2.37 problem 37
Internal
problem
ID
[8126]
Book
:
Second
order
enumerated
odes
Section
:
section
2
Problem
number
:
37
Date
solved
:
Monday, October 21, 2024 at 04:54:00 PM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} x y^{\prime \prime }+2 y^{\prime }-x y&=0 \end{align*}
2.37.1 Solved as second order ode using change of variable on y method 1
Time used: 0.290 (sec)
In normal form the given ode is written as
\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}
Where
\begin{align*} p \left (x \right )&=\frac {2}{x}\\ q \left (x \right )&=-1 \end{align*}
Calculating the Liouville ode invariant \(Q\) given by
\begin{align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= -1 - \frac {\left (\frac {2}{x}\right )'}{2}- \frac {\left (\frac {2}{x}\right )^2}{4} \\ &= -1 - \frac {\left (-\frac {2}{x^{2}}\right )}{2}- \frac {\left (\frac {4}{x^{2}}\right )}{4} \\ &= -1 - \left (-\frac {1}{x^{2}}\right )-\frac {1}{x^{2}}\\ &= -1 \end{align*}
Since the Liouville ode invariant does not depend on the independent variable \(x\) then the
transformation
\begin{align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end{align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given
by
\begin{align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {\frac {2}{x}}{2} }\\ &= \frac {1}{x}\tag {5} \end{align*}
Hence (3) becomes
\begin{align*} y = \frac {v \left (x \right )}{x}\tag {4} \end{align*}
Applying this change of variable to the original ode results in
\begin{align*} v^{\prime \prime }\left (x \right )-v \left (x \right ) = 0 \end{align*}
Which is now solved for \(v \left (x \right )\).
This is second order with constant coefficients homogeneous ODE. In standard form the
ODE is
\[ A v''(x) + B v'(x) + C v(x) = 0 \]
Where in the above \(A=1, B=0, C=-1\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives
\[ \lambda ^{2} {\mathrm e}^{x \lambda }-{\mathrm e}^{x \lambda } = 0 \tag {1} \]
Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\) gives
\[ \lambda ^{2}-1 = 0 \tag {2} \]
Equation (2) is the characteristic equation of the ODE. Its roots determine the
general solution form.Using the quadratic formula
\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]
Substituting \(A=1, B=0, C=-1\) into the above gives
\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (-1\right )}\\ &= \pm 1 \end{align*}
Hence
\begin{align*}
\lambda _1 &= + 1 \\
\lambda _2 &= - 1 \\
\end{align*}
Which simplifies to
\begin{align*}
\lambda _1 &= 1 \\
\lambda _2 &= -1 \\
\end{align*}
Since roots are real and distinct, then the solution is
\begin{align*}
v \left (x \right ) &= c_1 e^{\lambda _1 x} + c_2 e^{\lambda _2 x} \\
v \left (x \right ) &= c_1 e^{\left (1\right )x} +c_2 e^{\left (-1\right )x} \\
\end{align*}
Or
\[
v \left (x \right ) =c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}
\]
Will
add steps showing solving for IC soon.
Now that \(v \left (x \right )\) is known, then
\begin{align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}\right ) \left (z \left (x \right )\right )\tag {7} \end{align*}
But from (5)
\begin{align*} z \left (x \right )&= \frac {1}{x} \end{align*}
Hence (7) becomes
\begin{align*} y = \frac {c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}}{x} \end{align*}
Will add steps showing solving for IC soon.
2.37.2 Solved as second order Bessel ode
Time used: 0.061 (sec)
Writing the ode as
\begin{align*} x^{2} y^{\prime \prime }+2 x y^{\prime }-x^{2} y = 0\tag {1} \end{align*}
Bessel ode has the form
\begin{align*} x^{2} y^{\prime \prime }+x y^{\prime }+\left (-n^{2}+x^{2}\right ) y = 0\tag {2} \end{align*}
The generalized form of Bessel ode is given by Bowman (1958) as the following
\begin{align*} x^{2} y^{\prime \prime }+\left (1-2 \alpha \right ) x y^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) y = 0\tag {3} \end{align*}
With the standard solution
\begin{align*} y&=x^{\alpha } \left (c_1 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_2 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}
Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives
\begin{align*} \alpha &= -{\frac {1}{2}}\\ \beta &= i\\ n &= {\frac {1}{2}}\\ \gamma &= 1 \end{align*}
Substituting all the above into (4) gives the solution as
\begin{align*} y = \frac {i c_1 \sqrt {2}\, \sinh \left (x \right )}{\sqrt {x}\, \sqrt {\pi }\, \sqrt {i x}}-\frac {c_2 \sqrt {2}\, \cosh \left (x \right )}{\sqrt {x}\, \sqrt {\pi }\, \sqrt {i x}} \end{align*}
Will add steps showing solving for IC soon.
