28.34 problem 94

Internal problem ID [10917]
Internal file name [OUTPUT/10174_Sunday_December_31_2023_11_03_24_AM_32701816/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: 94.
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)]`]]

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

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

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

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

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

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

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

Summary

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

Verification of solutions

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

28.34.2 Solving as type second_order_integrable_as_is (not using ABC version)

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

Which is now solved for \(y\).

28.34.3 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^{n}+b\\ r(x) &= a n \,x^{n -1}\\ s(x) &= 0 \end {align*}

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

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

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

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

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

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

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

Summary

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

Verification of solutions

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

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
   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: 0.094 (sec). Leaf size: 53

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

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

✓ Solution by Mathematica

Time used: 0.244 (sec). Leaf size: 121

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

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