Internal problem ID [10445]
Internal file name [OUTPUT/9393_Monday_June_06_2022_02_21_27_PM_31744357/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: 38.
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]
\[ \boxed {y^{\prime }-a \,{\mathrm e}^{-\lambda \,x^{2}} y^{2}-y \lambda x=a \,b^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,{\mathrm e}^{-\lambda \,x^{2}} y^{2}+\lambda x y +a \,b^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,{\mathrm e}^{-\lambda \,x^{2}} y^{2}+\lambda x y +a \,b^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \,b^{2}\), \(f_1(x)=\lambda x\) and \(f_2(x)={\mathrm e}^{-\lambda \,x^{2}} a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{{\mathrm e}^{-\lambda \,x^{2}} a 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 \lambda x \,{\mathrm e}^{-\lambda \,x^{2}} a\\ f_1 f_2 &=\lambda x \,{\mathrm e}^{-\lambda \,x^{2}} a\\ f_2^2 f_0 &={\mathrm e}^{-2 \lambda \,x^{2}} a^{3} b^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} {\mathrm e}^{-\lambda \,x^{2}} a u^{\prime \prime }\left (x \right )+\lambda x \,{\mathrm e}^{-\lambda \,x^{2}} a u^{\prime }\left (x \right )+{\mathrm e}^{-2 \lambda \,x^{2}} a^{3} b^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )+c_{2} \cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = a b \,{\mathrm e}^{-\frac {\lambda \,x^{2}}{2}} \left (c_{1} \cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )-c_{2} \sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )\right ) \] Using the above in (1) gives the solution \[ y = -\frac {b \,{\mathrm e}^{-\frac {\lambda \,x^{2}}{2}} \left (c_{1} \cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )-c_{2} \sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )\right ) {\mathrm e}^{\lambda \,x^{2}}}{c_{1} \sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )+c_{2} \cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {b \,{\mathrm e}^{\frac {\lambda \,x^{2}}{2}} \left (-c_{3} \cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )+\sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )\right )}{c_{3} \sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )+\cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {b \,{\mathrm e}^{\frac {\lambda \,x^{2}}{2}} \left (-c_{3} \cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )+\sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )\right )}{c_{3} \sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )+\cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {b \,{\mathrm e}^{\frac {\lambda \,x^{2}}{2}} \left (-c_{3} \cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )+\sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )\right )}{c_{3} \sin \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )+\cos \left (\frac {\sqrt {2}\, a b \sqrt {\pi }\, \operatorname {erf}\left (\frac {x \sqrt {2}\, \sqrt {\lambda }}{2}\right )}{2 \sqrt {\lambda }}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,{\mathrm e}^{-\lambda \,x^{2}} y^{2}-y \lambda x =a \,b^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,{\mathrm e}^{-\lambda \,x^{2}} y^{2}+y \lambda x +a \,b^{2} \end {array} \]
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 <- Chini successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 45
dsolve(diff(y(x),x)=a*exp(-lambda*x^2)*y(x)^2+lambda*x*y(x)+a*b^2,y(x), singsol=all)
\[ y \left (x \right ) = \tan \left (\frac {a b \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {\lambda }\, x}{2}\right )-2 c_{1} \sqrt {\lambda }}{2 \sqrt {\lambda }}\right ) b \,{\mathrm e}^{\frac {x^{2} \lambda }{2}} \]
✓ Solution by Mathematica
Time used: 2.252 (sec). Leaf size: 63
DSolve[y'[x]==a*Exp[-\[Lambda]*x^2]*y[x]^2+\[Lambda]*x*y[x]+a*b^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {b^2} e^{\frac {\lambda x^2}{2}} \tan \left (\frac {\sqrt {\frac {\pi }{2}} a \sqrt {b^2} \text {erf}\left (\frac {\sqrt {\lambda } x}{\sqrt {2}}\right )}{\sqrt {\lambda }}+c_1\right ) \]