Internal problem ID [8353]
Internal file name [OUTPUT/7286_Sunday_June_05_2022_05_42_35_PM_81311901/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 16.
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]
\[ \boxed {y^{\prime }+y^{2}+\left (y x -1\right ) f \left (x \right )=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -f \left (x \right ) x y -y^{2}+f \left (x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -f \left (x \right ) x y -y^{2}+f \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=f \left (x \right )\), \(f_1(x)=-f \left (x \right ) x\) 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 &=f \left (x \right ) x\\ f_2^2 f_0 &=f \left (x \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )-f \left (x \right ) x u^{\prime }\left (x \right )+f \left (x \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x \left (\left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} +x \,{\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )} c_{1} \] Using the above in (1) gives the solution \[ y = \frac {\left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} +x \,{\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )} c_{1}}{x \left (\left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} +x \,{\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )} c_{1}}{x \left (\left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} +x \,{\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )} c_{1}}{x \left (\left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} +x \,{\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )} c_{1}}{x \left (\left (\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) c_{1} +c_{2} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+y^{2}+\left (y x -1\right ) f \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-y^{2}-\left (y x -1\right ) f \left (x \right ) \end {array} \]
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 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: 0.015 (sec). Leaf size: 75
dsolve(diff(y(x),x) + y(x)^2 +(x*y(x)-1)*f(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )} x +\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x -c_{1}}{\left (-c_{1} +\int {\mathrm e}^{-\left (\int \frac {f \left (x \right ) x^{2}+2}{x}d x \right )}d x \right ) x} \]
✓ Solution by Mathematica
Time used: 0.134 (sec). Leaf size: 114
DSolve[y'[x] + y[x]^2 +(x*y[x]-1)*f[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x \exp \left (-\int _1^x\left (f(K[1]) K[1]+\frac {2}{K[1]}\right )dK[1]\right )+\int _1^x\exp \left (-\int _1^{K[2]}\left (f(K[1]) K[1]+\frac {2}{K[1]}\right )dK[1]\right )dK[2]+c_1}{x \left (\int _1^x\exp \left (-\int _1^{K[2]}\left (f(K[1]) K[1]+\frac {2}{K[1]}\right )dK[1]\right )dK[2]+c_1\right )} \\ y(x)\to \frac {1}{x} \\ \end{align*}