Internal problem ID [9331]
Internal file name [OUTPUT/8267_Monday_June_06_2022_02_35_26_AM_20462012/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 998.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime }-\frac {\left (y-\ln \left (x \right )-\operatorname {Ci}\left (x \right )\right )^{2}+\cos \left (x \right )}{x}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\ln \left (x \right )^{2}+2 \ln \left (x \right ) \operatorname {Ci}\left (x \right )-2 y \ln \left (x \right )+\operatorname {Ci}\left (x \right )^{2}-2 y \,\operatorname {Ci}\left (x \right )+y^{2}+\cos \left (x \right )}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {\ln \left (x \right )^{2}}{x}+\frac {2 \ln \left (x \right ) \operatorname {Ci}\left (x \right )}{x}-\frac {2 y \ln \left (x \right )}{x}+\frac {\operatorname {Ci}\left (x \right )^{2}}{x}-\frac {2 y \,\operatorname {Ci}\left (x \right )}{x}+\frac {y^{2}}{x}+\frac {\cos \left (x \right )}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {\ln \left (x \right )^{2}+2 \ln \left (x \right ) \operatorname {Ci}\left (x \right )+\operatorname {Ci}\left (x \right )^{2}+\cos \left (x \right )}{x}\), \(f_1(x)=\frac {-2 \ln \left (x \right )-2 \,\operatorname {Ci}\left (x \right )}{x}\) and \(f_2(x)=\frac {1}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {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 {1}{x^{2}}\\ f_1 f_2 &=\frac {-2 \ln \left (x \right )-2 \,\operatorname {Ci}\left (x \right )}{x^{2}}\\ f_2^2 f_0 &=\frac {\ln \left (x \right )^{2}+2 \ln \left (x \right ) \operatorname {Ci}\left (x \right )+\operatorname {Ci}\left (x \right )^{2}+\cos \left (x \right )}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {1}{x^{2}}+\frac {-2 \ln \left (x \right )-2 \,\operatorname {Ci}\left (x \right )}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {\left (\ln \left (x \right )^{2}+2 \ln \left (x \right ) \operatorname {Ci}\left (x \right )+\operatorname {Ci}\left (x \right )^{2}+\cos \left (x \right )\right ) u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {{\mathrm e}^{-\frac {\left (\int \frac {1+2 \ln \left (x \right )+2 \,\operatorname {Ci}\left (x \right )}{x}d x \right )}{2}} \left (c_{2} x^{2}+c_{1} \right )}{\sqrt {x}} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\left (\left (c_{2} x^{2}+c_{1} \right ) \operatorname {Ci}\left (x \right )+\left (c_{2} x^{2}+c_{1} \right ) \ln \left (x \right )-c_{2} x^{2}+c_{1} \right ) {\mathrm e}^{-\frac {\left (\int \frac {1+2 \ln \left (x \right )+2 \,\operatorname {Ci}\left (x \right )}{x}d x \right )}{2}}}{x^{\frac {3}{2}}} \] Using the above in (1) gives the solution \[ y = \frac {\left (c_{2} x^{2}+c_{1} \right ) \operatorname {Ci}\left (x \right )+\left (c_{2} x^{2}+c_{1} \right ) \ln \left (x \right )-c_{2} x^{2}+c_{1}}{c_{2} x^{2}+c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (x^{2}+c_{3} \right ) \operatorname {Ci}\left (x \right )+\left (x^{2}+c_{3} \right ) \ln \left (x \right )-x^{2}+c_{3}}{x^{2}+c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (x^{2}+c_{3} \right ) \operatorname {Ci}\left (x \right )+\left (x^{2}+c_{3} \right ) \ln \left (x \right )-x^{2}+c_{3}}{x^{2}+c_{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (x^{2}+c_{3} \right ) \operatorname {Ci}\left (x \right )+\left (x^{2}+c_{3} \right ) \ln \left (x \right )-x^{2}+c_{3}}{x^{2}+c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \ln \left (x \right )^{2}+2 \ln \left (x \right ) \mathrm {Ci}\left (x \right )-2 y \ln \left (x \right )+\mathrm {Ci}\left (x \right )^{2}-2 y \,\mathrm {Ci}\left (x \right )-y^{\prime } x +y^{2}+\cos \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-\ln \left (x \right )^{2}-2 \ln \left (x \right ) \mathrm {Ci}\left (x \right )+2 y \ln \left (x \right )-\mathrm {Ci}\left (x \right )^{2}+2 y \,\mathrm {Ci}\left (x \right )-y^{2}-\cos \left (x \right )}{x} \end {array} \]
Maple trace Kovacic algorithm successful
`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 trying Riccati trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(2*ln(x)+2*Ci(x)+1)*(diff(y(x), x))/x-(ln(x)^2+2*ln(x)*Ci(x)+Ci(x)^2+ Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) <- Kovacics algorithm successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 27
dsolve(diff(y(x),x) = ((y(x)-ln(x)-Ci(x))^2+cos(x))/x,y(x), singsol=all)
\[ y \left (x \right ) = \ln \left (x \right )+\operatorname {Ci}\left (x \right )+\frac {-c_{1} x^{2}+1}{c_{1} x^{2}+1} \]
✓ Solution by Mathematica
Time used: 0.738 (sec). Leaf size: 36
DSolve[y'[x] == (Cos[x] + (-CosIntegral[x] - Log[x] + y[x])^2)/x,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \operatorname {CosIntegral}(x)-\frac {2 x^2}{x^2-2 c_1}+\log (x)+1 \\ y(x)\to \operatorname {CosIntegral}(x)+\log (x)+1 \\ \end{align*}