Internal problem ID [11073]
Internal file name [OUTPUT/10330_Wednesday_January_24_2024_10_07_18_PM_82271614/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 2.1.2-8. Other equations.
Problem number: 250.
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)]`]]
\[ \boxed {\left (a \,x^{n}+b x +c \right ) y^{\prime \prime }-a n \left (n -1\right ) x^{-2+n} y=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (a \,x^{n}+b x +c \right ) y^{\prime \prime }-a n \left (n -1\right ) x^{-2+n} y\right )d x &= 0 \\ -\frac {\left (a \,x^{n} n +b x \right ) y}{x}-\left (-a \,x^{n}-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 {a \,x^{n} n +b x}{\left (a \,x^{n}+b x +c \right ) x}\\ q(x) &=\frac {c_{1}}{a \,x^{n}+b x +c} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {\left (a \,x^{n} n +b x \right ) y}{\left (a \,x^{n}+b x +c \right ) x} = \frac {c_{1}}{a \,x^{n}+b x +c} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {a \,x^{n} n +b x}{\left (a \,x^{n}+b x +c \right ) x}d x} \\ &= \frac {1}{{\mathrm e}^{n \ln \left (x \right )} a +b x +c} \\ \end{align*} Which simplifies to \[ \mu = \frac {1}{a \,x^{n}+b x +c} \] 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^{n}+b x +c}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{a \,x^{n}+b x +c}\right ) &= \left (\frac {1}{a \,x^{n}+b x +c}\right ) \left (\frac {c_{1}}{a \,x^{n}+b x +c}\right )\\ \mathrm {d} \left (\frac {y}{a \,x^{n}+b x +c}\right ) &= \left (\frac {c_{1}}{\left (a \,x^{n}+b x +c \right )^{2}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{a \,x^{n}+b x +c} &= \int {\frac {c_{1}}{\left (a \,x^{n}+b x +c \right )^{2}}\,\mathrm {d} x}\\ \frac {y}{a \,x^{n}+b x +c} &= \frac {x c_{1}}{\left (x b n -b x +c n \right ) \left ({\mathrm e}^{n \ln \left (x \right )} a +b x +c \right )}+\left (\int \frac {n \left (x b n -b x +c n -c \right )}{\left (x b n -b x +c n \right )^{2} \left ({\mathrm e}^{n \ln \left (x \right )} a +b x +c \right )}d x \right ) c_{1} + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{a \,x^{n}+b x +c}\) results in \begin {align*} y &= \left (a \,x^{n}+b x +c \right ) \left (\frac {x c_{1}}{\left (x b n -b x +c n \right ) \left ({\mathrm e}^{n \ln \left (x \right )} a +b x +c \right )}+\left (\int \frac {n \left (x b n -b x +c n -c \right )}{\left (x b n -b x +c n \right )^{2} \left ({\mathrm e}^{n \ln \left (x \right )} a +b x +c \right )}d x \right ) c_{1} \right )+c_{2} \left (a \,x^{n}+b x +c \right ) \end {align*}
which simplifies to \begin {align*} y &= \left (a \,x^{n}+b x +c \right ) \left (c_{1} \left (\frac {x}{\left (a \,x^{n}+b x +c \right ) \left (x b n -b x +c n \right )}+n \left (\int \frac {\left (n -1\right ) \left (b x +c \right )}{\left (x \left (n -1\right ) b +c n \right )^{2} \left (a \,x^{n}+b x +c \right )}d x \right )\right )+c_{2} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (a \,x^{n}+b x +c \right ) \left (c_{1} \left (\frac {x}{\left (a \,x^{n}+b x +c \right ) \left (x b n -b x +c n \right )}+n \left (\int \frac {\left (n -1\right ) \left (b x +c \right )}{\left (x \left (n -1\right ) b +c n \right )^{2} \left (a \,x^{n}+b x +c \right )}d x \right )\right )+c_{2} \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (a \,x^{n}+b x +c \right ) \left (c_{1} \left (\frac {x}{\left (a \,x^{n}+b x +c \right ) \left (x b n -b x +c n \right )}+n \left (\int \frac {\left (n -1\right ) \left (b x +c \right )}{\left (x \left (n -1\right ) b +c n \right )^{2} \left (a \,x^{n}+b x +c \right )}d x \right )\right )+c_{2} \right ) \] Verified OK.
Writing the ode as \[ \left (a \,x^{n}+b x +c \right ) y^{\prime \prime }-a n \left (n -1\right ) x^{-2+n} y = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (a \,x^{n}+b x +c \right ) y^{\prime \prime }-a n \left (n -1\right ) x^{-2+n} y\right )d x &= 0 \\ \frac {-12 b y-6 x y x^{-2+n} a \,n^{3}-18 x y x^{-2+n} a \,n^{2}+6 x^{2} x^{-2+n} y^{\prime } a \,n^{2}+18 x^{2} x^{-2+n} y^{\prime } a n +6 x \,x^{n} y^{\prime \prime } a n -6 x^{3} x^{-2+n} y^{\prime \prime } a n -12 a n \,x^{-2+n} y x +6 y^{\prime } c \,n^{2}+18 y^{\prime } c n +12 x y^{\prime } b -6 y b \,n^{2}-18 y b n +6 x y^{\prime } b \,n^{2}+18 x y^{\prime } b n +12 y^{\prime } c -6 a y^{\prime \prime \prime } x^{n} x^{2}+6 a y^{\prime \prime \prime } x^{-2+n} x^{4}+12 a y^{\prime \prime } x^{n} x -12 a y^{\prime \prime } x^{-2+n} x^{3}+12 a y^{\prime } x^{-2+n} x^{2}}{6 \left (1+n \right ) \left (2+n \right )} = c_{1} \end {align*}
Which is now solved for \(y\).
