2.4.24 problem 24

Internal problem ID [8589]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 24
Date solved : Thursday, December 12, 2024 at 09:31:37 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solve

\begin{align*} 2 x^{2} y^{\prime \prime }-x y^{\prime }+\left (-x^{2}+1\right ) y&=\cos \left (x \right ) \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.

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

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

Table 2.76: Table \(p(x),q(x)\) singularites.

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

\(q(x)=-\frac {x^{2}-1}{2 x^{2}}\)
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

\[ 2 x^{2} y^{\prime \prime }-x y^{\prime }+\left (-x^{2}+1\right ) y = \cos \left (x \right ) \]

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

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

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

When \(n = 0\) the above becomes

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

Or

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

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

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

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

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

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

\begin{align*} r_1 &= 1\\ r_2 &= {\frac {1}{2}} \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 )-x^{m} m +x^{m}\right ) c_{0} = \cos \left (x \right ) \end{align*}

This equation will used later to find the particular solution.

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

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

Solving for \(r\) gives the roots of the indicial equation as \(\left [1, {\frac {1}{2}}\right ]\).

Since \(r_1 - r_2 = {\frac {1}{2}}\) is not 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 ) &= 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 +1}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}} \end{align*}

We start by finding \(y_{1}\left (x \right )\). 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives

\[ a_{1} = 0 \]

For \(2\le n\) the recursive equation is

\begin{equation} \tag{3} 2 a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n} \left (n +r \right )-a_{n -2}+a_{n} = 0 \end{equation}

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

\[ a_{n} = \frac {a_{n -2}}{2 n^{2}+4 n r +2 r^{2}-3 n -3 r +1}\tag {4} \]

Which for the root \(r = 1\) becomes

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

At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.

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

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

\[ a_{2}=\frac {1}{2 r^{2}+5 r +3} \]

Which for the root \(r = 1\) becomes

\[ a_{2}={\frac {1}{10}} \]

And the table now becomes

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

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

\[ a_{3}=0 \]

And the table now becomes

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

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

\[ a_{4}=\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63} \]

Which for the root \(r = 1\) becomes

\[ a_{4}={\frac {1}{360}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(\frac {1}{2 r^{2}+5 r +3}\)	\(\frac {1}{10}\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63}\)	\(\frac {1}{360}\)

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

\[ a_{5}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(\frac {1}{2 r^{2}+5 r +3}\)	\(\frac {1}{10}\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63}\)	\(\frac {1}{360}\)
\(a_{5}\)	\(0\)	\(0\)

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

\begin{align*} y_{1}\left (x \right )&= x \left (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 \right ) \\ &= x \left (1+\frac {x^{2}}{10}+\frac {x^{4}}{360}+O\left (x^{6}\right )\right ) \end{align*}

Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now 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. (2B) gives

\[ b_{1} = 0 \]

For \(2\le n\) the recursive equation is

\begin{equation} \tag{3} 2 b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n} \left (n +r \right )-b_{n -2}+b_{n} = 0 \end{equation}

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

\[ b_{n} = \frac {b_{n -2}}{2 n^{2}+4 n r +2 r^{2}-3 n -3 r +1}\tag {4} \]

Which for the root \(r = {\frac {1}{2}}\) becomes

\[ b_{n} = \frac {b_{n -2}}{n \left (2 n -1\right )}\tag {5} \]

At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.

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

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

\[ b_{2}=\frac {1}{2 r^{2}+5 r +3} \]

Which for the root \(r = {\frac {1}{2}}\) becomes

\[ b_{2}={\frac {1}{6}} \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(\frac {1}{2 r^{2}+5 r +3}\)	\(\frac {1}{6}\)

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

\[ b_{3}=0 \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(\frac {1}{2 r^{2}+5 r +3}\)	\(\frac {1}{6}\)
\(b_{3}\)	\(0\)	\(0\)

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

\[ b_{4}=\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63} \]

Which for the root \(r = {\frac {1}{2}}\) becomes

\[ b_{4}={\frac {1}{168}} \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(\frac {1}{2 r^{2}+5 r +3}\)	\(\frac {1}{6}\)
\(b_{3}\)	\(0\)	\(0\)
\(b_{4}\)	\(\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63}\)	\(\frac {1}{168}\)

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

\[ b_{5}=0 \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(\frac {1}{2 r^{2}+5 r +3}\)	\(\frac {1}{6}\)
\(b_{3}\)	\(0\)	\(0\)
\(b_{4}\)	\(\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63}\)	\(\frac {1}{168}\)
\(b_{5}\)	\(0\)	\(0\)

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

\begin{align*} y_{2}\left (x \right )&= x \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 ) \\ &= \sqrt {x}\, \left (1+\frac {x^{2}}{6}+\frac {x^{4}}{168}+O\left (x^{6}\right )\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 x \left (1+\frac {x^{2}}{10}+\frac {x^{4}}{360}+O\left (x^{6}\right )\right ) + c_2 \sqrt {x}\, \left (1+\frac {x^{2}}{6}+\frac {x^{4}}{168}+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 )-x^{m} m +x^{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 \(r\).

