4.60 problem 57

Internal problem ID [7929]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 57
Date solved : Monday, October 21, 2024 at 04:35:09 PM
CAS classification : [[_Emden, _Fowler]]

Solve

\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+4 x^{4} y&=0 \end{align*}

Using series expansion around \(x=0\)

The type of the expansion point is first determined. This is done on the homogeneous part of the ODE.

\[ x^{2} y^{\prime \prime }+3 x y^{\prime }+4 x^{4} y = 0 \]

The following is summary of singularities for the above ode. Writing the ode as

\begin{align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end{align*}

Where

\begin{align*} p(x) &= \frac {3}{x}\\ q(x) &= 4 x^{2}\\ \end{align*}

Table 106: Table \(p(x),q(x)\) singularites.

\(p(x)=\frac {3}{x}\)
singularity	type
\(x = 0\)	\(\text {``regular''}\)

\(q(x)=4 x^{2}\)
singularity	type
\(x = \infty \)	\(\text {``regular''}\)
\(x = -\infty \)	\(\text {``regular''}\)

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : \([0, \infty , -\infty ]\)

Irregular singular points : \([\infty ]\)

Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be

\[ x^{2} y^{\prime \prime }+3 x y^{\prime }+4 x^{4} y = 0 \]

Let the solution be represented as Frobenius power series of the form

\[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \]

Then

\begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*}

Substituting the above back into the ode gives

\begin{equation} \tag{1} x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+3 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+4 x^{4} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation}

Which simplifies to

\begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{4+n +r} a_{n}\right ) = 0 \end{equation}

The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives

\begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{4+n +r} a_{n} &= \moverset {\infty }{\munderset {n =4}{\sum }}4 a_{n -4} x^{n +r} \\ \end{align*}

Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\).

\begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}4 a_{n -4} x^{n +r}\right ) = 0 \end{equation}

The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives

\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 x^{n +r} a_{n} \left (n +r \right ) = 0 \]

When \(n = 0\) the above becomes

\[ x^{r} a_{0} r \left (-1+r \right )+3 x^{r} a_{0} r = 0 \]

Or

\[ \left (x^{r} r \left (-1+r \right )+3 x^{r} r \right ) a_{0} = 0 \]

Since \(a_{0}\neq 0\) then the above simplifies to

\[ x^{r} r \left (2+r \right ) = 0 \]

Since the above is true for all \(x\) then the indicial equation becomes

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

Solving for \(r\) gives the roots of the indicial equation as

\begin{align*} r_1 &= 0\\ r_2 &= -2 \end{align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes

\[ x^{r} r \left (2+r \right ) = 0 \]

Solving for \(r\) gives the roots of the indicial equation as \([0, -2]\).

Since \(r_1 - r_2 = 2\) is an integer, then we can construct two linearly independent solutions

\begin{align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end{align*}

Or

\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x^{2}} \end{align*}

Or

\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2}\right ) \end{align*}

Where \(C\) above can be zero. We start by finding \(y_{1}\). 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives

\[ a_{1} = 0 \]

Substituting \(n = 2\) in Eq. (2B) gives

\[ a_{2} = 0 \]

Substituting \(n = 3\) in Eq. (2B) gives

\[ a_{3} = 0 \]

For \(4\le n\) the recursive equation is

\begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n} \left (n +r \right )+4 a_{n -4} = 0 \end{equation}

Solving for \(a_{n}\) from recursive equation (4) gives

\[ a_{n} = -\frac {4 a_{n -4}}{n^{2}+2 n r +r^{2}+2 n +2 r}\tag {4} \]

Which for the root \(r = 0\) becomes

\[ a_{n} = -\frac {4 a_{n -4}}{n \left (n +2\right )}\tag {5} \]

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.

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(0\)	\(0\)
\(a_{3}\)	\(0\)	\(0\)

For \(n = 4\), using the above recursive equation gives

\[ a_{4}=-\frac {4}{r^{2}+10 r +24} \]

Which for the root \(r = 0\) becomes

\[ a_{4}=-{\frac {1}{6}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(0\)	\(0\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(-\frac {4}{r^{2}+10 r +24}\)	\(-{\frac {1}{6}}\)

For \(n = 5\), using the above recursive equation gives

\[ a_{5}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(0\)	\(0\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(-\frac {4}{r^{2}+10 r +24}\)	\(-{\frac {1}{6}}\)
\(a_{5}\)	\(0\)	\(0\)

Using the above table, then the solution \(y_{1}\left (x \right )\) is

\begin{align*} y_{1}\left (x \right )&= a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \\ &= 1-\frac {x^{4}}{6}+O\left (x^{6}\right ) \end{align*}

Now the second solution \(y_{2}\left (x \right )\) is found. Let

\[ r_{1}-r_{2} = N \]

Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=2\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{2}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that

\begin{align*} a_N &= a_{2} \\ &= 0 \end{align*}

Therefore

\begin{align*} \lim _{r\rightarrow r_{2}}0&= \lim _{r\rightarrow -2}0\\ &= 0 \end{align*}

The limit is \(0\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form

\begin{align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2} \end{align*}

Eq (3) derived above is used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq(3) gives

\[ b_{1} = 0 \]

Substituting \(n = 2\) in Eq(3) gives

\[ b_{2} = 0 \]

Substituting \(n = 3\) in Eq(3) gives

\[ b_{3} = 0 \]

For \(4\le n\) the recursive equation is

\begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )+3 b_{n} \left (n +r \right )+4 b_{n -4} = 0 \end{equation}

Which for for the root \(r = -2\) becomes

\begin{equation} \tag{4A} b_{n} \left (n -2\right ) \left (n -3\right )+3 b_{n} \left (n -2\right )+4 b_{n -4} = 0 \end{equation}

Solving for \(b_{n}\) from the recursive equation (4) gives

\[ b_{n} = -\frac {4 b_{n -4}}{n^{2}+2 n r +r^{2}+2 n +2 r}\tag {5} \]

Which for the root \(r = -2\) becomes

\[ b_{n} = -\frac {4 b_{n -4}}{n^{2}-2 n}\tag {6} \]

At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -2\) and after as more terms are found using the above recursive equation.

