30.15 problem 163

Internal problem ID [10986]
Internal file name [OUTPUT/10243_Sunday_December_31_2023_11_13_29_AM_55836174/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 2.1.2-5 Equation of form \((a x^2+b x+c) y''+f(x)y'+g(x)y=0\)
Problem number: 163.
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]]

\[ \boxed {\left (x^{2}+a \right ) y^{\prime \prime }+2 b x y^{\prime }+2 \left (b -1\right ) y=0} \]

30.15.1 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^{2}+a \right ) y^{\prime \prime }+2 b x y^{\prime }+\left (2 b -2\right ) y\right )d x &= 0 \\ \left (2 b x -2 x \right ) y+\left (x^{2}+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 {2 b x -2 x}{x^{2}+a}\\ q(x) &=\frac {c_{1}}{x^{2}+a} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {\left (2 b x -2 x \right ) y}{x^{2}+a} = \frac {c_{1}}{x^{2}+a} \end {align*}

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

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

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

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

Summary

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

Verification of solutions

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

30.15.2 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ \left (x^{2}+a \right ) y^{\prime \prime }+2 b x y^{\prime }+\left (2 b -2\right ) y = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (x^{2}+a \right ) y^{\prime \prime }+2 b x y^{\prime }+\left (2 b -2\right ) y\right )d x &= 0 \\ 2 b x y+y^{\prime } a -2 y x +y^{\prime } x^{2} = 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 {2 b x -2 x}{x^{2}+a}\\ q(x) &=\frac {c_{1}}{x^{2}+a} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {\left (2 b x -2 x \right ) y}{x^{2}+a} = \frac {c_{1}}{x^{2}+a} \end {align*}

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

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

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

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

Summary

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

Verification of solutions

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

30.15.3 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^{2}+a\\ q(x) &= 2 b x\\ r(x) &= 2 b -2\\ s(x) &= 0 \end {align*}

Hence \begin {align*} p''(x) &= 2\\ q'(x) &= 2 b \end {align*}

Therefore (1) becomes \begin {align*} 2- \left (2 b\right ) + \left (2 b -2\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 (2 b x -2 x \right ) y+\left (x^{2}+a \right ) y^{\prime }&=c_{1} \end {align*}

We now have a first order ode to solve which is \begin {align*} \left (2 b x -2 x \right ) y+\left (x^{2}+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 {2 b x -2 x}{x^{2}+a}\\ q(x) &=\frac {c_{1}}{x^{2}+a} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {\left (2 b x -2 x \right ) y}{x^{2}+a} = \frac {c_{1}}{x^{2}+a} \end {align*}

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

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

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

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

Summary

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

Verification of solutions

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

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 2F1 ODE 
   -> Trying to convert hypergeometric functions to elementary form... 
   <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals 
<- linear_1 successful`
 

✓ Solution by Maple

Time used: 0.047 (sec). Leaf size: 41

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

\[ y \left (x \right ) = c_{1} \left (\frac {x^{2}+a}{a}\right )^{-b +1}+c_{2} x \operatorname {hypergeom}\left (\left [1, b -\frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {x^{2}}{a}\right ) \]

✓ Solution by Mathematica

Time used: 0.426 (sec). Leaf size: 64

DSolve[(x^2+a)*y''[x]+2*b*x*y'[x]+2*(b-1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \left (a+x^2\right ) \left (\frac {c_2 x \left (\frac {a+x^2}{a}\right )^{-b} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},2-b,\frac {3}{2},-\frac {x^2}{a}\right )}{a^2}+c_1 \left (a+x^2\right )^{-b}\right ) \]