Internal problem ID [10447]
Internal file name [OUTPUT/9395_Monday_June_06_2022_02_21_29_PM_90333248/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. subsection 1.2.3-2. Equations with power and
exponential functions
Problem number: 40.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {x^{4} \left (y^{\prime }-y^{2}\right )=a +b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{4} y^{2}+b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}}+a}{x^{4}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+\frac {b \,{\mathrm e}^{\frac {k}{x}}}{x^{4}}+\frac {c \,{\mathrm e}^{\frac {2 k}{x}}}{x^{4}}+\frac {a}{x^{4}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a +b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}}}{x^{4}}\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {a +b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}}}{x^{4}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\frac {\left (a +b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}}\right ) u \left (x \right )}{x^{4}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x \,{\mathrm e}^{-\frac {k}{2 x}} \left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{1} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {{\mathrm e}^{-\frac {k}{2 x}} \left (\left (\left (i \sqrt {a}+\frac {k}{2}\right ) \sqrt {c}-\frac {i b}{2}\right ) c_{1} \operatorname {WhittakerM}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\operatorname {WhittakerW}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{2} k \sqrt {c}+\left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{1} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{2} \right ) \left (i c \,{\mathrm e}^{\frac {k}{x}}+\left (-\frac {k}{2}-x \right ) \sqrt {c}+\frac {i b}{2}\right )\right )}{\sqrt {c}\, x} \] Using the above in (1) gives the solution \[ y = \frac {\left (\left (i \sqrt {a}+\frac {k}{2}\right ) \sqrt {c}-\frac {i b}{2}\right ) c_{1} \operatorname {WhittakerM}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\operatorname {WhittakerW}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{2} k \sqrt {c}+\left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{1} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{2} \right ) \left (i c \,{\mathrm e}^{\frac {k}{x}}+\left (-\frac {k}{2}-x \right ) \sqrt {c}+\frac {i b}{2}\right )}{\sqrt {c}\, x^{2} \left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{1} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\frac {c_{3} \left (\left (-2 i \sqrt {a}-k \right ) \sqrt {c}+i b \right ) \operatorname {WhittakerM}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2}-\operatorname {WhittakerW}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) k \sqrt {c}+\left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right ) \left (i c \,{\mathrm e}^{\frac {k}{x}}+\left (-\frac {k}{2}-x \right ) \sqrt {c}+\frac {i b}{2}\right )}{\sqrt {c}\, x^{2} \left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\frac {c_{3} \left (\left (-2 i \sqrt {a}-k \right ) \sqrt {c}+i b \right ) \operatorname {WhittakerM}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2}-\operatorname {WhittakerW}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) k \sqrt {c}+\left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right ) \left (i c \,{\mathrm e}^{\frac {k}{x}}+\left (-\frac {k}{2}-x \right ) \sqrt {c}+\frac {i b}{2}\right )}{\sqrt {c}\, x^{2} \left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\frac {c_{3} \left (\left (-2 i \sqrt {a}-k \right ) \sqrt {c}+i b \right ) \operatorname {WhittakerM}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2}-\operatorname {WhittakerW}\left (-\frac {-2 k \sqrt {c}+i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) k \sqrt {c}+\left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right ) \left (i c \,{\mathrm e}^{\frac {k}{x}}+\left (-\frac {k}{2}-x \right ) \sqrt {c}+\frac {i b}{2}\right )}{\sqrt {c}\, x^{2} \left (\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{4} \left (y^{\prime }-y^{2}\right )=a +b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{4} y^{2}+b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}}+a}{x^{4}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati Special trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(a+b*exp(k/x)+c*exp(2*k/x))*y(x)/x^4, y(x)` *** Sublevel 2 *** Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying an equivalence, under non-integer power transformations, to LODEs admitting Liouvillian solutions. -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker successful <- special function solution successful Change of variables used: [x = 1/ln(t)] Linear ODE actually solved: (ln(t)*a+ln(t)*b*t^k+ln(t)*c*(t^k)^2)*u(t)+(t*ln(t)+2*t)*diff(u(t),t)+ln(t)*t^2*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 302
dsolve(x^4*(diff(y(x),x)-y(x)^2)=a+b*exp(k/x)+c*exp(2*k/x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\left (i \sqrt {a}+\frac {k}{2}\right ) \sqrt {c}-\frac {i b}{2}\right ) \operatorname {WhittakerM}\left (-\frac {i b -2 k \sqrt {c}}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-c_{1} k \operatorname {WhittakerW}\left (-\frac {i b -2 k \sqrt {c}}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) \sqrt {c}+\left (\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right ) \left (i {\mathrm e}^{\frac {k}{x}} c +\left (-\frac {k}{2}-x \right ) \sqrt {c}+\frac {i b}{2}\right )}{\sqrt {c}\, x^{2} \left (\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )} \]
✓ Solution by Mathematica
Time used: 4.039 (sec). Leaf size: 940
DSolve[x^4*(y'[x]-y[x]^2)==a+b*Exp[k/x]+c*Exp[2*k/x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {e^{k/x} \log \left (e^{k/x}\right ) \left (c_1 \left (b+\sqrt {c} \left (2 \sqrt {a}-i k\right )\right ) \operatorname {HypergeometricU}\left (\frac {\frac {i b}{\sqrt {c}}+3 k+2 i \sqrt {a}}{2 k},\frac {2 i \sqrt {a}}{k}+2,\frac {2 i \sqrt {c} e^{k/x}}{k}\right )-2 i \sqrt {c} k L_{-\frac {\frac {i b}{\sqrt {c}}+3 k+2 i \sqrt {a}}{2 k}}^{\frac {2 i \sqrt {a}}{k}+1}\left (\frac {2 i \sqrt {c} e^{k/x}}{k}\right )\right )-c_1 k \left (k-i \log \left (e^{k/x}\right ) \left (\sqrt {a}-\sqrt {c} e^{k/x}\right )\right ) \operatorname {HypergeometricU}\left (\frac {\frac {i b}{\sqrt {c}}+k+2 i \sqrt {a}}{2 k},\frac {2 i \sqrt {a}}{k}+1,\frac {2 i \sqrt {c} e^{k/x}}{k}\right )-k \left (k-i \log \left (e^{k/x}\right ) \left (\sqrt {a}-\sqrt {c} e^{k/x}\right )\right ) L_{-\frac {\frac {i b}{\sqrt {c}}+k+2 i \sqrt {a}}{2 k}}^{\frac {2 i \sqrt {a}}{k}}\left (\frac {2 i \sqrt {c} e^{k/x}}{k}\right )}{k x^2 \log \left (e^{k/x}\right ) \left (c_1 \operatorname {HypergeometricU}\left (\frac {\frac {i b}{\sqrt {c}}+k+2 i \sqrt {a}}{2 k},\frac {2 i \sqrt {a}}{k}+1,\frac {2 i \sqrt {c} e^{k/x}}{k}\right )+L_{-\frac {\frac {i b}{\sqrt {c}}+k+2 i \sqrt {a}}{2 k}}^{\frac {2 i \sqrt {a}}{k}}\left (\frac {2 i \sqrt {c} e^{k/x}}{k}\right )\right )} \\ y(x)\to \frac {\frac {e^{k/x} \left (b+\sqrt {c} \left (2 \sqrt {a}-i k\right )\right ) \operatorname {HypergeometricU}\left (\frac {\frac {i b}{\sqrt {c}}+3 k+2 i \sqrt {a}}{2 k},\frac {2 i \sqrt {a}}{k}+2,\frac {2 i \sqrt {c} e^{k/x}}{k}\right )}{k \operatorname {HypergeometricU}\left (\frac {\frac {i b}{\sqrt {c}}+k+2 i \sqrt {a}}{2 k},\frac {2 i \sqrt {a}}{k}+1,\frac {2 i \sqrt {c} e^{k/x}}{k}\right )}+i \left (\sqrt {a}-\sqrt {c} e^{k/x}\right )-\frac {k}{\log \left (e^{k/x}\right )}}{x^2} \\ y(x)\to \frac {\frac {e^{k/x} \left (b+\sqrt {c} \left (2 \sqrt {a}-i k\right )\right ) \operatorname {HypergeometricU}\left (\frac {\frac {i b}{\sqrt {c}}+3 k+2 i \sqrt {a}}{2 k},\frac {2 i \sqrt {a}}{k}+2,\frac {2 i \sqrt {c} e^{k/x}}{k}\right )}{k \operatorname {HypergeometricU}\left (\frac {\frac {i b}{\sqrt {c}}+k+2 i \sqrt {a}}{2 k},\frac {2 i \sqrt {a}}{k}+1,\frac {2 i \sqrt {c} e^{k/x}}{k}\right )}+i \left (\sqrt {a}-\sqrt {c} e^{k/x}\right )-\frac {k}{\log \left (e^{k/x}\right )}}{x^2} \\ \end{align*}