4.20 problem 20

Internal problem ID [7889]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 20
Date solved : Monday, October 21, 2024 at 04:31:26 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solve

\begin{align*} 2 x^{2} y^{\prime \prime }+2 x y^{\prime }-x y&=1 \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.

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

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 {1}{x}\\ q(x) &= -\frac {1}{2 x}\\ \end{align*}

Table 71: Table \(p(x),q(x)\) singularites.

\(p(x)=\frac {1}{x}\)
singularity	type
\(x = 0\)	\(\text {``regular''}\)

\(q(x)=-\frac {1}{2 x}\)
singularity	type
\(x = 0\)	\(\text {``regular''}\)

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

Regular singular points : \([0]\)

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

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

\[ 2 x^{2} y^{\prime \prime }+2 x y^{\prime }-x y = 1 \]

Since this is an inhomogeneous, then let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ode \(2 x^{2} y^{\prime \prime }+2 x y^{\prime }-x y = 0\), and \(y_p\) is a particular solution to the inhomogeneous ode.which is found using the balance equation generated from indicial equation

First, we solve for \(y_h\) 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} 2 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 )+2 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-x \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 }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+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 }}\left (-x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right ) \\ \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 }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right ) = 0 \end{equation}

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

\[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 x^{n +r} a_{n} \left (n +r \right ) = 0 \]

When \(n = 0\) the above becomes

\[ 2 x^{r} a_{0} r \left (-1+r \right )+2 x^{r} a_{0} r = 0 \]

Or

\[ \left (2 x^{r} r \left (-1+r \right )+2 x^{r} r \right ) a_{0} = 0 \]

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

\[ 2 x^{r} r^{2} = 0 \]

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

\[ 2 r^{2} = 0 \]

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

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

The corresponding balance equation is found by replacing \(r\) by \(m\) and \(a\) by \(c\) to avoid confusing terms between particular solution and the homogeneous solution. Hence the balance equation is

\begin{align*}\left (2 x^{m} m \left (-1+m \right )+2 x^{m} m \right ) c_{0} = 1 \end{align*}

This equation will used later to find the particular solution.

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

\[ 2 x^{r} r^{2} = 0 \]

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 (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end{align*}

Now the second solution \(y_{2}\) is found using

\begin{align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end{align*}

Then the general solution will be

\[ y = c_1 y_{1}\left (x \right )+c_2 y_{2}\left (x \right ) \]

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 \(y_{1}\left (x \right )\). 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} 2 a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n} \left (n +r \right )-a_{n -1} = 0 \end{equation}

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

\[ a_{n} = \frac {a_{n -1}}{2 n^{2}+4 n r +2 r^{2}}\tag {4} \]

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

\[ a_{n} = \frac {a_{n -1}}{2 n^{2}}\tag {5} \]

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

\[ a_{1}=\frac {1}{2 \left (r +1\right )^{2}} \]

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

\[ a_{1}={\frac {1}{2}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1}{2 \left (r +1\right )^{2}}\)	\(\frac {1}{2}\)

For \(n = 2\), using the above recursive equation gives

\[ a_{2}=\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}} \]

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

\[ a_{2}={\frac {1}{16}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1}{2 \left (r +1\right )^{2}}\)	\(\frac {1}{2}\)
\(a_{2}\)	\(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\)	\(\frac {1}{16}\)

For \(n = 3\), using the above recursive equation gives

\[ a_{3}=\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}} \]

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

\[ a_{3}={\frac {1}{288}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1}{2 \left (r +1\right )^{2}}\)	\(\frac {1}{2}\)
\(a_{2}\)	\(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\)	\(\frac {1}{16}\)
\(a_{3}\)	\(\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(\frac {1}{288}\)

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

\[ a_{4}=\frac {1}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \]

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

\[ a_{4}={\frac {1}{9216}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1}{2 \left (r +1\right )^{2}}\)	\(\frac {1}{2}\)
\(a_{2}\)	\(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\)	\(\frac {1}{16}\)
\(a_{3}\)	\(\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(\frac {1}{288}\)
\(a_{4}\)	\(\frac {1}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\)	\(\frac {1}{9216}\)

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

\[ a_{5}=\frac {1}{32 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \]

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

\[ a_{5}={\frac {1}{460800}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1}{2 \left (r +1\right )^{2}}\)	\(\frac {1}{2}\)
\(a_{2}\)	\(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\)	\(\frac {1}{16}\)
\(a_{3}\)	\(\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(\frac {1}{288}\)
\(a_{4}\)	\(\frac {1}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\)	\(\frac {1}{9216}\)
\(a_{5}\)	\(\frac {1}{32 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\)	\(\frac {1}{460800}\)

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

\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}{2}+\frac {x^{2}}{16}+\frac {x^{3}}{288}+\frac {x^{4}}{9216}+\frac {x^{5}}{460800}+O\left (x^{6}\right ) \\ \end{align*}

Now the second solution is found. The second solution is given by

\[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \]

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
\(b_{1}\)	\(\frac {1}{2 \left (r +1\right )^{2}}\)	\(\frac {1}{2}\)	\(-\frac {1}{\left (r +1\right )^{3}}\)	\(-1\)
\(b_{2}\)	\(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\)	\(\frac {1}{16}\)	\(\frac {-2 r -3}{2 \left (r +1\right )^{3} \left (r +2\right )^{3}}\)	\(-{\frac {3}{16}}\)
\(b_{3}\)	\(\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(\frac {1}{288}\)	\(\frac {-3 r^{2}-12 r -11}{4 \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3}}\)	\(-{\frac {11}{864}}\)
\(b_{4}\)	\(\frac {1}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\)	\(\frac {1}{9216}\)	\(\frac {-2 r^{3}-15 r^{2}-35 r -25}{4 \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\)	\(-{\frac {25}{55296}}\)
\(b_{5}\)	\(\frac {1}{32 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\)	\(\frac {1}{460800}\)	\(\frac {-5 r^{4}-60 r^{3}-255 r^{2}-450 r -274}{16 \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\)	\(-{\frac {137}{13824000}}\)

