Internal problem ID [8518]
Internal file name [OUTPUT/7451_Sunday_June_05_2022_10_55_23_PM_16771672/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 182.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {x \left (x^{3}-1\right ) y^{\prime }-2 x y^{2}+y=-x^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {2 x \,y^{2}-x^{2}-y}{x \left (x^{3}-1\right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {2 y^{2}}{x^{3}-1}-\frac {x}{x^{3}-1}-\frac {y}{x \left (x^{3}-1\right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {x}{x^{3}-1}\), \(f_1(x)=-\frac {1}{x \left (x^{3}-1\right )}\) and \(f_2(x)=\frac {2}{x^{3}-1}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {2 u}{x^{3}-1}} \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 {6 x^{2}}{\left (x^{3}-1\right )^{2}}\\ f_1 f_2 &=-\frac {2}{x \left (x^{3}-1\right )^{2}}\\ f_2^2 f_0 &=-\frac {4 x}{\left (x^{3}-1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {2 u^{\prime \prime }\left (x \right )}{x^{3}-1}-\left (-\frac {6 x^{2}}{\left (x^{3}-1\right )^{2}}-\frac {2}{x \left (x^{3}-1\right )^{2}}\right ) u^{\prime }\left (x \right )-\frac {4 x u \left (x \right )}{\left (x^{3}-1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {c_{2} x^{2}+c_{1}}{\left (x^{3}-1\right )^{\frac {2}{3}}} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {2 x \left (c_{1} x +c_{2} \right )}{\left (x^{3}-1\right )^{\frac {5}{3}}} \] Using the above in (1) gives the solution \[ y = \frac {x \left (c_{1} x +c_{2} \right )}{c_{2} x^{2}+c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (c_{3} x +1\right ) x}{x^{2}+c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3} x +1\right ) x}{x^{2}+c_{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3} x +1\right ) x}{x^{2}+c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x^{3}-1\right ) y^{\prime }-2 x y^{2}+y=-x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 x y^{2}-y-x^{2}}{x \left (x^{3}-1\right )} \end {array} \]
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 to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(3*x^3+1)*(diff(y(x), x))/(x*(x^3-1))+2*x*y(x)/(x^3-1)^2, y(x)` Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type 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: 0.0 (sec). Leaf size: 18
dsolve(x*(x^3-1)*diff(y(x),x) - 2*x*y(x)^2 + y(x) + x^2=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x \left (x +c_{1} \right )}{c_{1} x^{2}+1} \]
✓ Solution by Mathematica
Time used: 0.368 (sec). Leaf size: 31
DSolve[x*(x^3-1)*y'[x] - 2*x*y[x]^2 + y[x] + x^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x (1+2 c_1 x)}{x^2+2 c_1} \\ y(x)\to x^2 \\ \end{align*}