30.27 problem 175

Internal problem ID [10998]
Internal file name [OUTPUT/10255_Sunday_December_31_2023_11_24_17_AM_39314145/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 2.1.2-5 Equation of form \((a x^2+b x+c) y''+f(x)y'+g(x)y=0\)
Problem number: 175.
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]]

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

30.27.1 Solving as second order integrable as is ode

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

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

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {2 a x -d x +b -k}{a \,x^{2}+b x +c}d x} \\ &= {\mathrm e}^{-\frac {\left (2 a -d \right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}-\frac {2 \left (b -k -\frac {\left (2 a -d \right ) b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}} \\ \end{align*} Which simplifies to \[ \mu = {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{a \,x^{2}+b x +c}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y\right ) &= \left ({\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}}\right ) \left (\frac {c_{1}}{a \,x^{2}+b x +c}\right )\\ \mathrm {d} \left ({\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y\right ) &= \left (c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y &= \int {c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}\,\mathrm {d} x}\\ {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y &= \int c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}}\) results in \begin {align*} y &= \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (\int c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \end {align*}

which simplifies to \begin {align*} y &= \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (c_{1} \left (\int \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (c_{1} \left (\int \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \right ) \\ \end{align*}

Verification of solutions

\[ y = \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (c_{1} \left (\int \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \right ) \] Verified OK.

30.27.2 Solving as type second_order_integrable_as_is (not using ABC version)

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

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

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {2 a x -d x +b -k}{a \,x^{2}+b x +c}d x} \\ &= {\mathrm e}^{-\frac {\left (2 a -d \right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}-\frac {2 \left (b -k -\frac {\left (2 a -d \right ) b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}} \\ \end{align*} Which simplifies to \[ \mu = {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{a \,x^{2}+b x +c}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y\right ) &= \left ({\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}}\right ) \left (\frac {c_{1}}{a \,x^{2}+b x +c}\right )\\ \mathrm {d} \left ({\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y\right ) &= \left (c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y &= \int {c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}\,\mathrm {d} x}\\ {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y &= \int c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}}\) results in \begin {align*} y &= \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (\int c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \end {align*}

which simplifies to \begin {align*} y &= \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (c_{1} \left (\int \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (c_{1} \left (\int \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \right ) \\ \end{align*}

Verification of solutions

\[ y = \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (c_{1} \left (\int \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \right ) \] Verified OK.

30.27.3 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) &= a \,x^{2}+b x +c\\ q(x) &= d x +k\\ r(x) &= -2 a +d\\ s(x) &= 0 \end {align*}

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

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

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

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

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {2 a x -d x +b -k}{a \,x^{2}+b x +c}d x} \\ &= {\mathrm e}^{-\frac {\left (2 a -d \right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}-\frac {2 \left (b -k -\frac {\left (2 a -d \right ) b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}} \\ \end{align*} Which simplifies to \[ \mu = {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{a \,x^{2}+b x +c}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y\right ) &= \left ({\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}}\right ) \left (\frac {c_{1}}{a \,x^{2}+b x +c}\right )\\ \mathrm {d} \left ({\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y\right ) &= \left (c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y &= \int {c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}\,\mathrm {d} x}\\ {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}} y &= \int c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}} \left (a \,x^{2}+b x +c \right )^{\frac {-2 a +d}{2 a}}\) results in \begin {align*} y &= \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (\int c_{1} \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \end {align*}

which simplifies to \begin {align*} y &= \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (c_{1} \left (\int \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (c_{1} \left (\int \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \right ) \\ \end{align*}

Verification of solutions

\[ y = \left (a \,x^{2}+b x +c \right )^{\frac {2 a -d}{2 a}} {\mathrm e}^{-\frac {2 \left (a k -\frac {b d}{2}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}} \left (c_{1} \left (\int \left (a \,x^{2}+b x +c \right )^{\frac {-4 a +d}{2 a}} {\mathrm e}^{\frac {\left (2 a k -b d \right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{a \sqrt {4 a c -b^{2}}}}d x \right )+c_{2} \right ) \] Verified OK.

