Internal problem ID [10492]
Internal file name [OUTPUT/9440_Monday_June_06_2022_02_32_32_PM_95367239/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 1.2.5-2
Problem number: 18.
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]
\[ \boxed {y^{\prime } x -a \ln \left (\lambda x \right )^{m} y^{2}-k y=a \,b^{2} x^{2 k} \ln \left (\lambda x \right )^{m}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \ln \left (\lambda x \right )^{m} y^{2}+k y +a \,b^{2} x^{2 k} \ln \left (\lambda x \right )^{m}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,b^{2} x^{2 k} \ln \left (\lambda x \right )^{m}}{x}+\frac {a \ln \left (\lambda x \right )^{m} y^{2}}{x}+\frac {k y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a \,b^{2} x^{2 k} \ln \left (\lambda x \right )^{m}}{x}\), \(f_1(x)=\frac {k}{x}\) and \(f_2(x)=\frac {a \ln \left (\lambda x \right )^{m}}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {a \ln \left (\lambda x \right )^{m} 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 {a \ln \left (\lambda x \right )^{m} m}{x^{2} \ln \left (\lambda x \right )}-\frac {a \ln \left (\lambda x \right )^{m}}{x^{2}}\\ f_1 f_2 &=\frac {k a \ln \left (\lambda x \right )^{m}}{x^{2}}\\ f_2^2 f_0 &=\frac {a^{3} \ln \left (\lambda x \right )^{3 m} b^{2} x^{2 k}}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {a \ln \left (\lambda x \right )^{m} u^{\prime \prime }\left (x \right )}{x}-\left (\frac {a \ln \left (\lambda x \right )^{m} m}{x^{2} \ln \left (\lambda x \right )}-\frac {a \ln \left (\lambda x \right )^{m}}{x^{2}}+\frac {k a \ln \left (\lambda x \right )^{m}}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {a^{3} \ln \left (\lambda x \right )^{3 m} b^{2} x^{2 k} u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}+c_{2} {\mathrm e}^{-i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = i a b \,x^{k -1} \ln \left (\lambda x \right )^{m} {\mathrm e}^{-i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )} \left (c_{1} {\mathrm e}^{2 i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}-c_{2} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {i b \,x^{k -1} {\mathrm e}^{-i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )} \left (c_{1} {\mathrm e}^{2 i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}-c_{2} \right ) x}{c_{1} {\mathrm e}^{i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}+c_{2} {\mathrm e}^{-i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {i b \,x^{k} \left (c_{3} {\mathrm e}^{2 i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}-1\right )}{c_{3} {\mathrm e}^{2 i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}+1} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {i b \,x^{k} \left (c_{3} {\mathrm e}^{2 i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}-1\right )}{c_{3} {\mathrm e}^{2 i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}+1} \\ \end{align*}
Verification of solutions
\[ y = -\frac {i b \,x^{k} \left (c_{3} {\mathrm e}^{2 i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}-1\right )}{c_{3} {\mathrm e}^{2 i a b \left (\int x^{k -1} \ln \left (\lambda x \right )^{m}d x \right )}+1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a \ln \left (\lambda x \right )^{m} y^{2}-k y=a \,b^{2} x^{2 k} \ln \left (\lambda x \right )^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \ln \left (\lambda x \right )^{m} y^{2}+k y+a \,b^{2} x^{2 k} \ln \left (\lambda x \right )^{m}}{x} \end {array} \]
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: 0.047 (sec). Leaf size: 31
dsolve(x*diff(y(x),x)=a*(ln(lambda*x))^m*y(x)^2+k*y(x)+a*b^2*x^(2*k)*(ln(lambda*x))^m,y(x), singsol=all)
\[ y \left (x \right ) = -\tan \left (-a b \left (\int x^{-1+k} \ln \left (x \lambda \right )^{m}d x \right )+c_{1} \right ) b \,x^{k} \]
✓ Solution by Mathematica
Time used: 2.161 (sec). Leaf size: 70
DSolve[x*y'[x]==a*(Log[\[Lambda]*x])^m*y[x]^2+k*y[x]+a*b^2*x^(2*k)*(Log[\[Lambda]*x])^m,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {b^2} x^k \tan \left (\frac {a \sqrt {b^2} x^k (\lambda x)^{-k} \log ^m(\lambda x) (-k \log (\lambda x))^{-m} \Gamma (m+1,-k \log (x \lambda ))}{k}+c_1\right ) \]