Internal problem ID [3653]
Internal file name [OUTPUT/3146_Sunday_June_05_2022_08_53_39_AM_71950282/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 14
Problem number: 399.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, [_Riccati, _special]]
\[ \boxed {x^{\frac {3}{2}} y^{\prime }-b \,x^{\frac {3}{2}} y^{2}=a} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a +b \,x^{\frac {3}{2}} y^{2}}{x^{\frac {3}{2}}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = b \,y^{2}+\frac {a}{x^{\frac {3}{2}}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a}{x^{\frac {3}{2}}}\), \(f_1(x)=0\) and \(f_2(x)=b\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{b 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 {b^{2} a}{x^{\frac {3}{2}}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} b u^{\prime \prime }\left (x \right )+\frac {b^{2} a u \left (x \right )}{x^{\frac {3}{2}}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = -c_{1} x^{\frac {1}{4}} \operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )-c_{2} \operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) x^{\frac {1}{4}}+2 \sqrt {a}\, \sqrt {b}\, \sqrt {x}\, \left (\operatorname {BesselJ}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{1} +\operatorname {BesselY}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {2 a b \left (\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{2} +\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{1} \right )}{x^{\frac {1}{4}}} \] Using the above in (1) gives the solution \[ y = \frac {2 a \left (\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{2} +\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{1} \right )}{x^{\frac {1}{4}} \left (-c_{1} x^{\frac {1}{4}} \operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )-c_{2} \operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) x^{\frac {1}{4}}+2 \sqrt {a}\, \sqrt {b}\, \sqrt {x}\, \left (\operatorname {BesselJ}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{1} +\operatorname {BesselY}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{2} \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {2 a \left (\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )+\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{3} \right )}{\sqrt {x}\, \left (2 \operatorname {BesselJ}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{3} x^{\frac {1}{4}} \sqrt {a}\, \sqrt {b}+2 \operatorname {BesselY}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}-\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{3} -\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 a \left (\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )+\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{3} \right )}{\sqrt {x}\, \left (2 \operatorname {BesselJ}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{3} x^{\frac {1}{4}} \sqrt {a}\, \sqrt {b}+2 \operatorname {BesselY}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}-\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{3} -\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {2 a \left (\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )+\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{3} \right )}{\sqrt {x}\, \left (2 \operatorname {BesselJ}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{3} x^{\frac {1}{4}} \sqrt {a}\, \sqrt {b}+2 \operatorname {BesselY}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}-\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{3} -\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\frac {3}{2}} y^{\prime }-b \,x^{\frac {3}{2}} y^{2}=a \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a +b \,x^{\frac {3}{2}} y^{2}}{x^{\frac {3}{2}}} \end {array} \]
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 Special <- Riccati Special successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 119
dsolve(x^(3/2)*diff(y(x),x) = a+b*x^(3/2)*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {2 a \left (\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{1} +\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )\right )}{\sqrt {x}\, \left (-2 \sqrt {a}\, x^{\frac {1}{4}} \operatorname {BesselJ}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) \sqrt {b}\, c_{1} -2 \operatorname {BesselY}\left (0, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}+\operatorname {BesselJ}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right ) c_{1} +\operatorname {BesselY}\left (1, 4 \sqrt {a}\, \sqrt {b}\, x^{\frac {1}{4}}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.252 (sec). Leaf size: 373
DSolve[x^(3/2) y'[x]==a+ b x^(3/2) y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {a} \sqrt {b} \sqrt [4]{x} \operatorname {BesselY}\left (1,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )+\operatorname {BesselY}\left (2,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )-\sqrt {a} \sqrt {b} \sqrt [4]{x} \operatorname {BesselY}\left (3,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )-\sqrt {a} \sqrt {b} c_1 \sqrt [4]{x} \operatorname {BesselJ}\left (1,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )-c_1 \operatorname {BesselJ}\left (2,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )+\sqrt {a} \sqrt {b} c_1 \sqrt [4]{x} \operatorname {BesselJ}\left (3,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )}{2 b x \operatorname {BesselY}\left (2,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )-2 b c_1 x \operatorname {BesselJ}\left (2,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )} \\ y(x)\to -\frac {\sqrt {a} \sqrt {b} \sqrt [4]{x} \operatorname {BesselJ}\left (1,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )+\operatorname {BesselJ}\left (2,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )-\sqrt {a} \sqrt {b} \sqrt [4]{x} \operatorname {BesselJ}\left (3,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )}{2 b x \operatorname {BesselJ}\left (2,4 \sqrt {a} \sqrt {b} \sqrt [4]{x}\right )} \\ \end{align*}