2.1.48 problem 48

Internal problem ID [8186]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 48
Date solved : Sunday, November 10, 2024 at 03:08:18 AM
CAS classification : [[_2nd_order, _missing_y]]

Solve

\begin{align*} \left (t^{2}+9\right ) y^{\prime \prime }+2 t y^{\prime }&=0 \end{align*}

With initial conditions

\begin{align*} y \left (3\right )&=2 \pi \\ y^{\prime }\left (3\right )&={\frac {2}{3}} \end{align*}

Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as

\begin{align*} y^{\prime \prime } + p(t)y^{\prime } + q(t) y &= F \end{align*}

Where here

\begin{align*} p(t) &=\frac {2 t}{t^{2}+9}\\ q(t) &=0\\ F &=0 \end{align*}

Hence the ode is

\begin{align*} y^{\prime \prime }+\frac {2 t y^{\prime }}{t^{2}+9} = 0 \end{align*}

The domain of \(p(t)=\frac {2 t}{t^{2}+9}\) is

\[ \{-\infty <t <\infty \} \]

And the point \(t_0 = 3\) is inside this domain. Hence solution exists and is unique.

Solved as second order linear exact ode

Time used: 0.180 (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) &= t^{2}+9\\ q(x) &= 2 t\\ r(x) &= 0\\ s(x) &= 0 \end{align*}

Hence

\begin{align*} p''(x) &= 2\\ q'(x) &= 2 \end{align*}

Therefore (1) becomes

\begin{align*} 2- \left (2\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*} \left (t^{2}+9\right ) y^{\prime }&=c_1 \end{align*}

We now have a first order ode to solve which is

\begin{align*} \left (t^{2}+9\right ) y^{\prime } = 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 {\frac {c_1}{t^{2}+9}\, dt}\\ y &= \frac {c_1 \arctan \left (\frac {t}{3}\right )}{3} + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= 4 \arctan \left (\frac {t}{3}\right )+\pi \\ \end{align*}

Figure 2.107: Solution plot
\(y = 4 \arctan \left (\frac {t}{3}\right )+\pi \)

Solved as second order missing y ode

Time used: 0.075 (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*} \left (t^{2}+9\right ) p^{\prime }\left (t \right )+2 t p \left (t \right ) = 0 \end{align*}

Which is now solve for \(p(t)\) as first order ode. In canonical form a linear first order is

\begin{align*} p^{\prime }\left (t \right ) + q(t)p \left (t \right ) &= p(t) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(t) &=\frac {2 t}{t^{2}+9}\\ p(t) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int \frac {2 t}{t^{2}+9}d t}\\ &= t^{2}+9 \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \mu p &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (p \left (t^{2}+9\right )\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} p \left (t^{2}+9\right )&= \int {0 \,dt} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(t^{2}+9\) gives the final solution

\[ p \left (t \right ) = \frac {c_1}{t^{2}+9} \]

Solving for the constant of integration from initial conditions, the solution becomes

\begin{align*} p \left (t \right ) = \frac {12}{t^{2}+9} \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 } = \frac {12}{t^{2}+9} \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 {\frac {12}{t^{2}+9}\, dt}\\ y &= 4 \arctan \left (\frac {t}{3}\right ) + c_2 \end{align*}

Solving for the constant of integration from initial conditions, the solution becomes

\begin{align*} y = 4 \arctan \left (\frac {t}{3}\right )+\pi \end{align*}

Summary of solutions found

\begin{align*} y &= 4 \arctan \left (\frac {t}{3}\right )+\pi \\ \end{align*}

Figure 2.108: Solution plot
\(y = 4 \arctan \left (\frac {t}{3}\right )+\pi \)

Solved as second order integrable as is ode

Time used: 0.040 (sec)

Integrating both sides of the ODE w.r.t \(t\) gives

\begin{align*} \int \left (\left (t^{2}+9\right ) y^{\prime \prime }+2 t y^{\prime }\right )d t &= 0 \\ \left (t^{2}+9\right ) y^{\prime } = 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 {\frac {c_1}{t^{2}+9}\, dt}\\ y &= \frac {c_1 \arctan \left (\frac {t}{3}\right )}{3} + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= 4 \arctan \left (\frac {t}{3}\right )+\pi \\ \end{align*}

Figure 2.109: Solution plot
\(y = 4 \arctan \left (\frac {t}{3}\right )+\pi \)

Solved as second order integrable as is ode (ABC method)

Time used: 0.036 (sec)

Writing the ode as

\[ \left (t^{2}+9\right ) y^{\prime \prime }+2 t y^{\prime } = 0 \]

Integrating both sides of the ODE w.r.t \(t\) gives

\begin{align*} \int \left (\left (t^{2}+9\right ) y^{\prime \prime }+2 t y^{\prime }\right )d t &= 0 \\ \left (t^{2}+9\right ) y^{\prime } = 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 {\frac {c_1}{t^{2}+9}\, dt}\\ y &= \frac {c_1 \arctan \left (\frac {t}{3}\right )}{3} + c_2 \end{align*}

Will add steps showing solving for IC soon.

