Internal problem ID [7082]
Internal file name [OUTPUT/6068_Sunday_June_05_2022_04_17_23_PM_32546509/index.tex]
Book: Own collection of miscellaneous problems
Section: section 1.0
Problem number: 39.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {u^{\prime }+u^{2}=\frac {1}{x^{\frac {4}{5}}}} \]
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= -\frac {u^{2} x^{\frac {4}{5}}-1}{x^{\frac {4}{5}}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ u' = -u^{2}+\frac {1}{x^{\frac {4}{5}}} \] With Riccati ODE standard form \[ u' = f_0(x)+ f_1(x)u+f_2(x)u^{2} \] Shows that \(f_0(x)=\frac {1}{x^{\frac {4}{5}}}\), \(f_1(x)=0\) and \(f_2(x)=-1\). Let \begin {align*} u &= \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 {1}{x^{\frac {4}{5}}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )+\frac {u \left (x \right )}{x^{\frac {4}{5}}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {BesselK}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{2} +\operatorname {BesselI}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{1} \right ) \sqrt {x} \] The above shows that \[ u^{\prime }\left (x \right ) = \left (-\operatorname {BesselK}\left (\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{2} +\operatorname {BesselI}\left (-\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{1} \right ) x^{\frac {1}{10}} \] Using the above in (1) gives the solution \[ u = \frac {-\operatorname {BesselK}\left (\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{2} +\operatorname {BesselI}\left (-\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{1}}{x^{\frac {2}{5}} \left (\operatorname {BesselK}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{2} +\operatorname {BesselI}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ u = \frac {-\operatorname {BesselK}\left (\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right )+\operatorname {BesselI}\left (-\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{3}}{x^{\frac {2}{5}} \left (\operatorname {BesselK}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right )+\operatorname {BesselI}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} u &= \frac {-\operatorname {BesselK}\left (\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right )+\operatorname {BesselI}\left (-\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{3}}{x^{\frac {2}{5}} \left (\operatorname {BesselK}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right )+\operatorname {BesselI}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ u = \frac {-\operatorname {BesselK}\left (\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right )+\operatorname {BesselI}\left (-\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{3}}{x^{\frac {2}{5}} \left (\operatorname {BesselK}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right )+\operatorname {BesselI}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & u^{\prime }+u^{2}=\frac {1}{x^{\frac {4}{5}}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & u^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }=-u^{2}+\frac {1}{x^{\frac {4}{5}}} \end {array} \]
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 <- Riccati Special successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 46
dsolve(diff(u(x),x)+u(x)^2=x^(-4/5),u(x), singsol=all)
\[ u \left (x \right ) = \frac {\operatorname {BesselI}\left (-\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right ) c_{1} -\operatorname {BesselK}\left (\frac {1}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right )}{x^{\frac {2}{5}} \left (c_{1} \operatorname {BesselI}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right )+\operatorname {BesselK}\left (\frac {5}{6}, \frac {5 x^{\frac {3}{5}}}{3}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.293 (sec). Leaf size: 286
DSolve[u'[x]+u[x]^2==x^(-4/5),u[x],x,IncludeSingularSolutions -> True]
\begin{align*} u(x)\to \frac {(-1)^{5/6} x^{3/5} \operatorname {Gamma}\left (\frac {11}{6}\right ) \operatorname {BesselI}\left (-\frac {1}{6},\frac {5 x^{3/5}}{3}\right )+(-1)^{5/6} \operatorname {Gamma}\left (\frac {11}{6}\right ) \operatorname {BesselI}\left (\frac {5}{6},\frac {5 x^{3/5}}{3}\right )+(-1)^{5/6} x^{3/5} \operatorname {Gamma}\left (\frac {11}{6}\right ) \operatorname {BesselI}\left (\frac {11}{6},\frac {5 x^{3/5}}{3}\right )+c_1 x^{3/5} \operatorname {Gamma}\left (\frac {1}{6}\right ) \operatorname {BesselI}\left (-\frac {11}{6},\frac {5 x^{3/5}}{3}\right )+c_1 \operatorname {Gamma}\left (\frac {1}{6}\right ) \operatorname {BesselI}\left (-\frac {5}{6},\frac {5 x^{3/5}}{3}\right )+c_1 x^{3/5} \operatorname {Gamma}\left (\frac {1}{6}\right ) \operatorname {BesselI}\left (\frac {1}{6},\frac {5 x^{3/5}}{3}\right )}{2 x \left ((-1)^{5/6} \operatorname {Gamma}\left (\frac {11}{6}\right ) \operatorname {BesselI}\left (\frac {5}{6},\frac {5 x^{3/5}}{3}\right )+c_1 \operatorname {Gamma}\left (\frac {1}{6}\right ) \operatorname {BesselI}\left (-\frac {5}{6},\frac {5 x^{3/5}}{3}\right )\right )} \\ u(x)\to \frac {x^{3/5} \operatorname {BesselI}\left (-\frac {11}{6},\frac {5 x^{3/5}}{3}\right )+\operatorname {BesselI}\left (-\frac {5}{6},\frac {5 x^{3/5}}{3}\right )+x^{3/5} \operatorname {BesselI}\left (\frac {1}{6},\frac {5 x^{3/5}}{3}\right )}{2 x \operatorname {BesselI}\left (-\frac {5}{6},\frac {5 x^{3/5}}{3}\right )} \\ \end{align*}