Internal problem ID [10599]
Internal file name [OUTPUT/9547_Monday_June_06_2022_03_07_31_PM_8322341/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. subsection 1.2.8-1. Equations containing
arbitrary functions (but not containing their derivatives).
Problem number: 7.
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]
\[ \boxed {y^{\prime } x -f \left (x \right ) y^{2}-y n=f \left (x \right ) x^{2 n} a} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {f \left (x \right ) y^{2}+y n +f \left (x \right ) x^{2 n} a}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,x^{2 n} f \left (x \right )}{x}+\frac {f \left (x \right ) y^{2}}{x}+\frac {y n}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a \,x^{2 n} f \left (x \right )}{x}\), \(f_1(x)=\frac {n}{x}\) and \(f_2(x)=\frac {f \left (x \right )}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {f \left (x \right ) 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 {f \left (x \right )}{x^{2}}+\frac {f^{\prime }\left (x \right )}{x}\\ f_1 f_2 &=\frac {n f \left (x \right )}{x^{2}}\\ f_2^2 f_0 &=\frac {f \left (x \right )^{3} a \,x^{2 n}}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {f \left (x \right ) u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {f \left (x \right )}{x^{2}}+\frac {f^{\prime }\left (x \right )}{x}+\frac {n f \left (x \right )}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {f \left (x \right )^{3} a \,x^{2 n} u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}+c_{2} {\mathrm e}^{-i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = i x^{n -1} f \left (x \right ) \sqrt {a}\, {\mathrm e}^{-i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )} \left (c_{1} {\mathrm e}^{2 i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}-c_{2} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {i x^{n -1} \sqrt {a}\, {\mathrm e}^{-i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )} \left (c_{1} {\mathrm e}^{2 i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}-c_{2} \right ) x}{c_{1} {\mathrm e}^{i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}+c_{2} {\mathrm e}^{-i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {i x^{n} \sqrt {a}\, \left (c_{3} {\mathrm e}^{2 i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}-1\right )}{c_{3} {\mathrm e}^{2 i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}+1} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {i x^{n} \sqrt {a}\, \left (c_{3} {\mathrm e}^{2 i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}-1\right )}{c_{3} {\mathrm e}^{2 i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}+1} \\ \end{align*}
Verification of solutions
\[ y = -\frac {i x^{n} \sqrt {a}\, \left (c_{3} {\mathrm e}^{2 i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}-1\right )}{c_{3} {\mathrm e}^{2 i \sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )}+1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -f \left (x \right ) y^{2}-y n =f \left (x \right ) x^{2 n} a \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {f \left (x \right ) y^{2}+y n +f \left (x \right ) x^{2 n} a}{x} \end {array} \]
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 <- Chini successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 30
dsolve(x*diff(y(x),x)=f(x)*y(x)^2+n*y(x)+a*x^(2*n)*f(x),y(x), singsol=all)
\[ y \left (x \right ) = -\tan \left (-\sqrt {a}\, \left (\int f \left (x \right ) x^{n -1}d x \right )+c_{1} \right ) \sqrt {a}\, x^{n} \]
✓ Solution by Mathematica
Time used: 0.577 (sec). Leaf size: 41
DSolve[x*y'[x]==f[x]*y[x]^2+n*y[x]+a*x^(2*n)*f[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {a} x^n \tan \left (\sqrt {a} \int _1^xf(K[1]) K[1]^{n-1}dK[1]+c_1\right ) \]