Internal problem ID [10429]
Internal file name [OUTPUT/9377_Monday_June_06_2022_02_20_50_PM_93195118/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: 22.
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]
\[ \boxed {y^{\prime }-y^{2}-a x \,{\mathrm e}^{\lambda x} y={\mathrm e}^{\lambda x} a} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+{\mathrm e}^{\lambda x} a x y +{\mathrm e}^{\lambda x} a \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+{\mathrm e}^{\lambda x} a x y +{\mathrm e}^{\lambda x} a \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)={\mathrm e}^{\lambda x} a\), \(f_1(x)=a x \,{\mathrm e}^{\lambda x}\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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' &=0\\ f_1 f_2 &=a x \,{\mathrm e}^{\lambda x}\\ f_2^2 f_0 &={\mathrm e}^{\lambda x} a \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-a x \,{\mathrm e}^{\lambda x} u^{\prime }\left (x \right )+{\mathrm e}^{\lambda x} a u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {x \left (c_{2} \lambda ^{2}+\left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) c_{1} \right )}{\lambda ^{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {c_{2} \lambda ^{2} x +c_{1} \left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) x +{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}} c_{1}}{\lambda ^{2} x} \] Using the above in (1) gives the solution \[ y = -\frac {c_{2} \lambda ^{2} x +c_{1} \left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) x +{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}} c_{1}}{x^{2} \left (c_{2} \lambda ^{2}+\left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\lambda ^{2} x -c_{3} \left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) x -{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}} c_{3}}{x^{2} \left (\lambda ^{2}+\left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\lambda ^{2} x -c_{3} \left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) x -{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}} c_{3}}{x^{2} \left (\lambda ^{2}+\left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\lambda ^{2} x -c_{3} \left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) x -{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}} c_{3}}{x^{2} \left (\lambda ^{2}+\left (\int \frac {{\mathrm e}^{\frac {\left (\lambda x -1\right ) {\mathrm e}^{\lambda x} a}{\lambda ^{2}}}}{x^{2}}d x \right ) c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-a x \,{\mathrm e}^{\lambda x} y={\mathrm e}^{\lambda x} a \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a x \,{\mathrm e}^{\lambda x} y+{\mathrm e}^{\lambda x} a \end {array} \]
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 differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries found: 2 potential symmetries. Proceeding with integration step`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 85
dsolve(diff(y(x),x)=y(x)^2+a*x*exp(lambda*x)*y(x)+a*exp(lambda*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-c_{1} \lambda ^{2} x +\left (\int \frac {{\mathrm e}^{\frac {{\mathrm e}^{x \lambda } a \left (x \lambda -1\right )}{\lambda ^{2}}}}{x^{2}}d x \right ) x +{\mathrm e}^{\frac {{\mathrm e}^{x \lambda } a \left (x \lambda -1\right )}{\lambda ^{2}}}}{x^{2} \left (c_{1} \lambda ^{2}-\left (\int \frac {{\mathrm e}^{\frac {{\mathrm e}^{x \lambda } a \left (x \lambda -1\right )}{\lambda ^{2}}}}{x^{2}}d x \right )\right )} \]
✓ Solution by Mathematica
Time used: 2.132 (sec). Leaf size: 110
DSolve[y'[x]==y[x]^2+a*x*Exp[\[Lambda]*x]*y[x]+a*Exp[\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x \int _1^x\frac {e^{\frac {a e^{\lambda K[1]} (\lambda K[1]-1)}{\lambda ^2}}}{K[1]^2}dK[1]+e^{\frac {a e^{\lambda x} (\lambda x-1)}{\lambda ^2}}+c_1 x}{x^2 \left (\int _1^x\frac {e^{\frac {a e^{\lambda K[1]} (\lambda K[1]-1)}{\lambda ^2}}}{K[1]^2}dK[1]+c_1\right )} \\ y(x)\to -\frac {1}{x} \\ \end{align*}