2.37.3 Solved as second order ode using Kovacic algorithm
Time used: 0.054 (sec)
Writing the ode as
\begin{align*} x y^{\prime \prime }+2 y^{\prime }-x 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 &= x \\ B &= 2\tag {3} \\ C &= -x \end{align*}
Applying the Liouville transformation on the dependent variable gives
\begin{align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end{align*}
Then (2) becomes
\begin{align*} z''(x) = r z(x)\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 {1}{1}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= 1\\ t &= 1 \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(x) &= z \left (x \right ) \tag {7} \end{align*}
Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation
\begin{align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \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 49: 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) \\ &= 0 - 0 \\ &= 0 \end{align*}
There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole
larger than \(2\) and the order at \(\infty \) is \(0\) then the necessary conditions for case one are met.
Therefore
\begin{align*} L &= [1] \end{align*}
Since \(r = 1\) is not a function of \(x\), then there is no need run Kovacic algorithm to obtain a solution
for transformed ode \(z''=r z\) as one solution is
\[ z_1(x) = {\mathrm e}^{-x} \]
Using the above, the solution for the original ode can
now be found. 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} \,dx} \\
&= z_1 e^{ -\int \frac {1}{2} \frac {2}{x} \,dx} \\
&= z_1 e^{-\ln \left (x \right )} \\
&= z_1 \left (\frac {1}{x}\right ) \\
\end{align*}
Which simplifies to
\[
y_1 = \frac {{\mathrm e}^{-x}}{x}
\]
The second solution \(y_2\) to the original ode is found using reduction of order
\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \]
Substituting gives
\begin{align*}
y_2 &= y_1 \int \frac { e^{\int -\frac {2}{x} \,dx}}{\left (y_1\right )^2} \,dx \\
&= y_1 \int \frac { e^{-2 \ln \left (x \right )}}{\left (y_1\right )^2} \,dx \\
&= y_1 \left (\frac {{\mathrm e}^{2 x}}{2}\right ) \\
\end{align*}
Therefore the solution is
\begin{align*}
y &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left (\frac {{\mathrm e}^{-x}}{x}\right ) + c_2 \left (\frac {{\mathrm e}^{-x}}{x}\left (\frac {{\mathrm e}^{2 x}}{2}\right )\right ) \\
\end{align*}
Will add steps showing solving for IC soon.
2.37.4 Solved as second order ode adjoint method
Time used: 0.712 (sec)
In normal form the ode
\begin{align*} x y^{\prime \prime }+2 y^{\prime }-x y = 0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end{align*}
Where
\begin{align*} p \left (x \right )&=\frac {2}{x}\\ q \left (x \right )&=-1\\ r \left (x \right )&=0 \end{align*}
The Lagrange adjoint ode is given by
\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {2 \xi \left (x \right )}{x}\right )' + \left (-\xi \left (x \right )\right ) &= 0\\ \frac {\xi ^{\prime \prime }\left (x \right ) x^{2}-\xi \left (x \right ) x^{2}-2 \xi ^{\prime }\left (x \right ) x +2 \xi \left (x \right )}{x^{2}}&= 0 \end{align*}
Which is solved for \(\xi (x)\). In normal form the given ode is written as
\begin{align*} \xi ^{\prime \prime }+p \left (x \right ) \xi ^{\prime }+q \left (x \right ) \xi &=0 \tag {2} \end{align*}
Where
\begin{align*} p \left (x \right )&=-\frac {2}{x}\\ q \left (x \right )&=\frac {2}{x^{2}}-1 \end{align*}
Calculating the Liouville ode invariant \(Q\) given by
\begin{align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= \frac {2}{x^{2}}-1 - \frac {\left (-\frac {2}{x}\right )'}{2}- \frac {\left (-\frac {2}{x}\right )^2}{4} \\ &= \frac {2}{x^{2}}-1 - \frac {\left (\frac {2}{x^{2}}\right )}{2}- \frac {\left (\frac {4}{x^{2}}\right )}{4} \\ &= \frac {2}{x^{2}}-1 - \left (\frac {1}{x^{2}}\right )-\frac {1}{x^{2}}\\ &= -1 \end{align*}
Since the Liouville ode invariant does not depend on the independent variable \(x\) then the
transformation
\begin{align*} \xi = v \left (x \right ) z \left (x \right )\tag {3} \end{align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given
by
\begin{align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {-\frac {2}{x}}{2} }\\ &= x\tag {5} \end{align*}
Hence (3) becomes
\begin{align*} \xi = v \left (x \right ) x\tag {4} \end{align*}
Applying this change of variable to the original ode results in
\begin{align*} x \left (v^{\prime \prime }\left (x \right )-v \left (x \right )\right ) = 0 \end{align*}
Which is now solved for \(v \left (x \right )\).
