Internal problem ID [10442]
Internal file name [OUTPUT/9390_Monday_June_06_2022_02_21_21_PM_25838751/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 1.2.3-2. Equations with power and
exponential functions
Problem number: 35.
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]
\[ \boxed {y^{\prime } x -a \,{\mathrm e}^{\lambda x} y^{2}-k y=a \,b^{2} x^{2 k} {\mathrm e}^{\lambda x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {{\mathrm e}^{\lambda x} a \,y^{2}+k y +a \,b^{2} x^{2 k} {\mathrm e}^{\lambda x}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,b^{2} x^{2 k} {\mathrm e}^{\lambda x}}{x}+\frac {{\mathrm e}^{\lambda x} a \,y^{2}}{x}+\frac {k y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a \,b^{2} x^{2 k} {\mathrm e}^{\lambda x}}{x}\), \(f_1(x)=\frac {k}{x}\) and \(f_2(x)=\frac {a \,{\mathrm e}^{\lambda x}}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {a \,{\mathrm e}^{\lambda x} 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 \,{\mathrm e}^{\lambda x}}{x^{2}}+\frac {a \lambda \,{\mathrm e}^{\lambda x}}{x}\\ f_1 f_2 &=\frac {k a \,{\mathrm e}^{\lambda x}}{x^{2}}\\ f_2^2 f_0 &=\frac {a^{3} {\mathrm e}^{3 \lambda x} b^{2} x^{2 k}}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {a \,{\mathrm e}^{\lambda x} u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {a \,{\mathrm e}^{\lambda x}}{x^{2}}+\frac {a \lambda \,{\mathrm e}^{\lambda x}}{x}+\frac {k a \,{\mathrm e}^{\lambda x}}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {a^{3} {\mathrm e}^{3 \lambda x} 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
\[ u \left (x \right ) = c_{1} \sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )+c_{2} \cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (c_{1} \cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )-c_{2} \sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )\right ) a b \,x^{k -1} {\mathrm e}^{\lambda x} \] Using the above in (1) gives the solution \[ y = -\frac {\left (c_{1} \cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )-c_{2} \sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )\right ) b \,x^{k -1} x}{c_{1} \sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )+c_{2} \cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {b \,x^{k} \left (-c_{3} \cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )+\sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )\right )}{c_{3} \sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )+\cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {b \,x^{k} \left (-c_{3} \cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )+\sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )\right )}{c_{3} \sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )+\cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {b \,x^{k} \left (-c_{3} \cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )+\sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )\right )}{c_{3} \sin \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )+\cos \left (a b \,x^{k} \left (-\lambda x \right )^{-k} \left (\Gamma \left (k \right )-\Gamma \left (k , -\lambda x \right )\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a \,{\mathrm e}^{\lambda x} y^{2}-k y=a \,b^{2} x^{2 k} {\mathrm e}^{\lambda x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,{\mathrm e}^{\lambda x} y^{2}+k y+a \,b^{2} x^{2 k} {\mathrm e}^{\lambda x}}{x} \end {array} \]
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: 0.015 (sec). Leaf size: 38
dsolve(x*diff(y(x),x)=a*exp(lambda*x)*y(x)^2+k*y(x)+a*b^2*x^(2*k)*exp(lambda*x),y(x), singsol=all)
\[ y \left (x \right ) = -\tan \left (a b \,x^{k} \left (\Gamma \left (k , -x \lambda \right )-\Gamma \left (k \right )\right ) \left (-x \lambda \right )^{-k}+c_{1} \right ) b \,x^{k} \]
✓ Solution by Mathematica
Time used: 1.593 (sec). Leaf size: 47
DSolve[x*y'[x]==a*Exp[\[Lambda]*x]*y[x]^2+k*y[x]+a*b^2*x^(2*k)*Exp[\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {b^2} x^k \tan \left (-a \sqrt {b^2} x^k (\lambda (-x))^{-k} \Gamma (k,-x \lambda )+c_1\right ) \]