28.45 problem 105

Internal problem ID [10928]
Internal file name [OUTPUT/10185_Sunday_December_31_2023_11_03_43_AM_1225267/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: 105.
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]]

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

28.45.1 Solving as second order integrable as is ode

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

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

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

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

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

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

Summary

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

Verification of solutions

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

28.45.2 Solving as type second_order_integrable_as_is (not using ABC version)

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

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

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

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

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

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

Summary

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

Verification of solutions

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

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

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

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

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

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

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

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

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

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

Summary

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

Verification of solutions

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

28.45.4 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x +a \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (b x +c \right ) y^{\prime }+b 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 {b y}{x +a}-\frac {\left (b x +c \right ) y^{\prime }}{x +a} \\ \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 (b x +c \right ) y^{\prime }}{x +a}+\frac {b y}{x +a}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-a \hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {b x +c}{x +a}, P_{3}\left (x \right )=\frac {b}{x +a}\right ] \\ {} & \circ & \left (x +a \right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-a \\ {} & {} & \left (\left (x +a \right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-a}}}=-b a +c \\ {} & \circ & \left (x +a \right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-a \\ {} & {} & \left (\left (x +a \right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-a}}}=0 \\ {} & \circ & x =-a \textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-a \hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-a \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x +a \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (b x +c \right ) y^{\prime }+b y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -a \hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & u \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-b a +b u +c \right ) \left (\frac {d}{d u}y \left (u \right )\right )+b 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 \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d}{d u}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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (b a -c -r +1\right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (b a -c -k -r \right )+b a_{k} \left (k +1+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (b a -c -r +1\right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, b a -c +1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -\left (a_{k +1} \left (b a -c -k -r \right )-b a_{k}\right ) \left (k +1+r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {b a_{k}}{b a -c -k -r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {b a_{k}}{b a -c -k} \\ \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 {b a_{k}}{b a -c -k}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +a \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +a \right )^{k}, a_{k +1}=\frac {b a_{k}}{b a -c -k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =b a -c +1 \\ {} & {} & a_{k +1}=\frac {b a_{k}}{-k -1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =b a -c +1 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{b a -c +k +1}, a_{k +1}=\frac {b a_{k}}{-k -1}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +a \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +a \right )^{b a -c +k +1}, a_{k +1}=\frac {b a_{k}}{-k -1}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} \left (x +a \right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} \left (x +a \right )^{b a -c +k +1}\right ), d_{k +1}=\frac {b d_{k}}{b a -c -k}, e_{k +1}=\frac {b e_{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)] 
   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 1F1 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: 0.031 (sec). Leaf size: 78

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

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

✓ Solution by Mathematica

Time used: 0.559 (sec). Leaf size: 90

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

\[ y(x)\to e^{-b (a+x)} (a+x)^{1-c} (-b (a+x))^{-c} \left (c_1 e^{a b} (a+x)^{a b} (-b (a+x))^c+b c_2 (-b (a+x))^{a b} (a+x)^c \Gamma (-a b+c-1,-b (a+x))\right ) \]