28.49 problem 109

Internal problem ID [10932]
Internal file name [OUTPUT/10189_Sunday_December_31_2023_11_03_49_AM_24274775/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: 109.
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, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

\[ \boxed {\left (x +\gamma \right ) y^{\prime \prime }+\left (a \,x^{n}+b \,x^{m}+c \right ) y^{\prime }+\left (a n \,x^{n -1}+b m \,x^{m -1}\right ) y=0} \]

28.49.1 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 +\gamma \right ) y^{\prime \prime }+\left (a \,x^{n}+b \,x^{m}+c \right ) y^{\prime }+\frac {\left (a n \,x^{n}+b \,x^{m} m \right ) y}{x}\right )d x &= 0 \\ \frac {\left (x \,x^{n} a +x^{m} b x +c x -x \right ) y}{x}+\left (x +\gamma \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 {a \,x^{n}+b \,x^{m}+c -1}{x +\gamma }\\ q(x) &=\frac {c_{1}}{x +\gamma } \end {align*}

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

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

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

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

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

Summary

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

Verification of solutions

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

28.49.2 Solving as type second_order_integrable_as_is (not using ABC version)

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

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

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

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

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

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

Summary

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

Verification of solutions

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

28.49.3 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 +\gamma \\ q(x) &= a \,x^{n}+b \,x^{m}+c\\ r(x) &= \frac {a n \,x^{n}+b \,x^{m} m}{x}\\ s(x) &= 0 \end {align*}

Hence \begin {align*} p''(x) &= 0\\ q'(x) &= \frac {a n \,x^{n}}{x}+\frac {b \,x^{m} m}{x} \end {align*}

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

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

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

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

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

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

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

Summary

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

Verification of solutions

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

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
<- linear_1 successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 63

dsolve((x+gamma)*diff(y(x),x$2)+(a*x^n+b*x^m+c)*diff(y(x),x)+(a*n*x^(n-1)+b*m*x^(m-1))*y(x)=0,y(x), singsol=all)
 

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

✗ Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

DSolve[(x+\[Gamma])*y''[x]+(a*x^n+b*x^m+c)*y'[x]+(a*n*x^(n-1)+b*m*x^(m-1))*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

Not solved