1.49 problem 49

Internal problem ID [7741]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 49
Date solved : Monday, October 21, 2024 at 04:01:29 PM
CAS classification : [[_Emden, _Fowler]]

Solve

\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y&=0 \end{align*}

1.49.1 Solved as second order Euler type ode

Time used: 0.096 (sec)

This is Euler second order ODE. Let the solution be \(y = t^r\), then \(y'=r t^{r-1}\) and \(y''=r(r-1) t^{r-2}\). Substituting these back into the given ODE gives

\[ t^{2}(r(r-1))t^{r-2}-3 t r t^{r-1}+5 t^{r} = 0 \]

Simplifying gives

\[ r \left (r -1\right )t^{r}-3 r\,t^{r}+5 t^{r} = 0 \]

Since \(t^{r}\neq 0\) then dividing throughout by \(t^{r}\) gives

\[ r \left (r -1\right )-3 r+5 = 0 \]

Or

\[ r^{2}-4 r +5 = 0 \tag {1} \]

Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are

\begin{align*} r_1 &= 2-i\\ r_2 &= 2+i \end{align*}

The roots are complex conjugate of each others. Let the roots be

\begin{align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end{align*}

Where in this case \(\alpha =2\) and \(\beta =-1\). Hence the solution becomes

\begin{align*} y &= c_1 t^{r_1} + c_2 t^{r_2} \\ &= c_1 t^{\alpha + i \beta } + c_2 t^{\alpha - i \beta } \\ &= t^{\alpha } \left ( c_1 t^{i \beta } + c_2 t^{- i \beta }\right ) \\ &= t^{\alpha } \left ( c_1 e^{\ln \left (t^{i \beta }\right )} + c_2 e^{\ln \left (t^{-i \beta }\right )}\right ) \\ &= t^{\alpha } \left ( c_1 e^{i \left (\beta \ln {t}\right )} + c_2 e^{-i \left (\beta \ln {t}\right )}\right ) \end{align*}

Using the values for \(\alpha =2, \beta =-1\), the above becomes

\begin{align*} y&= t^{2} \left ( c_1 e^{-i \ln \left (t \right )} + c_2 e^{i \ln \left (t \right )} \right ) \end{align*}

Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to change. Applying this to the above result gives

\begin{align*} y&=t^{2}\left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right ) \end{align*}

Will add steps showing solving for IC soon.

1.49.2 Solved as second order ode using change of variable on x method 2

Time used: 0.357 (sec)

In normal form the ode

\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y&=0 \tag {1} \end{align*}

Becomes

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

Where

\begin{align*} p \left (t \right )&=-\frac {3}{t}\\ q \left (t \right )&=\frac {5}{t^{2}} \end{align*}

Applying change of variables \(\tau = g \left (t \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (t \right )d t \right )}d t\\ &= \int {\mathrm e}^{-\left (\int -\frac {3}{t}d t \right )}d t\\ &= \int e^{3 \ln \left (t \right )} \,dt\\ &= \int t^{3}d t\\ &= \frac {t^{4}}{4}\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\\ &= \frac {\frac {5}{t^{2}}}{t^{6}}\\ &= \frac {5}{t^{8}}\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {5 y \left (\tau \right )}{t^{8}}&=0 \\ \end{align*}

But in terms of \(\tau \)

\begin{align*} \frac {5}{t^{8}}&=\frac {5}{16 \tau ^{2}} \end{align*}

Hence the above ode becomes

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {5 y \left (\tau \right )}{16 \tau ^{2}}&=0 \end{align*}

The above ode is now solved for \(y \left (\tau \right )\). This is Euler second order ODE. Let the solution be \(y \left (\tau \right ) = \tau ^r\), then \(y'=r \tau ^{r-1}\) and \(y''=r(r-1) \tau ^{r-2}\). Substituting these back into the given ODE gives

\[ 16 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+5 \tau ^{r} = 0 \]

Simplifying gives

\[ 16 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+5 \tau ^{r} = 0 \]

Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives

\[ 16 r \left (r -1\right )+0+5 = 0 \]

Or

\[ 16 r^{2}-16 r +5 = 0 \tag {1} \]

Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are

\begin{align*} r_1 &= \frac {1}{2}-\frac {i}{4}\\ r_2 &= \frac {1}{2}+\frac {i}{4} \end{align*}