The above ode can be simplified to
\begin{align*} v^{\prime \prime }\left (x \right )-v \left (x \right ) = 0 \end{align*}
This is second order with constant coefficients homogeneous ODE. In standard form the
ODE is
\[ A v''(x) + B v'(x) + C v(x) = 0 \]
Where in the above \(A=1, B=0, C=-1\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives
\[ \lambda ^{2} {\mathrm e}^{x \lambda }-{\mathrm e}^{x \lambda } = 0 \tag {1} \]
Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\) gives
\[ \lambda ^{2}-1 = 0 \tag {2} \]
Equation (2) is the characteristic equation of the ODE. Its roots determine the
general solution form.Using the quadratic formula
\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]
Substituting \(A=1, B=0, C=-1\) into the above gives
\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (-1\right )}\\ &= \pm 1 \end{align*}
Hence
\begin{align*}
\lambda _1 &= + 1 \\
\lambda _2 &= - 1 \\
\end{align*}
Which simplifies to
\begin{align*}
\lambda _1 &= 1 \\
\lambda _2 &= -1 \\
\end{align*}
Since roots are real and distinct, then the solution is
\begin{align*}
v \left (x \right ) &= c_1 e^{\lambda _1 x} + c_2 e^{\lambda _2 x} \\
v \left (x \right ) &= c_1 e^{\left (1\right )x} +c_2 e^{\left (-1\right )x} \\
\end{align*}
Or
\[
v \left (x \right ) =c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}
\]
Will
add steps showing solving for IC soon.
Now that \(v \left (x \right )\) is known, then
\begin{align*} \xi &= v \left (x \right ) z \left (x \right )\\ &= \left (c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}\right ) \left (z \left (x \right )\right )\tag {7} \end{align*}
But from (5)
\begin{align*} z \left (x \right )&= x \end{align*}
Hence (7) becomes
\begin{align*} \xi = \left (c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}\right ) x \end{align*}
Will add steps showing solving for IC soon.
The original ode (2) now reduces to first order ode
\begin{align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )}\\ y^{\prime }+y \left (\frac {2}{x}-\frac {\left (c_1 \,{\mathrm e}^{x}-c_2 \,{\mathrm e}^{-x}\right ) x +c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}}{\left (c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}\right ) x}\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(x)y &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=-\frac {{\mathrm e}^{x} c_1 x -{\mathrm e}^{-x} c_2 x -c_1 \,{\mathrm e}^{x}-c_2 \,{\mathrm e}^{-x}}{\left (c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}\right ) x}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {{\mathrm e}^{x} c_1 x -{\mathrm e}^{-x} c_2 x -c_1 \,{\mathrm e}^{x}-c_2 \,{\mathrm e}^{-x}}{\left (c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}\right ) x}d x}\\ &= \frac {x \,{\mathrm e}^{x}}{c_1 \,{\mathrm e}^{2 x}+c_2} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y x \,{\mathrm e}^{x}}{c_1 \,{\mathrm e}^{2 x}+c_2}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} \frac {y x \,{\mathrm e}^{x}}{c_1 \,{\mathrm e}^{2 x}+c_2}&= \int {0 \,dx} + c_3 \\ &=c_3 \end{align*}
Dividing throughout by the integrating factor \(\frac {x \,{\mathrm e}^{x}}{c_1 \,{\mathrm e}^{2 x}+c_2}\) gives the final solution
\[ y = \frac {c_3 \left (c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}\right )}{x} \]
Hence, the solution
found using Lagrange adjoint equation method is
\begin{align*}
y &= \frac {c_3 \left (c_1 \,{\mathrm e}^{x}+c_2 \,{\mathrm e}^{-x}\right )}{x} \\
\end{align*}
Will add steps showing solving for IC
soon.
2.37.5 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+2 y^{\prime }-x 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 }=y-\frac {2 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 {2 y^{\prime }}{x}-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 {2}{x}, P_{3}\left (x \right )=-1\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}}}=2 \\ {} & \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} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+2 y^{\prime }-x y=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 \cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} 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 +r +1\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 +r +1\right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1+r \right ) x^{-1+r}+a_{1} \left (1+r \right ) \left (2+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )-a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (2+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )-a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +3+r \right )-a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {a_{k}}{\left (k +2+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=\frac {a_{k}}{\left (k +1\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +2}=\frac {a_{k}}{\left (k +1\right ) \left (k +2\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {a_{k}}{\left (k +2\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {a_{k}}{\left (k +2\right ) \left (k +3\right )}, 2 a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right ), a_{k +2}=\frac {a_{k}}{\left (k +1\right ) \left (k +2\right )}, 0=0, b_{k +2}=\frac {b_{k}}{\left (k +2\right ) \left (k +3\right )}, 2 b_{1}=0\right ] \end {array} \]
2.37.6 Maple trace
Methods for second order ODEs:
2.37.7 Maple dsolve solution
Solving time : 0.004
(sec)
Leaf size : 17
dsolve(x*diff(diff(y(x),x),x)+2*diff(y(x),x)-x*y(x) = 0,
y(x),singsol=all)
\[
y = \frac {c_1 \sinh \left (x \right )+c_2 \cosh \left (x \right )}{x}
\]
2.37.8 Mathematica DSolve solution
Solving time : 0.041
(sec)
Leaf size : 28
DSolve[{x*D[y[x],{x,2}]+2*D[y[x],x]-x*y[x]==0,{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to \frac {2 c_1 e^{-x}+c_2 e^x}{2 x}
\]