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(0\)	\(0\)
\(b_{3}\)	\(0\)	\(0\)

For \(n = 4\), using the above recursive equation gives

\[ b_{4}=-\frac {4}{r^{2}+10 r +24} \]

Which for the root \(r = -2\) becomes

\[ b_{4}=-{\frac {1}{2}} \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(0\)	\(0\)
\(b_{3}\)	\(0\)	\(0\)
\(b_{4}\)	\(-\frac {4}{r^{2}+10 r +24}\)	\(-{\frac {1}{2}}\)

For \(n = 5\), using the above recursive equation gives

\[ b_{5}=0 \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(0\)	\(0\)
\(b_{3}\)	\(0\)	\(0\)
\(b_{4}\)	\(-\frac {4}{r^{2}+10 r +24}\)	\(-{\frac {1}{2}}\)
\(b_{5}\)	\(0\)	\(0\)

Using the above table, then the solution \(y_{2}\left (x \right )\) is

\begin{align*} y_{2}\left (x \right )&= 1 \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \right ) \\ &= \frac {1-\frac {x^{4}}{2}+O\left (x^{6}\right )}{x^{2}} \end{align*}

Therefore the homogeneous solution is

\begin{align*} y_h(x) &= c_1 y_{1}\left (x \right )+c_2 y_{2}\left (x \right ) \\ &= c_1 \left (1-\frac {x^{4}}{6}+O\left (x^{6}\right )\right ) + \frac {c_2 \left (1-\frac {x^{4}}{2}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*}

Hence the final solution is

\begin{align*} y &= y_h \\ &= c_1 \left (1-\frac {x^{4}}{6}+O\left (x^{6}\right )\right )+\frac {c_2 \left (1-\frac {x^{4}}{2}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*}

4.60.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+3 x y^{\prime }+4 x^{4} y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-4 x^{2} y-\frac {3 y^{\prime }}{x} \\ \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 x}y^{\prime }+\frac {3 y^{\prime }}{x}+4 x^{2} y=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {3}{x}, P_{3}\left (x \right )=4 x^{2}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=3 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 4 y x^{3}+\left (\frac {d}{d x}y^{\prime }\right ) x +3 y^{\prime }=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{3}\cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +3} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -3 \\ {} & {} & x^{3}\cdot y=\moverset {\infty }{\munderset {k =3}{\sum }}a_{k -3} x^{k +r} \\ {} & \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 ) x^{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 ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (2+r \right ) x^{-1+r}+a_{1} \left (1+r \right ) \left (3+r \right ) x^{r}+a_{2} \left (2+r \right ) \left (4+r \right ) x^{1+r}+a_{3} \left (3+r \right ) \left (5+r \right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +3\right )+4 a_{k -3}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, 0\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) \left (3+r \right )=0, a_{2} \left (2+r \right ) \left (4+r \right )=0, a_{3} \left (3+r \right ) \left (5+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=0, a_{3}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +r +3\right )+4 a_{k -3}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +4} \left (k +4+r \right ) \left (k +6+r \right )+4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +4+r \right ) \left (k +6+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +2\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}, a_{k +4}=-\frac {4 a_{k}}{\left (k +2\right ) \left (k +4\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +4\right ) \left (k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=-\frac {4 a_{k}}{\left (k +4\right ) \left (k +6\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right ), a_{k +4}=-\frac {4 a_{k}}{\left (k +2\right ) \left (k +4\right )}, a_{1}=0, a_{2}=0, a_{3}=0, b_{k +4}=-\frac {4 b_{k}}{\left (k +4\right ) \left (k +6\right )}, b_{1}=0, b_{2}=0, b_{3}=0\right ] \end {array} \]

4.60.2 Maple trace

Methods for second order ODEs:
 

4.60.3 Maple dsolve solution

Solving time : 0.016 (sec)
Leaf size : 28

dsolve(x^2*diff(diff(y(x),x),x)+3*x*diff(y(x),x)+4*x^4*y(x) = 0,y(x), 
       series,x=0)
 

\[ y = c_1 \left (1-\frac {1}{6} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_2 \left (-2+x^{4}+\operatorname {O}\left (x^{6}\right )\right )}{x^{2}} \]

4.60.4 Mathematica DSolve solution

Solving time : 0.01 (sec)
Leaf size : 30

AsymptoticDSolveValue[{x^2*D[y[x],{x,2}]+3*x*D[y[x],x]+4*x^4*y[x]==0,{}}, 
       y[x],{x,0,5}]
 

\[ y(x)\to c_2 \left (1-\frac {x^4}{6}\right )+c_1 \left (\frac {1}{x^2}-\frac {x^2}{2}\right ) \]