The roots are complex conjugate of each others. Let the roots be

\begin{align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end{align*}

Where in this case \(\alpha ={\frac {1}{2}}\) and \(\beta =-{\frac {1}{4}}\). Hence the solution becomes

\begin{align*} y \left (\tau \right ) &= c_1 \tau ^{r_1} + c_2 \tau ^{r_2} \\ &= c_1 \tau ^{\alpha + i \beta } + c_2 \tau ^{\alpha - i \beta } \\ &= \tau ^{\alpha } \left ( c_1 \tau ^{i \beta } + c_2 \tau ^{- i \beta }\right ) \\ &= \tau ^{\alpha } \left ( c_1 e^{\ln \left (\tau ^{i \beta }\right )} + c_2 e^{\ln \left (\tau ^{-i \beta }\right )}\right ) \\ &= \tau ^{\alpha } \left ( c_1 e^{i \left (\beta \ln {\tau }\right )} + c_2 e^{-i \left (\beta \ln {\tau }\right )}\right ) \end{align*}

Using the values for \(\alpha ={\frac {1}{2}}, \beta =-{\frac {1}{4}}\), the above becomes

\begin{align*} y \left (\tau \right )&= \tau ^{{\frac {1}{2}}} \left ( c_1 e^{-\frac {i \ln \left (\tau \right )}{4}} + c_2 e^{\frac {i \ln \left (\tau \right )}{4}} \right ) \end{align*}

Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to change. Applying this to the above result gives

\begin{align*} y \left (\tau \right )&=\sqrt {\tau }\left (c_1 \cos \left (\frac {\ln \left (\tau \right )}{4}\right )+c_2 \sin \left (\frac {\ln \left (\tau \right )}{4}\right )\right ) \end{align*}

Will add steps showing solving for IC soon.

The above solution is now transformed back to \(y\) using (6) which results in

\[ y = \frac {\sqrt {4}\, \sqrt {t^{4}}\, \left (c_1 \cos \left (\frac {\ln \left (\frac {t^{4}}{4}\right )}{4}\right )+c_2 \sin \left (\frac {\ln \left (\frac {t^{4}}{4}\right )}{4}\right )\right )}{4} \]

Will add steps showing solving for IC soon.

1.49.3 Solved as second order ode using change of variable on x method 1

Time used: 0.105 (sec)

In normal form the ode

\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y&=0 \tag {1} \end{align*}

Becomes

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

Where

\begin{align*} p \left (t \right )&=-\frac {3}{t}\\ q \left (t \right )&=\frac {5}{t^{2}} \end{align*}

Applying change of variables \(\tau = g \left (t \right )\) to (2) results

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {5} \end{align*}

Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5)

\begin{align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {5}\, \sqrt {\frac {1}{t^{2}}}}{c}\tag {6} \\ \tau '' &= -\frac {\sqrt {5}}{c \sqrt {\frac {1}{t^{2}}}\, t^{3}} \end{align*}

Substituting the above into (4) results in

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\\ &=\frac {-\frac {\sqrt {5}}{c \sqrt {\frac {1}{t^{2}}}\, t^{3}}-\frac {3}{t}\frac {\sqrt {5}\, \sqrt {\frac {1}{t^{2}}}}{c}}{\left (\frac {\sqrt {5}\, \sqrt {\frac {1}{t^{2}}}}{c}\right )^2} \\ &=-\frac {4 c \sqrt {5}}{5} \end{align*}

Therefore ode (3) now becomes

\begin{align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )-\frac {4 c \sqrt {5}\, \left (\frac {d}{d \tau }y \left (\tau \right )\right )}{5}+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end{align*}

The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give

\begin{align*} y \left (\tau \right ) &= {\mathrm e}^{\frac {2 \sqrt {5}\, c \tau }{5}} \left (c_1 \cos \left (\frac {\sqrt {5}\, c \tau }{5}\right )+c_2 \sin \left (\frac {\sqrt {5}\, c \tau }{5}\right )\right ) \end{align*}

Now from (6)

\begin{align*} \tau &= \int \frac {1}{c} \sqrt q \,dt \\ &= \frac {\int \sqrt {5}\, \sqrt {\frac {1}{t^{2}}}d t}{c}\\ &= \frac {\sqrt {5}\, \ln \left (t \right )}{c} \end{align*}

Substituting the above into the solution obtained gives

\[ y = t^{2} \left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right ) \]

Will add steps showing solving for IC soon.

