Internal problem ID [9240]
Internal file name [OUTPUT/8176_Monday_June_06_2022_02_02_55_AM_67041402/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 907.
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]
\[ \boxed {y^{\prime }-\frac {-2 \cos \left (x \right ) x +2 \sin \left (x \right ) x^{2}+2 x +2 y^{2}+4 y \cos \left (x \right ) x -4 x y+x^{2} \cos \left (2 x \right )+3 x^{2}-4 x^{2} \cos \left (x \right )}{2 x}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-2 \cos \left (x \right ) x +2 \sin \left (x \right ) x^{2}+2 x +2 y^{2}+4 y \cos \left (x \right ) x -4 x y +x^{2} \cos \left (2 x \right )+3 x^{2}-4 x^{2} \cos \left (x \right )}{2 x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\cos \left (x \right )+x \sin \left (x \right )+1+\frac {y^{2}}{x}+2 y \cos \left (x \right )-2 y +\cos \left (x \right )^{2} x +x -2 \cos \left (x \right ) x \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {x^{2} \cos \left (2 x \right )+2 \sin \left (x \right ) x^{2}-4 x^{2} \cos \left (x \right )-2 \cos \left (x \right ) x +3 x^{2}+2 x}{2 x}\), \(f_1(x)=\frac {4 \cos \left (x \right ) x -4 x}{2 x}\) 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 {4 \cos \left (x \right ) x -4 x}{2 x^{2}}\\ f_2^2 f_0 &=\frac {x^{2} \cos \left (2 x \right )+2 \sin \left (x \right ) x^{2}-4 x^{2} \cos \left (x \right )-2 \cos \left (x \right ) x +3 x^{2}+2 x}{2 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 {4 \cos \left (x \right ) x -4 x}{2 x^{2}}\right ) u^{\prime }\left (x \right )+\frac {\left (x^{2} \cos \left (2 x \right )+2 \sin \left (x \right ) x^{2}-4 x^{2} \cos \left (x \right )-2 \cos \left (x \right ) x +3 x^{2}+2 x \right ) u \left (x \right )}{2 x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\sin \left (x \right )-x} \left (c_{1} +c_{2} \ln \left (x \right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (x \left (c_{1} +c_{2} \ln \left (x \right )\right ) \cos \left (x \right )-\ln \left (x \right ) c_{2} x -c_{1} x +c_{2} \right ) {\mathrm e}^{\sin \left (x \right )-x}}{x} \] Using the above in (1) gives the solution \[ y = -\frac {x \left (c_{1} +c_{2} \ln \left (x \right )\right ) \cos \left (x \right )-\ln \left (x \right ) c_{2} x -c_{1} x +c_{2}}{c_{1} +c_{2} \ln \left (x \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-x \left (-1+\cos \left (x \right )\right ) \ln \left (x \right )-\cos \left (x \right ) x c_{3} +c_{3} x -1}{c_{3} +\ln \left (x \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-x \left (-1+\cos \left (x \right )\right ) \ln \left (x \right )-\cos \left (x \right ) x c_{3} +c_{3} x -1}{c_{3} +\ln \left (x \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-x \left (-1+\cos \left (x \right )\right ) \ln \left (x \right )-\cos \left (x \right ) x c_{3} +c_{3} x -1}{c_{3} +\ln \left (x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \cos \left (2 x \right )+2 \sin \left (x \right ) x^{2}+4 y \cos \left (x \right ) x -4 x^{2} \cos \left (x \right )-2 \cos \left (x \right ) x -2 y^{\prime } x +2 y^{2}-4 x y+3 x^{2}+2 x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {2 \cos \left (x \right ) x -2 \sin \left (x \right ) x^{2}-2 x -2 y^{2}-4 y \cos \left (x \right ) x +4 x y-x^{2} \cos \left (2 x \right )-3 x^{2}+4 x^{2} \cos \left (x \right )}{2 x} \end {array} \]
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 trying Riccati trying Riccati sub-methods: <- Riccati particular case Kamke (d) successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 20
dsolve(diff(y(x),x) = 1/2*(-2*cos(x)*x+2*sin(x)*x^2+2*x+2*y(x)^2+4*y(x)*cos(x)*x-4*x*y(x)+x^2*cos(2*x)+3*x^2-4*x^2*cos(x))/x,y(x), singsol=all)
\[ y \left (x \right ) = -\left (\cos \left (x \right )-1\right ) x +\frac {1}{-\ln \left (x \right )+c_{1}} \]
✓ Solution by Mathematica
Time used: 0.433 (sec). Leaf size: 32
DSolve[y'[x] == (x + (3*x^2)/2 - x*Cos[x] - 2*x^2*Cos[x] + (x^2*Cos[2*x])/2 + x^2*Sin[x] - 2*x*y[x] + 2*x*Cos[x]*y[x] + y[x]^2)/x,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to x+x (-\cos (x))+\frac {1}{-\log (x)+c_1} \\ y(x)\to x-x \cos (x) \\ \end{align*}