28.22 problem 82

Internal problem ID [10905]
Internal file name [OUTPUT/10162_Sunday_December_31_2023_11_03_06_AM_31797229/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: 82.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact linear second order ode", "second_order_integrable_as_is"

Maple gives the following as the ode type

[[_2nd_order, _exact, _linear, _homogeneous]]

\[ \boxed {x y^{\prime \prime }+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (2 a x +b \right ) y=0} \]

28.22.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x y^{\prime \prime }+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (2 a x +b \right ) y\right )d x &= 0 \\ \left (a \,x^{2}+b x +c -1\right ) y+y^{\prime } x = 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^{2}+b x +c -1}{x}\\ q(x) &=\frac {c_{1}}{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {\left (a \,x^{2}+b x +c -1\right ) y}{x} = \frac {c_{1}}{x} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {a \,x^{2}+b x +c -1}{x}d x} \\ &= {\mathrm e}^{\frac {a \,x^{2}}{2}+b x +\left (c -1\right ) \ln \left (x \right )} \\ \end{align*} Which simplifies to \[ \mu = x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} \] 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 (x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y\right ) &= \left (x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\right ) \left (\frac {c_{1}}{x}\right )\\ \mathrm {d} \left (x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y\right ) &= \left (c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y &= \int {c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\,\mathrm {d} x}\\ x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y &= \int c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu =x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\) results in \begin {align*} y &= x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (\int c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}d x \right )+c_{2} x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \end {align*}

which simplifies to \begin {align*} y &= x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \\ \end{align*}

Verification of solutions

\[ y = x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \] Verified OK.

28.22.2 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ x y^{\prime \prime }+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (2 a x +b \right ) y = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x y^{\prime \prime }+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (2 a x +b \right ) y\right )d x &= 0 \\ -y+y c +y^{\prime } x +y a \,x^{2}+y b x = 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^{2}+b x +c -1}{x}\\ q(x) &=\frac {c_{1}}{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {\left (a \,x^{2}+b x +c -1\right ) y}{x} = \frac {c_{1}}{x} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {a \,x^{2}+b x +c -1}{x}d x} \\ &= {\mathrm e}^{\frac {a \,x^{2}}{2}+b x +\left (c -1\right ) \ln \left (x \right )} \\ \end{align*} Which simplifies to \[ \mu = x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} \] 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 (x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y\right ) &= \left (x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\right ) \left (\frac {c_{1}}{x}\right )\\ \mathrm {d} \left (x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y\right ) &= \left (c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y &= \int {c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\,\mathrm {d} x}\\ x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y &= \int c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu =x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\) results in \begin {align*} y &= x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (\int c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}d x \right )+c_{2} x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \end {align*}

which simplifies to \begin {align*} y &= x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \\ \end{align*}

Verification of solutions

\[ y = x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \] Verified OK.

28.22.3 Solving as exact linear second order ode ode

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^{2}+b x +c\\ r(x) &= 2 a x +b\\ s(x) &= 0 \end {align*}

Hence \begin {align*} p''(x) &= 0\\ q'(x) &= 2 a x +b \end {align*}

Therefore (1) becomes \begin {align*} 0- \left (2 a x +b\right ) + \left (2 a x +b\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^{2}+b x +c -1\right ) y+y^{\prime } x&=c_{1} \end {align*}

We now have a first order ode to solve which is \begin {align*} \left (a \,x^{2}+b x +c -1\right ) y+y^{\prime } x = 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^{2}+b x +c -1}{x}\\ q(x) &=\frac {c_{1}}{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {\left (a \,x^{2}+b x +c -1\right ) y}{x} = \frac {c_{1}}{x} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {a \,x^{2}+b x +c -1}{x}d x} \\ &= {\mathrm e}^{\frac {a \,x^{2}}{2}+b x +\left (c -1\right ) \ln \left (x \right )} \\ \end{align*} Which simplifies to \[ \mu = x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} \] 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 (x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y\right ) &= \left (x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\right ) \left (\frac {c_{1}}{x}\right )\\ \mathrm {d} \left (x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y\right ) &= \left (c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y &= \int {c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\,\mathrm {d} x}\\ x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}} y &= \int c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu =x^{c -1} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}\) results in \begin {align*} y &= x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (\int c_{1} x^{c -2} {\mathrm e}^{\frac {x \left (a x +2 b \right )}{2}}d x \right )+c_{2} x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \end {align*}

which simplifies to \begin {align*} y &= x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \\ \end{align*}

Verification of solutions

\[ y = x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \] Verified OK.

28.22.4 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (2 a x +b \right ) 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 }=-\frac {\left (2 a x +b \right ) y}{x}-\frac {\left (a \,x^{2}+b x +c \right ) 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 {\left (a \,x^{2}+b x +c \right ) y^{\prime }}{x}+\frac {\left (2 a x +b \right ) 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 )=\frac {a \,x^{2}+b x +c}{x}, P_{3}\left (x \right )=\frac {2 a x +b}{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}}}=c \\ {} & \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 )+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (2 a x +b \right ) 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^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +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 +c \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (r +c \right )+a_{0} b \left (1+r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +c \right )+a_{k} b \left (k +1+r \right )+a_{k -1} a \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 +c \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -c +1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (r +c \right )+a_{0} b \left (1+r \right )=0 \\ \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 +c \right )+a_{k} b +a_{k -1} a \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +r +2\right ) \left (a_{k +2} \left (k +1+r +c \right )+a_{k +1} b +a_{k} a \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k +1+r +c} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k +1+c} \\ \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} a +a_{k +1} b}{k +1+c}, b a_{0}+c a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-c +1 \\ {} & {} & a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k +2} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-c +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -c +1}, a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k +2}, a_{1} \left (-c +2\right )+a_{0} b \left (-c +2\right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k -c +1}\right ), d_{k +2}=-\frac {a d_{k}+b d_{k +1}}{k +1+c}, b d_{0}+c d_{1}=0, e_{k +2}=-\frac {a e_{k}+b e_{k +1}}{k +2}, e_{1} \left (-c +2\right )+e_{0} b \left (-c +2\right )=0\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)] 
   One independent solution has integrals. Trying a hypergeometric solution free of integrals... 
   -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
No hypergeometric solution was found. 
<- linear_1 successful`
 

✓ Solution by Maple

Time used: 0.109 (sec). Leaf size: 46

dsolve(x*diff(y(x),x$2)+(a*x^2+b*x+c)*diff(y(x),x)+(2*a*x+b)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = x^{-c +1} {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{\frac {1}{2} a \,x^{2}+b x}d x \right )+c_{2} \right ) \]

✓ Solution by Mathematica

Time used: 1.772 (sec). Leaf size: 63

DSolve[x*y''[x]+(a*x^2+b*x+c)*y'[x]+(2*a*x+b)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to x^{1-c} e^{-\frac {1}{2} x (a x+2 b)} \left (c_2 \int _1^xe^{\frac {1}{2} a K[1]^2+b K[1]} K[1]^{c-2}dK[1]+c_1\right ) \]