (t \right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\left (\int t f \left (t \right )d t \right )+t \left (\int f \left (t \right )d t \right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} +\mathit {C2} t -\left (\int t f \left (t \right )d t \right )+t \left (\int f \left (t \right )d t \right ) \end {array} \]
Maple trace
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful`
Maple dsolve solution
Solving time : 0.003 (sec)
Leaf size : 15
dsolve(diff(diff(y(t),t),t) = f(t),
y(t),singsol=all)
\[
y = \int \left (\int f \left (t \right )d t \right )d t +c_{1} t +c_{2}
\]
Mathematica DSolve solution
Solving time : 0.011 (sec)
Leaf size : 30
DSolve[{D[y[t],{t,2}]==f[t],{}},
y[t],t,IncludeSingularSolutions->True]
\[
y(t)\to \int _1^t\int _1^{K[2]}f(K[1])dK[1]dK[2]+c_2 t+c_1
\]