1.49.4 Solved as second order ode using change of variable on y method 2

Time used: 0.185 (sec)

In normal form the ode

\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y&=0 \tag {1} \end{align*}

Becomes

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

Where

\begin{align*} p \left (t \right )&=-\frac {3}{t}\\ q \left (t \right )&=\frac {5}{t^{2}} \end{align*}

Applying change of variables on the depndent variable \(y = v \left (t \right ) t^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (t \right )\) and not \(y\).

\begin{align*} v^{\prime \prime }\left (t \right )+\left (\frac {2 n}{t}+p \right ) v^{\prime }\left (t \right )+\left (\frac {n \left (n -1\right )}{t^{2}}+\frac {n p}{t}+q \right ) v \left (t \right )&=0 \tag {3} \end{align*}

Let the coefficient of \(v \left (t \right )\) above be zero. Hence

\begin{align*} \frac {n \left (n -1\right )}{t^{2}}+\frac {n p}{t}+q&=0 \tag {4} \end{align*}

Substituting the earlier values found for \(p \left (t \right )\) and \(q \left (t \right )\) into (4) gives

\begin{align*} \frac {n \left (n -1\right )}{t^{2}}-\frac {3 n}{t^{2}}+\frac {5}{t^{2}}&=0 \tag {5} \end{align*}

Solving (5) for \(n\) gives

\begin{align*} n&=2+i \tag {6} \end{align*}

Substituting this value in (3) gives

\begin{align*} v^{\prime \prime }\left (t \right )+\left (\frac {4+2 i}{t}-\frac {3}{t}\right ) v^{\prime }\left (t \right )&=0 \\ v^{\prime \prime }\left (t \right )+\frac {\left (1+2 i\right ) v^{\prime }\left (t \right )}{t}&=0 \tag {7} \\ \end{align*}

Using the substitution

\begin{align*} u \left (t \right ) = v^{\prime }\left (t \right ) \end{align*}

Then (7) becomes

\begin{align*} u^{\prime }\left (t \right )+\frac {\left (1+2 i\right ) u \left (t \right )}{t} = 0 \tag {8} \\ \end{align*}

The above is now solved for \(u \left (t \right )\). In canonical form a linear first order is

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

Comparing the above to the given ode shows that

\begin{align*} q(t) &=\frac {1+2 i}{t}\\ p(t) &=0 \end{align*}

The integrating factor \(\mu \) is

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

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (u \,t^{1+2 i}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,t^{1+2 i}&= \int {0 \,dt} + c_1 \\ &=c_1 \end{align*}

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

\[ u \left (t \right ) = t^{-1-2 i} c_1 \]

Now that \(u \left (t \right )\) is known, then

\begin{align*} v^{\prime }\left (t \right )&= u \left (t \right )\\ v \left (t \right )&= \int u \left (t \right )d t +c_2\\ &= \frac {i t^{-2 i} c_1}{2}+c_2 \end{align*}

Hence

\begin{align*} y&= v \left (t \right ) t^{n}\\ &= \left (\frac {i t^{-2 i} c_1}{2}+c_2 \right ) t^{2+i}\\ &= \frac {\left (i t^{-2 i} c_1 +2 c_2 \right ) t^{2+i}}{2}\\ \end{align*}

Will add steps showing solving for IC soon.

1.49.5 Solved as second order ode using Kovacic algorithm

Time used: 0.098 (sec)

Writing the ode as

\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y &= 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} \\ B &= -3 t\tag {3} \\ C &= 5 \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 {-5}{4 t^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= -5\\ t &= 4 t^{2} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(t) &= \left ( -\frac {5}{4 t^{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 7: 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) \\ &= 2 - 0 \\ &= 2 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 t^{2}\). There is a pole at \(t=0\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(2\) 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 \(2\) 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

\[ r = -\frac {5}{4 t^{2}} \]

For the pole at \(t=0\) let \(b\) be the coefficient of \(\frac {1}{ t^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {5}{4}}\). Hence

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

Since the order of \(r\) at \(\infty \) is 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{t^{2}}\) in the Laurent series expansion of \(r\) at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\) from

\begin{alignat*}{2} r &= \frac {s}{t} &&= -\frac {5}{4 t^{2}} \end{alignat*}

Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {5}{4}}\). Hence

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

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

\[ r=-\frac {5}{4 t^{2}} \]

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

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(2\)	\(0\)	\(\frac {1}{2}+i\)	\(\frac {1}{2}-i\)

