2.1.38 problem 39

Internal problem ID [8176]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 39
Date solved : Tuesday, November 12, 2024 at 11:07:34 PM
CAS classification : [_rational, _Riccati]

Solve

\begin{align*} u^{\prime }+u^{2}&=\frac {1}{x^{{4}/{5}}} \end{align*}

Solved as first order ode of type Riccati

Time used: 0.246 (sec)

In canonical form the ODE is

\begin{align*} u' &= F(x,u)\\ &= -\frac {u^{2} x^{{4}/{5}}-1}{x^{{4}/{5}}} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ u' = -u^{2}+\frac {1}{x^{{4}/{5}}} \]

With Riccati ODE standard form

\[ u' = f_0(x)+ f_1(x)u+f_2(x)u^{2} \]

Shows that \(f_0(x)=\frac {1}{x^{{4}/{5}}}\), \(f_1(x)=0\) 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^{{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^{{4}/{5}}} = 0 \end{align*}

Writing the ode as

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )-x^{{6}/{5}} u = 0\tag {1} \end{align*}

Bessel ode has the form

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )+\left (\frac {d u}{d x}\right ) x +\left (-n^{2}+x^{2}\right ) u = 0\tag {2} \end{align*}

The generalized form of Bessel ode is given by Bowman (1958) as the following

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )+\left (1-2 \alpha \right ) x \left (\frac {d u}{d x}\right )+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) u = 0\tag {3} \end{align*}

With the standard solution

\begin{align*} u&=x^{\alpha } \left (c_1 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_2 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}

Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives

\begin{align*} \alpha &= {\frac {1}{2}}\\ \beta &= \frac {5 i}{3}\\ n &= {\frac {5}{6}}\\ \gamma &= {\frac {3}{5}} \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} u = c_1 \sqrt {x}\, \operatorname {BesselJ}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )+c_2 \sqrt {x}\, \operatorname {BesselY}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right ) \end{align*}

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = \frac {c_1 \operatorname {BesselJ}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )}{2 \sqrt {x}}+i c_1 \,x^{{1}/{10}} \left (\operatorname {BesselJ}\left (-\frac {1}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )+\frac {i \operatorname {BesselJ}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )}{2 x^{{3}/{5}}}\right )+\frac {c_2 \operatorname {BesselY}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )}{2 \sqrt {x}}+i c_2 \,x^{{1}/{10}} \left (\operatorname {BesselY}\left (-\frac {1}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )+\frac {i \operatorname {BesselY}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )}{2 x^{{3}/{5}}}\right ) \]

Doing change of constants, the solution becomes

\[ u = \frac {\frac {c_3 \operatorname {BesselJ}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )}{2 \sqrt {x}}+i c_3 \,x^{{1}/{10}} \left (\operatorname {BesselJ}\left (-\frac {1}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )+\frac {i \operatorname {BesselJ}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )}{2 x^{{3}/{5}}}\right )+\frac {\operatorname {BesselY}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )}{2 \sqrt {x}}+i x^{{1}/{10}} \left (\operatorname {BesselY}\left (-\frac {1}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )+\frac {i \operatorname {BesselY}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )}{2 x^{{3}/{5}}}\right )}{c_3 \sqrt {x}\, \operatorname {BesselJ}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )+\sqrt {x}\, \operatorname {BesselY}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )} \]

Figure 2.86: Slope field plot
\(u^{\prime }+u^{2} = \frac {1}{x^{{4}/{5}}}\)

Summary of solutions found

\begin{align*} u &= \frac {i \left (\operatorname {BesselJ}\left (-\frac {1}{6}, \frac {5 i x^{{3}/{5}}}{3}\right ) c_3 +\operatorname {BesselY}\left (-\frac {1}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )\right )}{x^{{2}/{5}} \left (c_3 \operatorname {BesselJ}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )+\operatorname {BesselY}\left (\frac {5}{6}, \frac {5 i x^{{3}/{5}}}{3}\right )\right )} \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )+u \left (x \right )^{2}=\frac {1}{x^{{4}/{5}}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}u \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=-u \left (x \right )^{2}+\frac {1}{x^{{4}/{5}}} \end {array} \]

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`
 

Maple dsolve solution

Solving time : 0.014 (sec)
Leaf size : 46

dsolve(diff(u(x),x)+u(x)^2 = 1/x^(4/5), 
       u(x),singsol=all)
 

\[ u \left (x \right ) = \frac {\operatorname {BesselI}\left (-\frac {1}{6}, \frac {5 x^{{3}/{5}}}{3}\right ) c_{1} -\operatorname {BesselK}\left (\frac {1}{6}, \frac {5 x^{{3}/{5}}}{3}\right )}{x^{{2}/{5}} \left (c_{1} \operatorname {BesselI}\left (\frac {5}{6}, \frac {5 x^{{3}/{5}}}{3}\right )+\operatorname {BesselK}\left (\frac {5}{6}, \frac {5 x^{{3}/{5}}}{3}\right )\right )} \]

Mathematica DSolve solution

Solving time : 0.267 (sec)
Leaf size : 286

DSolve[{D[u[x],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*}