Internal problem ID [10607]
Internal file name [OUTPUT/9555_Monday_June_06_2022_03_08_45_PM_80987473/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: 15.
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]
\[ \boxed {y^{\prime }-f \left (x \right ) y^{2}-\lambda y=a^{2} {\mathrm e}^{2 \lambda x} f \left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f \left (x \right ) y^{2}+y \lambda +a^{2} {\mathrm e}^{2 \lambda x} f \left (x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = f \left (x \right ) y^{2}+y \lambda +a^{2} {\mathrm e}^{2 \lambda x} f \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a^{2} {\mathrm e}^{2 \lambda x} f \left (x \right )\), \(f_1(x)=\lambda \) and \(f_2(x)=f \left (x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{f \left (x \right ) 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' &=f^{\prime }\left (x \right )\\ f_1 f_2 &=\lambda f \left (x \right )\\ f_2^2 f_0 &={\mathrm e}^{2 \lambda x} f \left (x \right )^{3} a^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} f \left (x \right ) u^{\prime \prime }\left (x \right )-\left (f^{\prime }\left (x \right )+\lambda f \left (x \right )\right ) u^{\prime }\left (x \right )+{\mathrm e}^{2 \lambda x} f \left (x \right )^{3} a^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}+c_{2} {\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = i a \,{\mathrm e}^{\lambda x} f \left (x \right ) \left (c_{1} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}-c_{2} {\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {i a \,{\mathrm e}^{\lambda x} \left (c_{1} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}-c_{2} {\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}\right )}{c_{1} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}+c_{2} {\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {i a \,{\mathrm e}^{\lambda x} \left (c_{3} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}-{\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}\right )}{c_{3} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}+{\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {i a \,{\mathrm e}^{\lambda x} \left (c_{3} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}-{\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}\right )}{c_{3} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}+{\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {i a \,{\mathrm e}^{\lambda x} \left (c_{3} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}-{\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}\right )}{c_{3} {\mathrm e}^{i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}+{\mathrm e}^{-i a \left (\int {\mathrm e}^{\lambda x} f \left (x \right )d x \right )}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-f \left (x \right ) y^{2}-\lambda y=a^{2} {\mathrm e}^{2 \lambda x} f \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=f \left (x \right ) y^{2}+\lambda y+a^{2} {\mathrm e}^{2 \lambda x} f \left (x \right ) \end {array} \]
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: 0.032 (sec). Leaf size: 26
dsolve(diff(y(x),x)=f(x)*y(x)^2+lambda*y(x)+a^2*exp(2*lambda*x)*f(x),y(x), singsol=all)
\[ y \left (x \right ) = -\tan \left (-a \left (\int {\mathrm e}^{x \lambda } f \left (x \right )d x \right )+c_{1} \right ) a \,{\mathrm e}^{x \lambda } \]
✓ Solution by Mathematica
Time used: 0.615 (sec). Leaf size: 47
DSolve[y'[x]==f[x]*y[x]^2+\[Lambda]*y[x]+a^2*Exp[2*\[Lambda]*x]*f[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {a^2} e^{\lambda x} \tan \left (\sqrt {a^2} \int _1^xe^{\lambda K[1]} f(K[1])dK[1]+c_1\right ) \]