The above table gives all values of \(b_{n}\) needed. Hence the second solution is

\begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \left (x \right )+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 \\ &= \left (1+\frac {x}{2}+\frac {x^{2}}{16}+\frac {x^{3}}{288}+\frac {x^{4}}{9216}+\frac {x^{5}}{460800}+O\left (x^{6}\right )\right ) \ln \left (x \right )-x -\frac {3 x^{2}}{16}-\frac {11 x^{3}}{864}-\frac {25 x^{4}}{55296}-\frac {137 x^{5}}{13824000}+O\left (x^{6}\right ) \\ \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}{2}+\frac {x^{2}}{16}+\frac {x^{3}}{288}+\frac {x^{4}}{9216}+\frac {x^{5}}{460800}+O\left (x^{6}\right )\right ) + c_2 \left (\left (1+\frac {x}{2}+\frac {x^{2}}{16}+\frac {x^{3}}{288}+\frac {x^{4}}{9216}+\frac {x^{5}}{460800}+O\left (x^{6}\right )\right ) \ln \left (x \right )-x -\frac {3 x^{2}}{16}-\frac {11 x^{3}}{864}-\frac {25 x^{4}}{55296}-\frac {137 x^{5}}{13824000}+O\left (x^{6}\right )\right ) \\ \end{align*}

The particular solution is found by solving for \(c,m\) the balance equation

\begin{align*} \left (2 x^{m} m \left (-1+m \right )+2 x^{m} m \right ) c_{0}&=F \end{align*}

Where \(F(x)\) is the RHS of the ode. If \(F(x)\) has more than one term, then this is done for each term one at a time and then all the particular solutions are added. The function \(F(x)\) will be converted to series if needed. in order to solve for \(c_n,m\) for each term, the same recursive relation used to find \(y_h(x)\) is used to find \(c_n,m\) which is used to find the particular solution \(\sum _{n=0} c_n x^{n+m}\) by replacing \(a_n\) by \(c_n\) and \(r\) by \(m\).

The following are the values of \(a_n\) found in terms of the indicial root \(r\).

\(a_{1} = \frac {a_{0}}{2 \left (r +1\right )^{2}}\)

\(a_{2} = \frac {a_{0}}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\)

\(a_{3} = \frac {a_{0}}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)

\(a_{4} = \frac {a_{0}}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\)

\(a_{5} = \frac {a_{0}}{32 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\)

Unable to solve the balance equation \(\left (2 x^{m} m \left (-1+m \right )+2 x^{m} m \right ) c_{0}\) for \(c_{0}\) and \(x\). No particular solution exists.

Failed to convert RHS \(1\) to series in order to find particular solution. Unable to solve. Terminating Unable to find the particular solution or no solution exists.

4.20.1 Maple step by step solution

4.20.2 Maple trace

Methods for second order ODEs:
 

4.20.3 Maple dsolve solution

Solving time : 0.100 (sec)
Leaf size : maple_leaf_size

dsolve(2*x^2*diff(diff(y(x),x),x)+2*x*diff(y(x),x)-x*y(x) = 1,y(x), 
       series,x=0)
 

\[ \text {No solution found} \]

4.20.4 Mathematica DSolve solution

Solving time : 0.154 (sec)
Leaf size : 360

AsymptoticDSolveValue[{2*x^2*D[y[x],{x,2}]+2*x*D[y[x],x]-x*y[x]==1,{}}, 
       y[x],{x,0,5}]
 

\[ y(x)\to c_2 \left (\frac {x^5}{460800}+\frac {x^4}{9216}+\frac {x^3}{288}+\frac {x^2}{16}+\frac {x}{2}+1\right )+c_1 \left (x^5 \left (\frac {\log (x)}{460800}-\frac {107}{13824000}\right )+x^4 \left (\frac {\log (x)}{9216}-\frac {19}{55296}\right )+x^3 \left (\frac {\log (x)}{288}-\frac {1}{108}\right )+x^2 \left (\frac {\log (x)}{16}-\frac {1}{8}\right )+x \left (\frac {\log (x)}{2}-\frac {1}{2}\right )+\log (x)+1\right )+\left (-\frac {137 x^6}{1990656000}+\frac {x^5}{4608000}+\frac {x^4}{73728}+\frac {x^3}{1728}+\frac {x^2}{64}+\frac {x}{4}+\frac {\log (x)}{2}\right ) \left (x^5 \left (\frac {\log (x)}{460800}-\frac {107}{13824000}\right )+x^4 \left (\frac {\log (x)}{9216}-\frac {19}{55296}\right )+x^3 \left (\frac {\log (x)}{288}-\frac {1}{108}\right )+x^2 \left (\frac {\log (x)}{16}-\frac {1}{8}\right )+x \left (\frac {\log (x)}{2}-\frac {1}{2}\right )+\log (x)+1\right )+\left (\frac {x^5}{460800}+\frac {x^4}{9216}+\frac {x^3}{288}+\frac {x^2}{16}+\frac {x}{2}+1\right ) \left (\frac {137 x^6 (6 \log (x)+5)}{11943936000}+\frac {x^5 (113-30 \log (x))}{138240000}+\frac {x^4 (41-12 \log (x))}{884736}+\frac {x^3 (3-\log (x))}{1728}+\frac {1}{128} x^2 (5-2 \log (x))+\frac {1}{4} x (2-\log (x))-\frac {1}{4} \log (x) (\log (x)+2)\right ) \]