27.46 problem 56

Internal problem ID [10879]
Internal file name [OUTPUT/10136_Sunday_December_24_2023_05_14_06_PM_36119654/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 2.1.2-2 Equation of form \(y''+f(x)y'+g(x)y=0\)
Problem number: 56.
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)]`]]

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

27.46.1 Solving as second order integrable as is ode

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

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

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

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

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

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

Summary

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

Verification of solutions

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

27.46.2 Solving as type second_order_integrable_as_is (not using ABC version)

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

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

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

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

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

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

Summary

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

Verification of solutions

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

27.46.3 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) &= 1\\ q(x) &= a \,x^{n}+x^{m} b\\ r(x) &= \frac {x^{m} b m +x^{n} a n}{x}\\ s(x) &= 0 \end {align*}

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

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

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

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

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

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

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

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

Summary

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

Verification of solutions

\[ y = {\mathrm e}^{-\frac {\left (a \,x^{n} \left (1+m \right )+x^{m} b \left (n +1\right )\right ) x}{\left (n +1\right ) \left (1+m \right )}} \left (c_{1} \left (\int {\mathrm e}^{\frac {\left (a \,x^{n} \left (1+m \right )+x^{m} b \left (n +1\right )\right ) x}{\left (n +1\right ) \left (1+m \right )}}d x \right )+c_{2} \right ) \] 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 
No hypergeometric solution was found. 
<- linear_1 successful`
 

✓ Solution by Maple

Time used: 0.063 (sec). Leaf size: 72

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

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

✓ Solution by Mathematica

Time used: 0.139 (sec). Leaf size: 74

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

\[ y(x)\to e^{x \left (-\frac {a x^n}{n+1}-\frac {b x^m}{m+1}\right )} \left (\int _1^x\exp \left (K[1] \left (\frac {b K[1]^m}{m+1}+\frac {a K[1]^n}{n+1}\right )\right ) c_1dK[1]+c_2\right ) \]