\(a_{1} = 0\)

\(a_{2} = \frac {a_{0}}{2 r^{2}+5 r +3}\)

\(a_{3} = 0\)

\(a_{4} = \frac {a_{0}}{\left (2 r^{2}+5 r +3\right ) \left (2 r^{2}+13 r +21\right )}\)

\(a_{5} = 0\)

Expanding the rhs of the ode \(\cos \left (x \right )\) in series gives

\[ \cos \left (x \right ) = 1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4} \]

Since the \(F=1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4}\) has more than one term then we find a particular solution for each term and add the result to find the particular solution to the ode.

Now we determine the particular solution \(y_p\) associated with \(F=1\) by solving the balance equation

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

For \(c_{0}\) and \(x\). This results in

\begin{align*} c_{0}&=1\\ m&=0 \end{align*}

The particular solution is therefore

\begin{align*} y_p &= \sum _{n=0}^{\infty } c_n x^{n+m}\\ &= \sum _{n=0}^{\infty } c_n x^{n+0} \end{align*}

Where in the above \(c_0 = 1\).

The remaining \(c_n\) values are found using the same recurrence relation given in the earlier table which was used to find the homogeneous solution but using \(c_0\) in place of \(a_{0}\) and using \(m=0\) in place of the root of the indicial equation used to find the homogeneous solution. By letting \(a_{0} = c_{0}\) or \(a_{0} = 1\) and \(r = m\) or \(r = 0\). The following table gives the resulting \(c_n\) values. These values will be used to find the particular solution. Values of \(c_n\) found not defined when doing the substitution will be discarded and not used

\(c_{0} = 1\)

\(c_{1} = 0\)

\(c_{2} = {\frac {1}{3}}\)

\(c_{3} = 0\)

\(c_{4} = {\frac {1}{63}}\)

\(c_{5} = 0\)

The particular solution is now found using

\begin{align*} y_p &= x^{m} \sum _{n=0}^{\infty } c_n x^n \\ &= 1 \sum _{n=0}^{\infty } c_n x^n \\ \end{align*}

Using the values found above for \(c_n\) into the above sum gives

\begin{align*} y_p &= 1\left (1+\frac {1}{3} x^{2}+\frac {1}{63} x^{4}\right ) \\ &= 1+\frac {1}{3} x^{2}+\frac {1}{63} x^{4} \\ \end{align*}

Now we determine the particular solution \(y_p\) associated with \(F=-\frac {x^{2}}{2}\) by solving the balance equation

\[ \left (2 x^{m} m \left (-1+m \right )-x^{m} m +x^{m}\right ) c_{0} = -\frac {x^{2}}{2} \]

For \(c_{0}\) and \(x\). This results in

\begin{align*} c_{0}&=-{\frac {1}{6}}\\ m&=2 \end{align*}

The particular solution is therefore

\begin{align*} y_p &= \sum _{n=0}^{\infty } c_n x^{n+m}\\ &= \sum _{n=0}^{\infty } c_n x^{n+2} \end{align*}

Where in the above \(c_0 = -{\frac {1}{6}}\).

The remaining \(c_n\) values are found using the same recurrence relation given in the earlier table which was used to find the homogeneous solution but using \(c_0\) in place of \(a_{0}\) and using \(m=2\) in place of the root of the indicial equation used to find the homogeneous solution. By letting \(a_{0} = c_{0}\) or \(a_{0} = -{\frac {1}{6}}\) and \(r = m\) or \(r = 2\). The following table gives the resulting \(c_n\) values. These values will be used to find the particular solution. Values of \(c_n\) found not defined when doing the substitution will be discarded and not used

\(c_{0} = -{\frac {1}{6}}\)

\(c_{1} = 0\)

\(c_{2} = -{\frac {1}{126}}\)

\(c_{3} = 0\)

\(c_{4} = -{\frac {1}{6930}}\)

\(c_{5} = 0\)

The particular solution is now found using

\begin{align*} y_p &= x^{m} \sum _{n=0}^{\infty } c_n x^n \\ &= x^{2} \sum _{n=0}^{\infty } c_n x^n \\ \end{align*}

