1.37 problem 38
Internal
problem
ID
[7729]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
38
Date
solved
:
Friday, October 25, 2024 at 12:04:29 PM
CAS
classification
:
[_rational, _Riccati]
Solve
\begin{align*} x y^{\prime }-y+y^{2}&=x^{{2}/{3}} \end{align*}
1.37.1 Solved as first order ode of type Riccati
Time used: 0.226 (sec)
In canonical form the ODE is
\begin{align*} y' &= F(x,y)\\ &= \frac {x^{{2}/{3}}-y^{2}+y}{x} \end{align*}
This is a Riccati ODE. Comparing the ODE to solve
\[ y' = -\frac {y^{2}}{x}+\frac {1}{x^{{1}/{3}}}+\frac {y}{x} \]
With Riccati ODE standard form
\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows that \(f_0(x)=\frac {1}{x^{{1}/{3}}}\) , \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=-\frac {1}{x}\) . Let
\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {u}{x}} \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' &=\frac {1}{x^{2}}\\ f_1 f_2 &=-\frac {1}{x^{2}}\\ f_2^2 f_0 &=\frac {1}{x^{{7}/{3}}} \end{align*}
Substituting the above terms back in equation (2) gives
\begin{align*} -\frac {u^{\prime \prime }\left (x \right )}{x}+\frac {u \left (x \right )}{x^{{7}/{3}}} = 0 \end{align*}
Writing the ode as
\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )-x^{{2}/{3}} 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 &= 3 i\\ n &= {\frac {3}{2}}\\ \gamma &= {\frac {1}{3}} \end{align*}
Substituting all the above into (4) gives the solution as
\begin{align*} u = -\frac {c_1 \,x^{{1}/{6}} \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{9 \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}+\frac {i c_2 \,x^{{1}/{6}} \sqrt {2}\, \sqrt {3}\, \left (-3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}+\cosh \left (3 x^{{1}/{3}}\right )\right )}{9 \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}} \end{align*}
Will add steps showing solving for IC soon.
Taking derivative gives
\[
u^{\prime }\left (x \right ) = -\frac {c_1 \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{54 x^{{5}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {c_1 \sqrt {2}\, \sqrt {3}\, \sinh \left (3 x^{{1}/{3}}\right )}{3 x^{{1}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}+\frac {i c_1 \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{54 \sqrt {x}\, \sqrt {\pi }\, \left (i x^{{1}/{3}}\right )^{{3}/{2}}}+\frac {i c_2 \sqrt {2}\, \sqrt {3}\, \left (-3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}+\cosh \left (3 x^{{1}/{3}}\right )\right )}{54 x^{{5}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {i c_2 \sqrt {2}\, \sqrt {3}\, \cosh \left (3 x^{{1}/{3}}\right )}{3 x^{{1}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}+\frac {c_2 \sqrt {2}\, \sqrt {3}\, \left (-3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}+\cosh \left (3 x^{{1}/{3}}\right )\right )}{54 \sqrt {x}\, \sqrt {\pi }\, \left (i x^{{1}/{3}}\right )^{{3}/{2}}}
\]
Doing change of constants, the solution becomes
\[
y = \frac {\left (-\frac {c_3 \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{54 x^{{5}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {c_3 \sqrt {2}\, \sqrt {3}\, \sinh \left (3 x^{{1}/{3}}\right )}{3 x^{{1}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}+\frac {i c_3 \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{54 \sqrt {x}\, \sqrt {\pi }\, \left (i x^{{1}/{3}}\right )^{{3}/{2}}}+\frac {i \sqrt {2}\, \sqrt {3}\, \left (-3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}+\cosh \left (3 x^{{1}/{3}}\right )\right )}{54 x^{{5}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {i \sqrt {2}\, \sqrt {3}\, \cosh \left (3 x^{{1}/{3}}\right )}{3 x^{{1}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}+\frac {\sqrt {2}\, \sqrt {3}\, \left (-3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}+\cosh \left (3 x^{{1}/{3}}\right )\right )}{54 \sqrt {x}\, \sqrt {\pi }\, \left (i x^{{1}/{3}}\right )^{{3}/{2}}}\right ) x}{-\frac {c_3 \,x^{{1}/{6}} \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{9 \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}+\frac {i x^{{1}/{6}} \sqrt {2}\, \sqrt {3}\, \left (-3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}+\cosh \left (3 x^{{1}/{3}}\right )\right )}{9 \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}}
\]
Figure 81: Slope field plot
\(x y^{\prime }-y+y^{2} = x^{{2}/{3}}\)
1.37.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }-y+y^{2}=x^{{2}/{3}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-y^{2}+x^{{2}/{3}}+y}{x} \end {array} \]
1.37.3 Maple trace
Methods for first order ODEs:
1.37.4 Maple dsolve solution
Solving time : 0.003
(sec)
Leaf size : 72
dsolve ( x * diff ( y ( x ), x )- y ( x )+ y ( x )^2 = x^(2/3),
y(x),singsol=all)
\[
y = \frac {x^{{1}/{3}} \left (c_1 \,{\mathrm e}^{6 x^{{1}/{3}}} \operatorname {abs}\left (1, 3 x^{{1}/{3}}-1\right )+c_1 \,{\mathrm e}^{6 x^{{1}/{3}}} {| 3 x^{{1}/{3}}-1|}-3 x^{{1}/{3}}\right )}{c_1 \,{\mathrm e}^{6 x^{{1}/{3}}} {| 3 x^{{1}/{3}}-1|}+3 x^{{1}/{3}}+1}
\]
1.37.5 Mathematica DSolve solution
Solving time : 0.193
(sec)
Leaf size : 131
DSolve [{ x * D [ y [ x ], x ]- y [ x ]+ y [ x ]^2== x ^(2/3),{}},
y[x],x,IncludeSingularSolutions-> True ]
\begin{align*}
y(x)\to \frac {3 x^{2/3} \left (c_1 \cosh \left (3 \sqrt [3]{x}\right )-i \sinh \left (3 \sqrt [3]{x}\right )\right )}{\left (-3 i \sqrt [3]{x}-c_1\right ) \cosh \left (3 \sqrt [3]{x}\right )+\left (3 c_1 \sqrt [3]{x}+i\right ) \sinh \left (3 \sqrt [3]{x}\right )} \\
y(x)\to \frac {3 x^{2/3} \cosh \left (3 \sqrt [3]{x}\right )}{3 \sqrt [3]{x} \sinh \left (3 \sqrt [3]{x}\right )-\cosh \left (3 \sqrt [3]{x}\right )} \\
\end{align*}