30.27.4 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \,x^{2}+b x +c \right ) y^{\prime \prime }+\left (d x +k \right ) y^{\prime }+\left (-2 a +d \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (2 a -d \right ) y}{a \,x^{2}+b x +c}-\frac {\left (d x +k \right ) y^{\prime }}{a \,x^{2}+b x +c} \\ \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} \\ {} & {} & y^{\prime \prime }+\frac {\left (d x +k \right ) y^{\prime }}{a \,x^{2}+b x +c}-\frac {\left (2 a -d \right ) y}{a \,x^{2}+b x +c}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {d x +k}{a \,x^{2}+b x +c}, P_{3}\left (x \right )=-\frac {2 a -d}{a \,x^{2}+b x +c}\right ] \\ {} & \circ & \left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left (\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}}}=0 \\ {} & \circ & {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left ({\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}}}=0 \\ {} & \circ & x =\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (a \,x^{2}+b x +c \right ) y^{\prime \prime }+\left (d x +k \right ) y^{\prime }+\left (-2 a +d \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (a \,u^{2}+u \sqrt {-4 a c +b^{2}}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (d u -\frac {d b}{2 a}+\frac {d \sqrt {-4 a c +b^{2}}}{2 a}+k \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-2 a +d \right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & \frac {a_{0} r \left (2 \sqrt {-4 a c +b^{2}}\, a r -2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, d +2 a k -b d \right ) u^{-1+r}}{2 a}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (\frac {a_{k +1} \left (k +r +1\right ) \left (2 \sqrt {-4 a c +b^{2}}\, a \left (k +1\right )+2 \sqrt {-4 a c +b^{2}}\, a r -2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, d +2 a k -b d \right )}{2 a}+a_{k} \left (k +r +1\right ) \left (a k +a r -2 a +d \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \frac {r \left (2 \sqrt {-4 a c +b^{2}}\, a r -2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, d +2 a k -b d \right )}{2 a}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}\, a}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {\left (k +r +1\right ) \left (a_{k +1} \left (\left (k +r \right ) a +\frac {d}{2}\right ) \sqrt {-4 a c +b^{2}}+a_{k} \left (k +r -2\right ) a^{2}+\left (a_{k} d +k a_{k +1}\right ) a -\frac {b d a_{k +1}}{2}\right )}{a}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a k +a r -2 a +d \right )}{2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a r +\sqrt {-4 a c +b^{2}}\, d +2 a k -b d} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a k -2 a +d \right )}{2 \sqrt {-4 a c +b^{2}}\, a k +\sqrt {-4 a c +b^{2}}\, d +2 a k -b d} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=-\frac {2 a a_{k} \left (a k -2 a +d \right )}{2 \sqrt {-4 a c +b^{2}}\, a k +\sqrt {-4 a c +b^{2}}\, d +2 a k -b d}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{k}, a_{k +1}=-\frac {2 a a_{k} \left (a k -2 a +d \right )}{2 \sqrt {-4 a c +b^{2}}\, a k +\sqrt {-4 a c +b^{2}}\, d +2 a k -b d}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}\, a} \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a k +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}}-2 a +d \right )}{2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}\, a} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}\, a}}, a_{k +1}=-\frac {2 a a_{k} \left (a k +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}}-2 a +d \right )}{2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{k +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}\, a}}, a_{k +1}=-\frac {2 a a_{k} \left (a k +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}}-2 a +d \right )}{2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {m =0}{\sum }}e_{m} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{m}\right )+\left (\moverset {\infty }{\munderset {m =0}{\sum }}f_{m} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{m +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}\, a}}\right ), e_{1+m}=-\frac {2 a e_{m} \left (a m -2 a +d \right )}{2 \sqrt {-4 a c +b^{2}}\, a m +\sqrt {-4 a c +b^{2}}\, d +2 a k -b d}, f_{1+m}=-\frac {2 a f_{m} \left (a m +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, d -2 a k +b d}{2 \sqrt {-4 a c +b^{2}}}-2 a +d \right )}{2 \sqrt {-4 a c +b^{2}}\, a m +2 \sqrt {-4 a c +b^{2}}\, a}\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 
   <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
   -> Trying to convert hypergeometric functions to elementary form... 
   <- elementary form is not straightforward to achieve - returning hypergeometric solution free of uncomputed integrals 
<- linear_1 successful`
 

✓ Solution by Maple

Time used: 0.047 (sec). Leaf size: 1412

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

\[ \text {Expression too large to display} \]

✓ Solution by Mathematica

Time used: 15.225 (sec). Leaf size: 164

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

\[ y(x)\to (x (a x+b)+c)^{1-\frac {d}{2 a}} \exp \left (\frac {(b d-2 a k) \arctan \left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{a \sqrt {4 a c-b^2}}\right ) \left (c_2 \int _1^x\exp \left (\frac {(d-4 a) \log (c+K[1] (b+a K[1]))-\frac {2 (b d-2 a k) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}}{2 a}\right )dK[1]+c_1\right ) \]