Using the values found above for \(c_n\) into the above sum gives

\begin{align*} y_p &= x^{2}\left (-\frac {1}{6}-\frac {1}{126} x^{2}-\frac {1}{6930} x^{4}\right ) \\ &= -\frac {1}{6} x^{2}-\frac {1}{126} x^{4}-\frac {1}{6930} x^{6} \\ \end{align*}

Now we determine the particular solution \(y_p\) associated with \(F=\frac {x^{4}}{24}\) by solving the balance equation

\[ \left (2 x^{m} m \left (-1+m \right )-x^{m} m +x^{m}\right ) c_{0} = \frac {x^{4}}{24} \]

For \(c_{0}\) and \(x\). This results in

\begin{align*} c_{0}&={\frac {1}{504}}\\ m&=4 \end{align*}

The particular solution is therefore

\begin{align*} y_p &= \sum _{n=0}^{\infty } c_n x^{n+m}\\ &= \sum _{n=0}^{\infty } c_n x^{n+4} \end{align*}

Where in the above \(c_0 = {\frac {1}{504}}\).

The remaining \(c_n\) values are found using the same recurrence relation given in the earlier table which was used to find the homogeneous solution but using \(c_0\) in place of \(a_{0}\) and using \(m=4\) in place of the root of the indicial equation used to find the homogeneous solution. By letting \(a_{0} = c_{0}\) or \(a_{0} = {\frac {1}{504}}\) and \(r = m\) or \(r = 4\). The following table gives the resulting \(c_n\) values. These values will be used to find the particular solution. Values of \(c_n\) found not defined when doing the substitution will be discarded and not used

\(c_{0} = {\frac {1}{504}}\)

\(c_{1} = 0\)

\(c_{2} = {\frac {1}{27720}}\)

\(c_{3} = 0\)

\(c_{4} = {\frac {1}{2910600}}\)

\(c_{5} = 0\)

The particular solution is now found using

\begin{align*} y_p &= x^{m} \sum _{n=0}^{\infty } c_n x^n \\ &= x^{4} \sum _{n=0}^{\infty } c_n x^n \\ \end{align*}

Using the values found above for \(c_n\) into the above sum gives

\begin{align*} y_p &= x^{4}\left (\frac {1}{504}+\frac {1}{27720} x^{2}+\frac {1}{2910600} x^{4}\right ) \\ &= \frac {1}{504} x^{4}+\frac {1}{27720} x^{6}+\frac {1}{2910600} x^{8} \\ \end{align*}

Adding all the above particular solution(s) gives

\[ y_p = 1+\frac {x^{2}}{6}+\frac {5 x^{4}}{504}-\frac {x^{6}}{9240}+\frac {x^{8}}{2910600}+O\left (x^{6}\right ) \]

Truncating the particular solution to the order of series requested gives

\[ y_p = 1+\frac {x^{2}}{6}+\frac {5 x^{4}}{504}+O\left (x^{6}\right ) \]

Hence the final solution is

\begin{align*} y &= y_h + y_p \\ &= 1+\frac {x^{2}}{6}+\frac {5 x^{4}}{504}+O\left (x^{6}\right )+c_1 x \left (1+\frac {x^{2}}{10}+\frac {x^{4}}{360}+O\left (x^{6}\right )\right )+c_2 \sqrt {x}\, \left (1+\frac {x^{2}}{6}+\frac {x^{4}}{168}+O\left (x^{6}\right )\right ) \\ \end{align*}

Maple step by step solution

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   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 
<- solving first the homogeneous part of the ODE successful`
 

Maple dsolve solution

Solving time : 0.018 (sec)
Leaf size : 43

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

\[ y = c_{1} \sqrt {x}\, \left (1+\frac {1}{6} x^{2}+\frac {1}{168} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x \left (1+\frac {1}{10} x^{2}+\frac {1}{360} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+\left (1+\frac {1}{6} x^{2}+\frac {5}{504} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) \]

Mathematica DSolve solution

Solving time : 0.049 (sec)
Leaf size : 176

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

\[ y(x)\to c_2 x \left (\frac {x^6}{28080}+\frac {x^4}{360}+\frac {x^2}{10}+1\right )+c_1 \sqrt {x} \left (\frac {x^6}{11088}+\frac {x^4}{168}+\frac {x^2}{6}+1\right )+\sqrt {x} \left (-\frac {x^{11/2}}{3861}+\frac {x^{7/2}}{630}+\frac {4 x^{3/2}}{15}+\frac {2}{\sqrt {x}}\right ) \left (\frac {x^6}{11088}+\frac {x^4}{168}+\frac {x^2}{6}+1\right )+x \left (\frac {37 x^5}{69300}-\frac {x^3}{84}-\frac {x}{3}-\frac {1}{x}\right ) \left (\frac {x^6}{28080}+\frac {x^4}{360}+\frac {x^2}{10}+1\right ) \]