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 \(\alpha _\infty ^{-} = \frac {1}{2}-i\) then

\begin{align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= \frac {1}{2}-i - \left ( \frac {1}{2}-i \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 ) + (-) [\sqrt r]_\infty \\ &= \frac {\frac {1}{2}-i}{t} + (-) \left ( 0 \right ) \\ &= \frac {\frac {1}{2}-i}{t}\\ &= \frac {\frac {1}{2}-i}{t} \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 {\frac {1}{2}-i}{t}\right ) \left (0\right ) + \left ( \left (\frac {-\frac {1}{2}+i}{t^{2}}\right ) + \left (\frac {\frac {1}{2}-i}{t}\right )^2 - \left (-\frac {5}{4 t^{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 \frac {\frac {1}{2}-i}{t}d t}\\ &= t^{\frac {1}{2}-i} \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 {-3 t}{t^{2}} \,dt} \\ &= z_1 e^{\frac {3 \ln \left (t \right )}{2}} \\ &= z_1 \left (t^{{3}/{2}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = t^{2-i} \]

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 {-3 t}{t^{2}} \,dt}}{\left (y_1\right )^2} \,dt \\ &= y_1 \int \frac { e^{3 \ln \left (t \right )}}{\left (y_1\right )^2} \,dt \\ &= y_1 \left (-\frac {i t^{4} t^{-4+2 i}}{2}\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (t^{2-i}\right ) + c_2 \left (t^{2-i}\left (-\frac {i t^{4} t^{-4+2 i}}{2}\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

1.49.6 Solved as second order ode adjoint method

Time used: 0.816 (sec)

In normal form the ode

\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y = 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 {3}{t}\\ q \left (t \right )&=\frac {5}{t^{2}}\\ 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 {3 \xi \left (t \right )}{t}\right )' + \left (\frac {5 \xi \left (t \right )}{t^{2}}\right ) &= 0\\ \xi ^{\prime \prime }\left (t \right )+\frac {2 \xi \left (t \right )}{t^{2}}+\frac {3 \xi ^{\prime }\left (t \right )}{t}&= 0 \end{align*}

Which is solved for \(\xi (t)\). This is Euler second order ODE. Let the solution be \(\xi = t^r\), then \(\xi '=r t^{r-1}\) and \(\xi ''=r(r-1) t^{r-2}\). Substituting these back into the given ODE gives

\[ t^{2}(r(r-1))t^{r-2}+3 t r t^{r-1}+2 t^{r} = 0 \]

Simplifying gives

\[ r \left (r -1\right )t^{r}+3 r\,t^{r}+2 t^{r} = 0 \]

Since \(t^{r}\neq 0\) then dividing throughout by \(t^{r}\) gives

\[ r \left (r -1\right )+3 r+2 = 0 \]

Or

\[ r^{2}+2 r +2 = 0 \tag {1} \]

Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are

\begin{align*} r_1 &= -1-i\\ r_2 &= -1+i \end{align*}

The roots are complex conjugate of each others. Let the roots be

\begin{align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end{align*}

Where in this case \(\alpha =-1\) and \(\beta =-1\). Hence the solution becomes

\begin{align*} \xi &= c_1 t^{r_1} + c_2 t^{r_2} \\ &= c_1 t^{\alpha + i \beta } + c_2 t^{\alpha - i \beta } \\ &= t^{\alpha } \left ( c_1 t^{i \beta } + c_2 t^{- i \beta }\right ) \\ &= t^{\alpha } \left ( c_1 e^{\ln \left (t^{i \beta }\right )} + c_2 e^{\ln \left (t^{-i \beta }\right )}\right ) \\ &= t^{\alpha } \left ( c_1 e^{i \left (\beta \ln {t}\right )} + c_2 e^{-i \left (\beta \ln {t}\right )}\right ) \end{align*}

Using the values for \(\alpha =-1, \beta =-1\), the above becomes

\begin{align*} \xi &= t^{-1} \left ( c_1 e^{-i \ln \left (t \right )} + c_2 e^{i \ln \left (t \right )} \right ) \end{align*}

Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to change. Applying this to the above result gives

\begin{align*} \xi &=\frac {1}{t}\left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right ) \end{align*}

Will add steps showing solving for IC soon.

