Internal problem ID [10892]
Internal file name [OUTPUT/10149_Sunday_December_24_2023_05_15_33_PM_49026519/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-3 Equation of form
\((a x + b)y''+f(x)y'+g(x)y=0\)
Problem number: 69.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "exact linear second order ode", "second_order_integrable_as_is"
Maple gives the following as the ode type
[[_2nd_order, _exact, _linear, _homogeneous]]
\[ \boxed {x y^{\prime \prime }+x a y^{\prime }+a y=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x y^{\prime \prime }+x a y^{\prime }+a y\right )d x &= 0 \\ \left (a x -1\right ) y+x y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=\frac {a x -1}{x}\\ q(x) &=\frac {c_{1}}{x} \end {align*}
Hence the ode is \begin {align*} y^{\prime }+\frac {\left (a x -1\right ) y}{x} = \frac {c_{1}}{x} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {a x -1}{x}d x} \\ &= {\mathrm e}^{a x -\ln \left (x \right )} \\ \end{align*} Which simplifies to \[ \mu = \frac {{\mathrm e}^{a x}}{x} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {{\mathrm e}^{a x} y}{x}\right ) &= \left (\frac {{\mathrm e}^{a x}}{x}\right ) \left (\frac {c_{1}}{x}\right )\\ \mathrm {d} \left (\frac {{\mathrm e}^{a x} y}{x}\right ) &= \left (\frac {c_{1} {\mathrm e}^{a x}}{x^{2}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {{\mathrm e}^{a x} y}{x} &= \int {\frac {c_{1} {\mathrm e}^{a x}}{x^{2}}\,\mathrm {d} x}\\ \frac {{\mathrm e}^{a x} y}{x} &= c_{1} a \left (-\frac {{\mathrm e}^{a x}}{x a}-\operatorname {expIntegral}_{1}\left (-a x \right )\right ) + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {{\mathrm e}^{a x}}{x}\) results in \begin {align*} y &= x \,{\mathrm e}^{-a x} c_{1} a \left (-\frac {{\mathrm e}^{a x}}{x a}-\operatorname {expIntegral}_{1}\left (-a x \right )\right )+c_{2} x \,{\mathrm e}^{-a x} \end {align*}
which simplifies to \begin {align*} y &= -\operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}-c_{1} +c_{2} x \,{\mathrm e}^{-a x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}-c_{1} +c_{2} x \,{\mathrm e}^{-a x} \\ \end{align*}
Verification of solutions
\[ y = -\operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}-c_{1} +c_{2} x \,{\mathrm e}^{-a x} \] Verified OK.
Writing the ode as \[ x y^{\prime \prime }+x a y^{\prime }+a y = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x y^{\prime \prime }+x a y^{\prime }+a y\right )d x &= 0 \\ -y+x a y+x y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=\frac {a x -1}{x}\\ q(x) &=\frac {c_{1}}{x} \end {align*}
Hence the ode is \begin {align*} y^{\prime }+\frac {\left (a x -1\right ) y}{x} = \frac {c_{1}}{x} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {a x -1}{x}d x} \\ &= {\mathrm e}^{a x -\ln \left (x \right )} \\ \end{align*} Which simplifies to \[ \mu = \frac {{\mathrm e}^{a x}}{x} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {{\mathrm e}^{a x} y}{x}\right ) &= \left (\frac {{\mathrm e}^{a x}}{x}\right ) \left (\frac {c_{1}}{x}\right )\\ \mathrm {d} \left (\frac {{\mathrm e}^{a x} y}{x}\right ) &= \left (\frac {c_{1} {\mathrm e}^{a x}}{x^{2}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {{\mathrm e}^{a x} y}{x} &= \int {\frac {c_{1} {\mathrm e}^{a x}}{x^{2}}\,\mathrm {d} x}\\ \frac {{\mathrm e}^{a x} y}{x} &= c_{1} a \left (-\frac {{\mathrm e}^{a x}}{x a}-\operatorname {expIntegral}_{1}\left (-a x \right )\right ) + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {{\mathrm e}^{a x}}{x}\) results in \begin {align*} y &= x \,{\mathrm e}^{-a x} c_{1} a \left (-\frac {{\mathrm e}^{a x}}{x a}-\operatorname {expIntegral}_{1}\left (-a x \right )\right )+c_{2} x \,{\mathrm e}^{-a x} \end {align*}
which simplifies to \begin {align*} y &= -\operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}-c_{1} +c_{2} x \,{\mathrm e}^{-a x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}-c_{1} +c_{2} x \,{\mathrm e}^{-a x} \\ \end{align*}
Verification of solutions
\[ y = -\operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}-c_{1} +c_{2} x \,{\mathrm e}^{-a x} \] Verified OK.
