Internal problem ID [1820]
Internal file name [OUTPUT/1821_Sunday_June_05_2022_02_33_58_AM_74117384/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 2.8.3, The method of Frobenius. Equal roots, and roots differering by an integer.
Page 223
Problem number: 1.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_Emden, _Fowler]]
\[ \boxed {t y^{\prime \prime }+y^{\prime }-4 y=0} \] With the expansion point for the power series method at \(t = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ t y^{\prime \prime }+y^{\prime }-4 y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(t) y^{\prime } + q(t) y &=0 \end {align*}
Where \begin {align*} p(t) &= \frac {1}{t}\\ q(t) &= -\frac {4}{t}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ t y^{\prime \prime }+y^{\prime }-4 y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )-4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 a_{n} t^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} t^{n +r -1}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ t^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} t^{-1+r} = 0 \] Or \[ \left (t^{-1+r} r \left (-1+r \right )+r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ t^{-1+r} r^{2} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ t^{-1+r} r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, 0]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (t \right ) &= y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. We start by finding the first solution \(y_{1}\left (t \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} \left (n +r \right )-4 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {4 a_{n -1}}{n^{2}+2 n r +r^{2}}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {4 a_{n -1}}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {4}{\left (r +1\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{1}=4 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{\left (r +1\right )^{2}}\) | \(4\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {16}{\left (r +1\right )^{2} \left (2+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{2}=4 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{\left (r +1\right )^{2}}\) | \(4\) |
| \(a_{2}\) | \(\frac {16}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(4\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {64}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {16}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{\left (r +1\right )^{2}}\) | \(4\) |
| \(a_{2}\) | \(\frac {16}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(4\) |
| \(a_{3}\) | \(\frac {64}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {16}{9}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {256}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {4}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{\left (r +1\right )^{2}}\) | \(4\) |
| \(a_{2}\) | \(\frac {16}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(4\) |
| \(a_{3}\) | \(\frac {64}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {16}{9}\) |
| \(a_{4}\) | \(\frac {256}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {4}{9}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1024}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {16}{225}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{\left (r +1\right )^{2}}\) | \(4\) |
| \(a_{2}\) | \(\frac {16}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(4\) |
| \(a_{3}\) | \(\frac {64}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {16}{9}\) |
| \(a_{4}\) | \(\frac {256}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {4}{9}\) |
| \(a_{5}\) | \(\frac {1024}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {16}{225}\) |
Using the above table, then the first solution \(y_{1}\left (t \right )\) becomes \begin{align*} y_{1}\left (t \right )&= a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}\dots \\ &= 4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (t \right ) = y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = 0\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =0\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {4}{\left (r +1\right )^{2}}\) | \(4\) | \(-\frac {8}{\left (r +1\right )^{3}}\) | \(-8\) |
| \(b_{2}\) | \(\frac {16}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(4\) | \(\frac {-96-64 r}{\left (r +1\right )^{3} \left (2+r \right )^{3}}\) | \(-12\) |
| \(b_{3}\) | \(\frac {64}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {16}{9}\) | \(\frac {-384 r^{2}-1536 r -1408}{\left (r +1\right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3}}\) | \(-{\frac {176}{27}}\) |
| \(b_{4}\) | \(\frac {256}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {4}{9}\) | \(-\frac {2048 \left (r^{2}+5 r +5\right ) \left (r +\frac {5}{2}\right )}{\left (r +1\right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\) | \(-{\frac {50}{27}}\) |
| \(b_{5}\) | \(\frac {1024}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {16}{225}\) | \(-\frac {2048 \left (5 r^{4}+60 r^{3}+255 r^{2}+450 r +274\right )}{\left (r +1\right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\) | \(-{\frac {1096}{3375}}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (t \right )&=y_{1}\left (t \right ) \ln \left (t \right )+b_{0}+b_{1} t +b_{2} t^{2}+b_{3} t^{3}+b_{4} t^{4}+b_{5} t^{5}+b_{6} t^{6}\dots \\ &= \left (4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right )\right ) \ln \left (t \right )-12 t^{2}-8 t -\frac {176 t^{3}}{27}-\frac {50 t^{4}}{27}-\frac {1096 t^{5}}{3375}+O\left (t^{6}\right ) \\ \end{align*} Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} \left (4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right )\right ) + c_{2} \left (\left (4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right )\right ) \ln \left (t \right )-12 t^{2}-8 t -\frac {176 t^{3}}{27}-\frac {50 t^{4}}{27}-\frac {1096 t^{5}}{3375}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right )\right )+c_{2} \left (\left (4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right )\right ) \ln \left (t \right )-12 t^{2}-8 t -\frac {176 t^{3}}{27}-\frac {50 t^{4}}{27}-\frac {1096 t^{5}}{3375}+O\left (t^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right )\right )+c_{2} \left (\left (4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right )\right ) \ln \left (t \right )-12 t^{2}-8 t -\frac {176 t^{3}}{27}-\frac {50 t^{4}}{27}-\frac {1096 t^{5}}{3375}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right )\right )+c_{2} \left (\left (4 t^{2}+4 t +1+\frac {16 t^{3}}{9}+\frac {4 t^{4}}{9}+\frac {16 t^{5}}{225}+O\left (t^{6}\right )\right ) \ln \left (t \right )-12 t^{2}-8 t -\frac {176 t^{3}}{27}-\frac {50 t^{4}}{27}-\frac {1096 t^{5}}{3375}+O\left (t^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t y^{\prime \prime }+y^{\prime }-4 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {4 y}{t}-\frac {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} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime }}{t}-\frac {4 y}{t}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (t \right )=\frac {1}{t}, P_{3}\left (t \right )=-\frac {4}{t}\right ] \\ {} & \circ & t \cdot P_{2}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t \cdot P_{2}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=1 \\ {} & \circ & t^{2}\cdot P_{3}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t^{2}\cdot P_{3}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=0 \\ {} & \circ & t =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & t y^{\prime \prime }+y^{\prime }-4 y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) t^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & t \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r^{2} t^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right )^{2}-4 a_{k}\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1\right )^{2}-4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {4 a_{k}}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {4 a_{k}}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +1}=\frac {4 a_{k}}{\left (k +1\right )^{2}}\right ] \end {array} \]
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 -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 59
Order:=6; dsolve(t*diff(y(t),t$2)+diff(y(t),t)-4*y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = \left (c_{2} \ln \left (t \right )+c_{1} \right ) \left (1+4 t +4 t^{2}+\frac {16}{9} t^{3}+\frac {4}{9} t^{4}+\frac {16}{225} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+\left (\left (-8\right ) t -12 t^{2}-\frac {176}{27} t^{3}-\frac {50}{27} t^{4}-\frac {1096}{3375} t^{5}+\operatorname {O}\left (t^{6}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 105
AsymptoticDSolveValue[t*y''[t]+y'[t]-4*y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_1 \left (\frac {16 t^5}{225}+\frac {4 t^4}{9}+\frac {16 t^3}{9}+4 t^2+4 t+1\right )+c_2 \left (-\frac {1096 t^5}{3375}-\frac {50 t^4}{27}-\frac {176 t^3}{27}-12 t^2+\left (\frac {16 t^5}{225}+\frac {4 t^4}{9}+\frac {16 t^3}{9}+4 t^2+4 t+1\right ) \log (t)-8 t\right ) \]