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^{n}+b x +c\\ q(x) &= 0\\ r(x) &= -a n \left (n -1\right ) x^{-2+n}\\ s(x) &= 0 \end {align*}
Hence \begin {align*} p''(x) &= \frac {a \,x^{n} n^{2}}{x^{2}}-\frac {a n \,x^{n}}{x^{2}}\\ q'(x) &= 0 \end {align*}
Therefore (1) becomes \begin {align*} \frac {a \,x^{n} n^{2}}{x^{2}}-\frac {a n \,x^{n}}{x^{2}}- \left (0\right ) + \left (-a n \left (n -1\right ) x^{-2+n}\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 x +c \right ) y^{\prime }+\left (-\frac {a \,x^{n} n}{x}-b \right ) y&=c_{1} \end {align*}
We now have a first order ode to solve which is \begin {align*} \left (a \,x^{n}+b x +c \right ) y^{\prime }+\left (-\frac {a \,x^{n} n}{x}-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) &=-\frac {a \,x^{n} n +b x}{\left (a \,x^{n}+b x +c \right ) x}\\ q(x) &=\frac {c_{1}}{a \,x^{n}+b x +c} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {\left (a \,x^{n} n +b x \right ) y}{\left (a \,x^{n}+b x +c \right ) x} = \frac {c_{1}}{a \,x^{n}+b x +c} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {a \,x^{n} n +b x}{\left (a \,x^{n}+b x +c \right ) x}d x} \\ &= \frac {1}{{\mathrm e}^{n \ln \left (x \right )} a +b x +c} \\ \end{align*} Which simplifies to \[ \mu = \frac {1}{a \,x^{n}+b x +c} \] 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^{n}+b x +c}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{a \,x^{n}+b x +c}\right ) &= \left (\frac {1}{a \,x^{n}+b x +c}\right ) \left (\frac {c_{1}}{a \,x^{n}+b x +c}\right )\\ \mathrm {d} \left (\frac {y}{a \,x^{n}+b x +c}\right ) &= \left (\frac {c_{1}}{\left (a \,x^{n}+b x +c \right )^{2}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{a \,x^{n}+b x +c} &= \int {\frac {c_{1}}{\left (a \,x^{n}+b x +c \right )^{2}}\,\mathrm {d} x}\\ \frac {y}{a \,x^{n}+b x +c} &= \frac {x c_{1}}{\left (x b n -b x +c n \right ) \left ({\mathrm e}^{n \ln \left (x \right )} a +b x +c \right )}+\left (\int \frac {n \left (x b n -b x +c n -c \right )}{\left (x b n -b x +c n \right )^{2} \left ({\mathrm e}^{n \ln \left (x \right )} a +b x +c \right )}d x \right ) c_{1} + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{a \,x^{n}+b x +c}\) results in \begin {align*} y &= \left (a \,x^{n}+b x +c \right ) \left (\frac {x c_{1}}{\left (x b n -b x +c n \right ) \left ({\mathrm e}^{n \ln \left (x \right )} a +b x +c \right )}+\left (\int \frac {n \left (x b n -b x +c n -c \right )}{\left (x b n -b x +c n \right )^{2} \left ({\mathrm e}^{n \ln \left (x \right )} a +b x +c \right )}d x \right ) c_{1} \right )+c_{2} \left (a \,x^{n}+b x +c \right ) \end {align*}
which simplifies to \begin {align*} y &= \left (a \,x^{n}+b x +c \right ) \left (c_{1} \left (\frac {x}{\left (a \,x^{n}+b x +c \right ) \left (x b n -b x +c n \right )}+n \left (\int \frac {\left (n -1\right ) \left (b x +c \right )}{\left (x \left (n -1\right ) b +c n \right )^{2} \left (a \,x^{n}+b x +c \right )}d x \right )\right )+c_{2} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (a \,x^{n}+b x +c \right ) \left (c_{1} \left (\frac {x}{\left (a \,x^{n}+b x +c \right ) \left (x b n -b x +c n \right )}+n \left (\int \frac {\left (n -1\right ) \left (b x +c \right )}{\left (x \left (n -1\right ) b +c n \right )^{2} \left (a \,x^{n}+b x +c \right )}d x \right )\right )+c_{2} \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (a \,x^{n}+b x +c \right ) \left (c_{1} \left (\frac {x}{\left (a \,x^{n}+b x +c \right ) \left (x b n -b x +c n \right )}+n \left (\int \frac {\left (n -1\right ) \left (b x +c \right )}{\left (x \left (n -1\right ) b +c n \right )^{2} \left (a \,x^{n}+b x +c \right )}d x \right )\right )+c_{2} \right ) \] 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: 33
dsolve((a*x^n+b*x+c)*diff(y(x),x$2)=a*n*(n-1)*x^(n-2)*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (\left (\int \frac {1}{\left (a \,x^{n}+b x +c \right )^{2}}d x \right ) c_{1} +c_{2} \right ) \left (a \,x^{n}+b x +c \right ) \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[(a*x^n+b*x+c)*y''[x]==a*n*(n-1)*x^(n-2)*y[x],y[x],x,IncludeSingularSolutions -> True]
Not solved