Figure 2.110: Solution plot
\(y = 4 \arctan \left (\frac {t}{3}\right )+\pi \)

Solved as second order ode using Kovacic algorithm

Time used: 0.225 (sec)

Writing the ode as

\begin{align*} \left (t^{2}+9\right ) y^{\prime \prime }+2 t y^{\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 &= t^{2}+9 \\ B &= 2 t\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 {9}{\left (t^{2}+9\right )^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 9\\ t &= \left (t^{2}+9\right )^{2} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(t) &= \left ( \frac {9}{\left (t^{2}+9\right )^{2}}\right ) z(t)\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 2.6: 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) \\ &= 4 - 0 \\ &= 4 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=\left (t^{2}+9\right )^{2}\). There is a pole at \(t=3 i\) of order \(2\). There is a pole at \(t=-3 i\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(4\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then necessary conditions for case two are met. Since pole order is not larger than \(2\) and the order at \(\infty \) is \(4\) then the necessary conditions for case three are met. Therefore

\begin{align*} L &= [1, 2, 4, 6, 12] \end{align*}

Attempting to find a solution using case \(n=1\).

Looking at poles of order 2. The partial fractions decomposition of \(r\) is

\[ r = -\frac {1}{4 \left (t -3 i\right )^{2}}-\frac {1}{4 \left (t +3 i\right )^{2}}-\frac {i}{12 \left (t -3 i\right )}+\frac {i}{12 t +36 i} \]

For the pole at \(t=3 i\) let \(b\) be the coefficient of \(\frac {1}{ \left (t -3 i\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {1}{4}}\). Hence

\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {1}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{2}} \end{alignat*}

For the pole at \(t=-3 i\) let \(b\) be the coefficient of \(\frac {1}{ \left (t +3 i\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {1}{4}}\). Hence

\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {1}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{2}} \end{alignat*}

Since the order of \(r\) at \(\infty \) is \(4 > 2\) then

\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= 0 \\ \alpha _{\infty }^{-} &= 1 \end{alignat*}

The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is

\[ r=\frac {9}{\left (t^{2}+9\right )^{2}} \]

pole \(c\) location	pole order	\([\sqrt r]_c\)	\(\alpha _c^{+}\)	\(\alpha _c^{-}\)
\(3 i\)	\(2\)	\(0\)	\(\frac {1}{2}\)	\(\frac {1}{2}\)
\(-3 i\)	\(2\)	\(0\)	\(\frac {1}{2}\)	\(\frac {1}{2}\)

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(4\)	\(0\)	\(0\)	\(1\)

Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using

\begin{align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end{align*}

Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{-} = 1\) then

\begin{align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{+}+\alpha _{c_2}^{+} \right ) \\ &= 1 - \left ( 1 \right ) \\ &= 0 \end{align*}

Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using

\begin{align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{t-c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}

The above gives

\begin{align*} \omega &= \left ( (+) [\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{+} }{t- c_1}\right )+\left ( (+) [\sqrt r]_{c_2} + \frac { \alpha _{c_2}^{+} }{t- c_2}\right ) + (-) [\sqrt r]_\infty \\ &= \frac {1}{2 t -6 i}+\frac {1}{2 t +6 i} + (-) \left ( 0 \right ) \\ &= \frac {1}{2 t -6 i}+\frac {1}{2 t +6 i}\\ &= \frac {t}{t^{2}+9} \end{align*}

Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(t)\) of degree \(d=0\) to solve the ode. The polynomial \(p(t)\) needs to satisfy the equation

\begin{align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end{align*}

Let

\begin{align*} p(t) &= 1\tag {2A} \end{align*}

Substituting the above in eq. (1A) gives

\begin{align*} \left (0\right ) + 2 \left (\frac {1}{2 t -6 i}+\frac {1}{2 t +6 i}\right ) \left (0\right ) + \left ( \left (-\frac {1}{2 \left (t -3 i\right )^{2}}-\frac {1}{2 \left (t +3 i\right )^{2}}\right ) + \left (\frac {1}{2 t -6 i}+\frac {1}{2 t +6 i}\right )^2 - \left (\frac {9}{\left (t^{2}+9\right )^{2}}\right ) \right ) &= 0\\ 0 = 0 \end{align*}

