Internal problem ID [13349]
Internal file name [OUTPUT/12521_Wednesday_February_14_2024_11_54_38_PM_22238963/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 5. LINEAR FIRST ORDER EQUATIONS. Additional exercises. page
103
Problem number: 5.1 (d).
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]
\[ \boxed {y^{\prime }-\left (y x +3 y\right )^{2}=1} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2} x^{2}+6 y^{2} x +9 y^{2}+1 \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2} x^{2}+6 y^{2} x +9 y^{2}+1 \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=1\), \(f_1(x)=0\) and \(f_2(x)=x^{2}+6 x +9\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\left (x^{2}+6 x +9\right ) 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' &=2 x +6\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\left (x^{2}+6 x +9\right )^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \left (x^{2}+6 x +9\right ) u^{\prime \prime }\left (x \right )-\left (2 x +6\right ) u^{\prime }\left (x \right )+\left (x^{2}+6 x +9\right )^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {BesselJ}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{1} +\operatorname {BesselY}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{2} \right ) \left (3+x \right )^{\frac {3}{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = \left (3+x \right )^{\frac {5}{2}} \left (\operatorname {BesselJ}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{2} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {\left (3+x \right ) \left (\operatorname {BesselJ}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{2} \right )}{\left (x^{2}+6 x +9\right ) \left (\operatorname {BesselJ}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{1} +\operatorname {BesselY}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\operatorname {BesselJ}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{3} -\operatorname {BesselY}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right )}{\left (\operatorname {BesselJ}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{3} +\operatorname {BesselY}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right )\right ) \left (3+x \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\operatorname {BesselJ}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{3} -\operatorname {BesselY}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right )}{\left (\operatorname {BesselJ}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{3} +\operatorname {BesselY}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right )\right ) \left (3+x \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\operatorname {BesselJ}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{3} -\operatorname {BesselY}\left (-\frac {1}{4}, \frac {\left (3+x \right )^{2}}{2}\right )}{\left (\operatorname {BesselJ}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right ) c_{3} +\operatorname {BesselY}\left (\frac {3}{4}, \frac {\left (3+x \right )^{2}}{2}\right )\right ) \left (3+x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\left (y x +3 y\right )^{2}=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=1+\left (y x +3 y\right )^{2} \end {array} \]
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 sub-methods: trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = 2*(diff(y(x), x))/(x+3)+(-x^2-6*x-9)*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 <- Bessel successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 57
dsolve(diff(y(x),x)=1+(x*y(x)+3*y(x))^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-\operatorname {BesselY}\left (-\frac {1}{4}, \frac {\left (x +3\right )^{2}}{2}\right ) c_{1} -\operatorname {BesselJ}\left (-\frac {1}{4}, \frac {\left (x +3\right )^{2}}{2}\right )}{\left (\operatorname {BesselY}\left (\frac {3}{4}, \frac {\left (x +3\right )^{2}}{2}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {3}{4}, \frac {\left (x +3\right )^{2}}{2}\right )\right ) \left (x +3\right )} \]
✓ Solution by Mathematica
Time used: 0.45 (sec). Leaf size: 351
DSolve[y'[x]==1+(x*y[x]+3*y[x])^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\left ((x+3)^3\right )^{2/3} \operatorname {Gamma}\left (\frac {7}{4}\right ) \operatorname {BesselJ}\left (-\frac {1}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )+3 \operatorname {Gamma}\left (\frac {7}{4}\right ) \operatorname {BesselJ}\left (\frac {3}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )-\left ((x+3)^3\right )^{2/3} \operatorname {Gamma}\left (\frac {7}{4}\right ) \operatorname {BesselJ}\left (\frac {7}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )+4 c_1 \left ((x+3)^3\right )^{2/3} \operatorname {Gamma}\left (\frac {5}{4}\right ) \operatorname {BesselJ}\left (-\frac {7}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )+12 c_1 \operatorname {Gamma}\left (\frac {5}{4}\right ) \operatorname {BesselJ}\left (-\frac {3}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )-4 c_1 \left ((x+3)^3\right )^{2/3} \operatorname {Gamma}\left (\frac {5}{4}\right ) \operatorname {BesselJ}\left (\frac {1}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )}{2 (x+3)^3 \left (\operatorname {Gamma}\left (\frac {7}{4}\right ) \operatorname {BesselJ}\left (\frac {3}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )+4 c_1 \operatorname {Gamma}\left (\frac {5}{4}\right ) \operatorname {BesselJ}\left (-\frac {3}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )\right )} \\ y(x)\to \frac {-\left ((x+3)^3\right )^{2/3} \operatorname {BesselJ}\left (-\frac {7}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )-3 \operatorname {BesselJ}\left (-\frac {3}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )+\left ((x+3)^3\right )^{2/3} \operatorname {BesselJ}\left (\frac {1}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )}{2 (x+3)^3 \operatorname {BesselJ}\left (-\frac {3}{4},\frac {1}{2} \left ((x+3)^3\right )^{2/3}\right )} \\ \end{align*}