Internal problem ID [7129]
Internal file name [OUTPUT/6115_Sunday_June_05_2022_04_23_19_PM_88734211/index.tex]
Book: Own collection of miscellaneous problems
Section: section 1.0
Problem number: 84.
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]
\[ \boxed {y^{\prime }-y^{2}=x^{2}-1} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= x^{2}+y^{2}-1 \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{2}+y^{2}-1 \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x^{2}-1\), \(f_1(x)=0\) 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 &=x^{2}-1 \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\left (x^{2}-1\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {c_{1} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+c_{2} \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{\sqrt {x}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (\frac {3}{2}+\frac {i}{2}\right ) c_{1} \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )-2 \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right ) c_{2} +\left (c_{1} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+c_{2} \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )\right ) \left (i x^{2}-\frac {1}{2}-\frac {1}{2} i\right )}{x^{\frac {3}{2}}} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\frac {3}{2}+\frac {i}{2}\right ) c_{1} \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )-2 \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right ) c_{2} +\left (c_{1} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+c_{2} \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )\right ) \left (i x^{2}-\frac {1}{2}-\frac {1}{2} i\right )}{x \left (c_{1} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+c_{2} \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-3-i\right ) c_{3} \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+4 \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) c_{3} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{2 x \left (c_{3} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-3-i\right ) c_{3} \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+4 \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) c_{3} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{2 x \left (c_{3} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-3-i\right ) c_{3} \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+4 \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) c_{3} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{2 x \left (c_{3} \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}=x^{2}-1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=x^{2}+y^{2}-1 \end {array} \]
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 to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (-x^2+1)*y(x), y(x)` *** Sublevel 2 *** 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)] checking if the LODE is missing y -> 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 <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 107
dsolve(diff(y(x),x)=x^2+y(x)^2-1,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-3-i\right ) \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+4 \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right ) c_{1} +\left (-2 i x^{2}+i+1\right ) \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) c_{1} \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{2 x \left (c_{1} \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.236 (sec). Leaf size: 153
DSolve[y'[x]==x^2+y[x]^2-1,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {i \left (x \operatorname {ParabolicCylinderD}\left (-\frac {1}{2}-\frac {i}{2},(-1+i) x\right )+(1+i) \operatorname {ParabolicCylinderD}\left (\frac {1}{2}-\frac {i}{2},(-1+i) x\right )-c_1 x \operatorname {ParabolicCylinderD}\left (-\frac {1}{2}+\frac {i}{2},(1+i) x\right )+(1-i) c_1 \operatorname {ParabolicCylinderD}\left (\frac {1}{2}+\frac {i}{2},(1+i) x\right )\right )}{\operatorname {ParabolicCylinderD}\left (-\frac {1}{2}-\frac {i}{2},(-1+i) x\right )+c_1 \operatorname {ParabolicCylinderD}\left (-\frac {1}{2}+\frac {i}{2},(1+i) x\right )} \\ y(x)\to \frac {(1+i) \operatorname {ParabolicCylinderD}\left (\frac {1}{2}+\frac {i}{2},(1+i) x\right )}{\operatorname {ParabolicCylinderD}\left (-\frac {1}{2}+\frac {i}{2},(1+i) x\right )}-i x \\ \end{align*}