The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is

\begin{align*} z_1(t) &= p e^{ \int \omega \,dt} \\ &= {\mathrm e}^{\int \left (\frac {1}{2 t -6 i}+\frac {1}{2 t +6 i}\right )d t}\\ &= \sqrt {-t^{2}-9} \end{align*}

The first solution to the original ode in \(y\) is found from

\begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {2 t}{t^{2}+9} \,dt} \\ &= z_1 e^{-\frac {\ln \left (t^{2}+9\right )}{2}} \\ &= z_1 \left (\frac {1}{\sqrt {t^{2}+9}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = i \]

The second solution \(y_2\) to the original ode is found using reduction of order

\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{y_1^2} \,dt \]

Substituting gives

\begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {2 t}{t^{2}+9} \,dt}}{\left (y_1\right )^2} \,dt \\ &= y_1 \int \frac { e^{-\ln \left (t^{2}+9\right )}}{\left (y_1\right )^2} \,dt \\ &= y_1 \left (-\frac {\arctan \left (\frac {t}{3}\right )}{3}\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (i\right ) + c_2 \left (i\left (-\frac {\arctan \left (\frac {t}{3}\right )}{3}\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= 4 \arctan \left (\frac {t}{3}\right )+\pi \\ \end{align*}

Figure 2.111: Solution plot
\(y = 4 \arctan \left (\frac {t}{3}\right )+\pi \)

Solved as second order ode adjoint method

Time used: 0.512 (sec)

In normal form the ode

\begin{align*} \left (t^{2}+9\right ) y^{\prime \prime }+2 t y^{\prime } = 0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=r \left (t \right ) \tag {2} \end{align*}

Where

\begin{align*} p \left (t \right )&=\frac {2 t}{t^{2}+9}\\ q \left (t \right )&=0\\ r \left (t \right )&=0 \end{align*}

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {2 t \xi \left (t \right )}{t^{2}+9}\right )' + \left (0\right ) &= 0\\ \xi ^{\prime \prime }\left (t \right )-\frac {2 t \xi ^{\prime }\left (t \right )}{t^{2}+9}+\frac {\left (2 t^{2}-18\right ) \xi \left (t \right )}{\left (t^{2}+9\right )^{2}}&= 0 \end{align*}

Which is solved for \(\xi (t)\). An ode of the form

\begin{align*} p \left (t \right ) \xi ^{\prime \prime }+q \left (t \right ) \xi ^{\prime }+r \left (t \right ) \xi &=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) &= -\frac {2 t}{t^{2}+9}\\ r(x) &= \frac {2 t^{2}-18}{\left (t^{2}+9\right )^{2}}\\ s(x) &= 0 \end{align*}

Hence

\begin{align*} p''(x) &= 0\\ q'(x) &= -\frac {2}{t^{2}+9}+\frac {4 t^{2}}{\left (t^{2}+9\right )^{2}} \end{align*}

Therefore (1) becomes

\begin{align*} 0- \left (-\frac {2}{t^{2}+9}+\frac {4 t^{2}}{\left (t^{2}+9\right )^{2}}\right ) + \left (\frac {2 t^{2}-18}{\left (t^{2}+9\right )^{2}}\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 ) \xi ^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) \xi \right )' &= s(x) \end{align*}

Integrating gives

\begin{align*} p \left (t \right ) \xi ^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) \xi &=\int {s \left (t \right )\, dt} \end{align*}

Substituting the above values for \(p,q,r,s\) gives

\begin{align*} \xi ^{\prime }-\frac {2 t \xi }{t^{2}+9}&=c_1 \end{align*}

We now have a first order ode to solve which is

\begin{align*} \xi ^{\prime }-\frac {2 t \xi }{t^{2}+9} = c_1 \end{align*}

In canonical form a linear first order is

\begin{align*} \xi ^{\prime } + q(t)\xi &= p(t) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(t) &=-\frac {2 t}{t^{2}+9}\\ p(t) &=\frac {c_1 \,t^{2}+9 c_1}{t^{2}+9} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -\frac {2 t}{t^{2}+9}d t}\\ &= \frac {1}{t^{2}+9} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu \xi \right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu \xi \right ) &= \left (\mu \right ) \left (\frac {c_1 \,t^{2}+9 c_1}{t^{2}+9}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (\frac {\xi }{t^{2}+9}\right ) &= \left (\frac {1}{t^{2}+9}\right ) \left (\frac {c_1 \,t^{2}+9 c_1}{t^{2}+9}\right ) \\ \mathrm {d} \left (\frac {\xi }{t^{2}+9}\right ) &= \left (\frac {c_1 \,t^{2}+9 c_1}{\left (t^{2}+9\right )^{2}}\right )\, \mathrm {d} t \\ \end{align*}