Writing the ode as \begin {align*} x y^{\prime \prime }+x a y^{\prime }+a 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 &= a x\tag {3} \\ C &= a \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 {a \left (a x -4\right )}{4 x}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= a \left (a x -4\right )\\ t &= 4 x \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {a \left (a x -4\right )}{4 x}\right ) z(x)\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 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
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) \\ &= 1 - 1 \\ &= 0 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 x\). There is a pole at \(x=0\) of order \(1\). 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*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 1. For the pole at \(x = 0\) of order 1 then \begin {align*} [\sqrt r]_c &= 0 \\ \alpha _c^+ &= 1 \\ \alpha _c^- &= 1 \end {align*}
Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = 0\) then \begin {alignat*} {3} v &= \frac {-O_r(\infty )}{2} &&= \frac {0}{2} &&= 0 \end {alignat*}
\([\sqrt r]_\infty \) is the sum of terms involving \(x^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{v} a_i x^i \\ &= \sum _{i=0}^{0} a_i x^i \tag {8} \end {align*}
Let \(a\) be the coefficient of \(x^v=x^0\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is \[ \sqrt r \approx \frac {a}{2}-\frac {1}{x}-\frac {1}{a \,x^{2}}-\frac {2}{a^{2} x^{3}}-\frac {5}{a^{3} x^{4}}-\frac {14}{a^{4} x^{5}}-\frac {42}{a^{5} x^{6}}-\frac {132}{a^{6} x^{7}} + \dots \tag {9} \] Comparing Eq. (9) with Eq. (8) shows that \[ a = \frac {a}{2} \] From Eq. (9) the sum up to \(v=0\) gives \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{0} a_i x^i \\ &= \frac {a}{2} \tag {10} \end {align*}
Now we need to find \(b\), where \(b\) be the coefficient of \(x^{v-1} = x^{-1}=\frac {1}{x}\) in \(r\) minus the coefficient of same term but in \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence \[ \left ( [\sqrt r]_\infty \right )^2 = \frac {a^{2}}{4} \] This shows that the coefficient of \(\frac {1}{x}\) in the above is \(0\). Now we need to find the coefficient of \(\frac {1}{x}\) in \(r\). How this is done depends on if \(v=0\) or not. Since \(v=0\) then starting from \(r=\frac {s}{t}\) and doing long division in the form \[ r = Q + \frac {R}{t} \] Where \(Q\) is the quotient and \(R\) is the remainder. Then the coefficient of \(\frac {1}{x}\) in \(r\) will be the coefficient in \(R\) of the term in \(x\) of degree of \(t\) minus one, divided by the leading coefficient in \(t\). Doing long division gives \begin {align*} r &= \frac {s}{t} \\ &= \frac {a \left (a x -4\right )}{4 x} \\ &= Q + \frac {R}{4 x}\\ &= \left (\frac {a^{2}}{4}\right ) + \left ( -\frac {a}{x}\right ) \\ &= \frac {a^{2}}{4}-\frac {a}{x} \end {align*}
Since the degree of \(t\) is \(1\), then we see that the coefficient of the term \(1\) in the remainder \(R\) is \(-4 a\). Dividing this by leading coefficient in \(t\) which is \(4\) gives \(-a\). Now \(b\) can be found. \begin {align*} b &= \left (-a\right )-\left (0\right )\\ &= -a \end {align*}
Hence \begin {alignat*} {3} [\sqrt r]_\infty &= \frac {a}{2}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {-a}{\frac {a}{2}} - 0 \right ) &&= -1\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {-a}{\frac {a}{2}} - 0 \right ) &&= 1 \end {alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=\frac {a \left (a x -4\right )}{4 x} \]
| pole \(c\) location | pole order | \([\sqrt r]_c\) | \(\alpha _c^{+}\) | \(\alpha _c^{-}\) |
| \(0\) | \(1\) | \(0\) | \(0\) | \(1\) |
| Order of \(r\) at \(\infty \) | \([\sqrt r]_\infty \) | \(\alpha _\infty ^{+}\) | \(\alpha _\infty ^{-}\) |
| \(0\) | \(\frac {a}{2}\) | \(-1\) | \(1\) |
Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using \begin {align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end {align*}
Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{-} = 1\) then \begin {align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= 1 - \left ( 1 \right ) \\ &= 0 \end {align*}
Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using \begin {align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end {align*}
The above gives \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (-) [\sqrt r]_\infty \\ &= \frac {1}{x} + (-) \left ( \frac {a}{2} \right ) \\ &= \frac {1}{x}-\frac {a}{2}\\ &= \frac {1}{x}-\frac {a}{2} \end {align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation \begin {align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end {align*}
Let \begin {align*} p(x) &= 1\tag {2A} \end {align*}
Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (\frac {1}{x}-\frac {a}{2}\right ) \left (0\right ) + \left ( \left (-\frac {1}{x^{2}}\right ) + \left (\frac {1}{x}-\frac {a}{2}\right )^2 - \left (\frac {a \left (a x -4\right )}{4 x}\right ) \right ) &= 0\\ 0 = 0 \end {align*}
The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \left (\frac {1}{x}-\frac {a}{2}\right )d x}\\ &= x \,{\mathrm e}^{-\frac {a x}{2}} \end {align*}
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 {a x}{x} \,dx} \\ &= z_1 e^{-\frac {a x}{2}} \\ &= z_1 \left ({\mathrm e}^{-\frac {a x}{2}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = x \,{\mathrm e}^{-a x} \] 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 {a x}{x} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-a x}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\frac {-\operatorname {expIntegral}_{1}\left (-a x \right ) a x -{\mathrm e}^{a x}}{x}\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x \,{\mathrm e}^{-a x}\right ) + c_{2} \left (x \,{\mathrm e}^{-a x}\left (\frac {-\operatorname {expIntegral}_{1}\left (-a x \right ) a x -{\mathrm e}^{a x}}{x}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \,{\mathrm e}^{-a x}+c_{2} \left (-\operatorname {expIntegral}_{1}\left (-a x \right ) x \,{\mathrm e}^{-a x} a -1\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \,{\mathrm e}^{-a x}+c_{2} \left (-\operatorname {expIntegral}_{1}\left (-a x \right ) x \,{\mathrm e}^{-a x} a -1\right ) \] Verified OK.