The original ode (2) 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 )}\\ y^{\prime }+y \left (-\frac {3}{t}-\frac {\left (-\frac {c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )}{t^{2}}+\frac {-\frac {c_1 \sin \left (\ln \left (t \right )\right )}{t}+\frac {c_2 \cos \left (\ln \left (t \right )\right )}{t}}{t}\right ) t}{c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\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 {2 c_1 \cos \left (\ln \left (t \right )\right )+c_2 \cos \left (\ln \left (t \right )\right )-c_1 \sin \left (\ln \left (t \right )\right )+2 c_2 \sin \left (\ln \left (t \right )\right )}{t \left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right )}\\ p(t) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -\frac {2 c_1 \cos \left (\ln \left (t \right )\right )+c_2 \cos \left (\ln \left (t \right )\right )-c_1 \sin \left (\ln \left (t \right )\right )+2 c_2 \sin \left (\ln \left (t \right )\right )}{t \left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right )}d t}\\ &= {\mathrm e}^{\frac {\ln \left (\tan \left (\ln \left (t \right )\right )^{2}+1\right )}{2}-\ln \left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right )-2 \ln \left (t \right )} \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 (y \,{\mathrm e}^{\frac {\ln \left (\tan \left (\ln \left (t \right )\right )^{2}+1\right )}{2}-\ln \left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right )-2 \ln \left (t \right )}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} y \,{\mathrm e}^{\frac {\ln \left (\tan \left (\ln \left (t \right )\right )^{2}+1\right )}{2}-\ln \left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right )-2 \ln \left (t \right )}&= \int {0 \,dt} + c_3 \\ &=c_3 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {\ln \left (\tan \left (\ln \left (t \right )\right )^{2}+1\right )}{2}-\ln \left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right )-2 \ln \left (t \right )}\) gives the final solution

\[ y = \frac {\left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right ) t^{2} c_3}{\sqrt {\tan \left (\ln \left (t \right )\right )^{2}+1}} \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \frac {\left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right ) t^{2} c_3}{\sqrt {\tan \left (\ln \left (t \right )\right )^{2}+1}} \\ \end{align*}

Will add steps showing solving for IC soon.

1.49.7 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{2} \left (\frac {d}{d t}y^{\prime }\right )-3 t y^{\prime }+5 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-\frac {5 y}{t^{2}}+\frac {3 y^{\prime }}{t} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }-\frac {3 y^{\prime }}{t}+\frac {5 y}{t^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & t^{2} \left (\frac {d}{d t}y^{\prime }\right )-3 t y^{\prime }+5 y=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & s =\ln \left (t \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\frac {d}{d s}y \left (s \right )\right ) s^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\frac {d}{d s}y \left (s \right )}{t} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=\left (\frac {d}{d s}\frac {d}{d s}y \left (s \right )\right ) {s^{\prime }\left (t \right )}^{2}+\left (\frac {d}{d t}s^{\prime }\left (t \right )\right ) \left (\frac {d}{d s}y \left (s \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=\frac {\frac {d}{d s}\frac {d}{d s}y \left (s \right )}{t^{2}}-\frac {\frac {d}{d s}y \left (s \right )}{t^{2}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & t^{2} \left (\frac {\frac {d}{d s}\frac {d}{d s}y \left (s \right )}{t^{2}}-\frac {\frac {d}{d s}y \left (s \right )}{t^{2}}\right )-3 \frac {d}{d s}y \left (s \right )+5 y \left (s \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d s}\frac {d}{d s}y \left (s \right )-4 \frac {d}{d s}y \left (s \right )+5 y \left (s \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-4 r +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {4\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (2-\mathrm {I}, 2+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (s \right )={\mathrm e}^{2 s} \cos \left (s \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (s \right )={\mathrm e}^{2 s} \sin \left (s \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (s \right )=\mathit {C1} y_{1}\left (s \right )+\mathit {C2} y_{2}\left (s \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (s \right )=\mathit {C1} \,{\mathrm e}^{2 s} \cos \left (s \right )+\mathit {C2} \,{\mathrm e}^{2 s} \sin \left (s \right ) \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} s =\ln \left (t \right ) \\ {} & {} & y=\mathit {C1} \,t^{2} \cos \left (\ln \left (t \right )\right )+\mathit {C2} \,t^{2} \sin \left (\ln \left (t \right )\right ) \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=t^{2} \left (\mathit {C1} \cos \left (\ln \left (t \right )\right )+\mathit {C2} \sin \left (\ln \left (t \right )\right )\right ) \end {array} \]

1.49.8 Maple trace

Methods for second order ODEs:
 

1.49.9 Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 19

dsolve(t^2*diff(diff(y(t),t),t)-3*t*diff(y(t),t)+5*y(t) = 0, 
       y(t),singsol=all)
 

\[ y = t^{2} \left (c_1 \sin \left (\ln \left (t \right )\right )+c_2 \cos \left (\ln \left (t \right )\right )\right ) \]

1.49.10 Mathematica DSolve solution

Solving time : 0.027 (sec)
Leaf size : 22

DSolve[{t^2*D[y[t],{t,2}]-3*t*D[y[t],t]+5*y[t]==0,{}}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to t^2 (c_2 \cos (\log (t))+c_1 \sin (\log (t))) \]