Integrating gives

\begin{align*} \frac {\xi }{t^{2}+9}&= \int {\frac {c_1 \,t^{2}+9 c_1}{\left (t^{2}+9\right )^{2}} \,dt} \\ &=\frac {c_1 \arctan \left (\frac {t}{3}\right )}{3} + c_2 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{t^{2}+9}\) gives the final solution

\[ \xi = \left (t^{2}+9\right ) \left (\frac {c_1 \arctan \left (\frac {t}{3}\right )}{3}+c_2 \right ) \]

Will add steps showing solving for IC soon.

The original ode now reduces to first order ode

\begin{align*} \xi \left (t \right ) y^{\prime }-y \xi ^{\prime }\left (t \right )+\xi \left (t \right ) p \left (t \right ) y&=\int \xi \left (t \right ) r \left (t \right )d t\\ y^{\prime }+y \left (p \left (t \right )-\frac {\xi ^{\prime }\left (t \right )}{\xi \left (t \right )}\right )&=\frac {\int \xi \left (t \right ) r \left (t \right )d t}{\xi \left (t \right )} \end{align*}

Or

\begin{align*} y^{\prime }+y \left (\frac {2 t}{t^{2}+9}-\frac {2 t \left (\frac {c_1 \arctan \left (\frac {t}{3}\right )}{3}+c_2 \right )+c_1}{\left (t^{2}+9\right ) \left (\frac {c_1 \arctan \left (\frac {t}{3}\right )}{3}+c_2 \right )}\right )&=0 \end{align*}

Which is now a first order ode. This is now solved for \(y\). In canonical form a linear first order is

\begin{align*} y^{\prime } + q(t)y &= p(t) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(t) &=-\frac {3 c_1}{\left (c_1 \arctan \left (\frac {t}{3}\right )+3 c_2 \right ) \left (t^{2}+9\right )}\\ p(t) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -\frac {3 c_1}{\left (c_1 \arctan \left (\frac {t}{3}\right )+3 c_2 \right ) \left (t^{2}+9\right )}d t}\\ &= \frac {1}{c_1 \arctan \left (\frac {t}{3}\right )+3 c_2} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (\frac {y}{c_1 \arctan \left (\frac {t}{3}\right )+3 c_2}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} \frac {y}{c_1 \arctan \left (\frac {t}{3}\right )+3 c_2}&= \int {0 \,dt} + c_3 \\ &=c_3 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{c_1 \arctan \left (\frac {t}{3}\right )+3 c_2}\) gives the final solution

\[ y = \left (c_1 \arctan \left (\frac {t}{3}\right )+3 c_2 \right ) c_3 \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \left (c_1 \arctan \left (\frac {t}{3}\right )+3 c_2 \right ) c_3 \\ \end{align*}

The constants can be merged to give

\[ y = c_1 \arctan \left (\frac {t}{3}\right )+3 c_2 \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= 4 \arctan \left (\frac {t}{3}\right )+\pi \\ \end{align*}

Figure 2.112: Solution plot
\(y = 4 \arctan \left (\frac {t}{3}\right )+\pi \)

Maple step by step solution

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
<- LODE missing y successful`
 

Maple dsolve solution

Solving time : 0.075 (sec)
Leaf size : 12

dsolve([(t^2+9)*diff(diff(y(t),t),t)+2*t*diff(y(t),t) = 0, 
       op([y(3) = 2*Pi, D(y)(3) = 2/3])],y(t),singsol=all)
 

\[ y = \pi +4 \arctan \left (\frac {t}{3}\right ) \]

Mathematica DSolve solution

Solving time : 0.02 (sec)
Leaf size : 15

DSolve[{(t^2+9)*D[y[t],{t,2}]+2*t*D[y[t],t]==0,{y[3]==2*Pi,Derivative[1][y][3 ]==2/3}}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to 4 \arctan \left (\frac {t}{3}\right )+\pi \]