An ode of the form \begin {align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end {align*}
is exact if \begin {align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end {align*}
For the given ode we have \begin {align*} p(x) &= x\\ q(x) &= a x\\ r(x) &= a\\ s(x) &= 0 \end {align*}
Hence \begin {align*} p''(x) &= 0\\ q'(x) &= a \end {align*}
Therefore (1) becomes \begin {align*} 0- \left (a\right ) + \left (a\right )&=0 \end {align*}
Hence the ode is exact. Since we now know the ode is exact, it can be written as \begin {align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end {align*}
Integrating gives \begin {align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end {align*}
Substituting the above values for \(p,q,r,s\) gives \begin {align*} \left (a x -1\right ) y+x y^{\prime }&=c_{1} \end {align*}
We now have a first order ode to solve which is \begin {align*} \left (a x -1\right ) y+x y^{\prime } = c_{1} \end {align*}
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=\frac {a x -1}{x}\\ q(x) &=\frac {c_{1}}{x} \end {align*}
Hence the ode is \begin {align*} y^{\prime }+\frac {\left (a x -1\right ) y}{x} = \frac {c_{1}}{x} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {a x -1}{x}d x} \\ &= {\mathrm e}^{a x -\ln \left (x \right )} \\ \end{align*} Which simplifies to \[ \mu = \frac {{\mathrm e}^{a x}}{x} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {{\mathrm e}^{a x} y}{x}\right ) &= \left (\frac {{\mathrm e}^{a x}}{x}\right ) \left (\frac {c_{1}}{x}\right )\\ \mathrm {d} \left (\frac {{\mathrm e}^{a x} y}{x}\right ) &= \left (\frac {c_{1} {\mathrm e}^{a x}}{x^{2}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {{\mathrm e}^{a x} y}{x} &= \int {\frac {c_{1} {\mathrm e}^{a x}}{x^{2}}\,\mathrm {d} x}\\ \frac {{\mathrm e}^{a x} y}{x} &= c_{1} a \left (-\frac {{\mathrm e}^{a x}}{x a}-\operatorname {expIntegral}_{1}\left (-a x \right )\right ) + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {{\mathrm e}^{a x}}{x}\) results in \begin {align*} y &= x \,{\mathrm e}^{-a x} c_{1} a \left (-\frac {{\mathrm e}^{a x}}{x a}-\operatorname {expIntegral}_{1}\left (-a x \right )\right )+c_{2} x \,{\mathrm e}^{-a x} \end {align*}
which simplifies to \begin {align*} y &= -\operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}-c_{1} +c_{2} x \,{\mathrm e}^{-a x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}-c_{1} +c_{2} x \,{\mathrm e}^{-a x} \\ \end{align*}
Verification of solutions
\[ y = -\operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}-c_{1} +c_{2} x \,{\mathrm e}^{-a x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+x a y^{\prime }+a 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 }=-a y^{\prime }-\frac {a y}{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 }+a y^{\prime }+\frac {a y}{x}=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 )=a , P_{3}\left (x \right )=\frac {a}{x}\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}}}=0 \\ {} & \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 )+x a y^{\prime }+a 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^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +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 (-1+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r \right )+a a_{k} \left (k +1+r \right )\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 \{0, 1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +1+r \right ) \left (a_{k +1} \left (k +r \right )+a a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a a_{k}}{k +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a a_{k}}{k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {a a_{k}}{k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=-\frac {a a_{k}}{k +1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=-\frac {a a_{k}}{k +1}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +1}\right ), b_{k +1}=-\frac {a b_{k}}{k}, c_{k +1}=-\frac {a c_{k}}{k +1}\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)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 29
dsolve(x*diff(y(x),x$2)+a*x*diff(y(x),x)+a*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \operatorname {expIntegral}_{1}\left (-a x \right ) c_{1} a x \,{\mathrm e}^{-a x}+c_{1} +c_{2} x \,{\mathrm e}^{-a x} \]
✓ Solution by Mathematica
Time used: 0.177 (sec). Leaf size: 35
DSolve[x*y''[x]+a*x*y'[x]+a*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-a x} \left (a c_2 x \operatorname {ExpIntegralEi}(a x)-c_2 e^{a x}+c_1 x\right ) \]