Internal problem ID [9280]
Internal file name [OUTPUT/8216_Monday_June_06_2022_02_19_34_AM_89926573/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 947.
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]
\[ \boxed {y^{\prime }-\frac {2 x^{2} \cos \left (x \right )+2 x^{3} \sin \left (x \right )-2 x \sin \left (x \right )+2 x +2 y^{2} x^{2}-4 y \sin \left (x \right ) x +4 y \cos \left (x \right ) x^{2}+4 x y+3-\cos \left (2 x \right )-2 \sin \left (2 x \right ) x -4 \sin \left (x \right )+\cos \left (2 x \right ) x^{2}+x^{2}+4 x \cos \left (x \right )}{2 x^{3}}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {2 x^{2} \cos \left (x \right )+2 x^{3} \sin \left (x \right )-2 x \sin \left (x \right )+2 x +2 x^{2} y^{2}-4 x y \sin \left (x \right )+4 \cos \left (x \right ) y \,x^{2}+4 x y +3-\cos \left (2 x \right )-2 \sin \left (2 x \right ) x -4 \sin \left (x \right )+\cos \left (2 x \right ) x^{2}+x^{2}+4 x \cos \left (x \right )}{2 x^{3}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {\cos \left (x \right )}{x}+\sin \left (x \right )-\frac {\sin \left (x \right )}{x^{2}}+\frac {1}{x^{2}}+\frac {y^{2}}{x}-\frac {2 y \sin \left (x \right )}{x^{2}}+\frac {2 \cos \left (x \right ) y}{x}+\frac {2 y}{x^{2}}+\frac {2}{x^{3}}-\frac {\cos \left (x \right )^{2}}{x^{3}}-\frac {2 \sin \left (x \right ) \cos \left (x \right )}{x^{2}}-\frac {2 \sin \left (x \right )}{x^{3}}+\frac {\cos \left (x \right )^{2}}{x}+\frac {2 \cos \left (x \right )}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {2 x^{3} \sin \left (x \right )+\cos \left (2 x \right ) x^{2}+2 x^{2} \cos \left (x \right )-2 \sin \left (2 x \right ) x -2 x \sin \left (x \right )+4 x \cos \left (x \right )+x^{2}-\cos \left (2 x \right )-4 \sin \left (x \right )+2 x +3}{2 x^{3}}\), \(f_1(x)=\frac {4 x^{2} \cos \left (x \right )-4 x \sin \left (x \right )+4 x}{2 x^{3}}\) 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 x^{2} \cos \left (x \right )-4 x \sin \left (x \right )+4 x}{2 x^{4}}\\ f_2^2 f_0 &=\frac {2 x^{3} \sin \left (x \right )+\cos \left (2 x \right ) x^{2}+2 x^{2} \cos \left (x \right )-2 \sin \left (2 x \right ) x -2 x \sin \left (x \right )+4 x \cos \left (x \right )+x^{2}-\cos \left (2 x \right )-4 \sin \left (x \right )+2 x +3}{2 x^{5}} \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 x^{2} \cos \left (x \right )-4 x \sin \left (x \right )+4 x}{2 x^{4}}\right ) u^{\prime }\left (x \right )+\frac {\left (2 x^{3} \sin \left (x \right )+\cos \left (2 x \right ) x^{2}+2 x^{2} \cos \left (x \right )-2 \sin \left (2 x \right ) x -2 x \sin \left (x \right )+4 x \cos \left (x \right )+x^{2}-\cos \left (2 x \right )-4 \sin \left (x \right )+2 x +3\right ) u \left (x \right )}{2 x^{5}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {-1+\sin \left (x \right )}{x}} \left (c_{1} +\ln \left (x \right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {{\mathrm e}^{\frac {-1+\sin \left (x \right )}{x}} \left (c_{2} \left (x \cos \left (x \right )-\sin \left (x \right )+1\right ) \ln \left (x \right )+\cos \left (x \right ) c_{1} x +c_{2} x -\sin \left (x \right ) c_{1} +c_{1} \right )}{x^{2}} \] Using the above in (1) gives the solution \[ y = -\frac {c_{2} \left (x \cos \left (x \right )-\sin \left (x \right )+1\right ) \ln \left (x \right )+\cos \left (x \right ) c_{1} x +c_{2} x -\sin \left (x \right ) c_{1} +c_{1}}{x \left (c_{1} +\ln \left (x \right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-x \cos \left (x \right )+\sin \left (x \right )-1\right ) \ln \left (x \right )-\cos \left (x \right ) x c_{3} +\sin \left (x \right ) c_{3} -x -c_{3}}{\left (\ln \left (x \right )+c_{3} \right ) x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-x \cos \left (x \right )+\sin \left (x \right )-1\right ) \ln \left (x \right )-\cos \left (x \right ) x c_{3} +\sin \left (x \right ) c_{3} -x -c_{3}}{\left (\ln \left (x \right )+c_{3} \right ) x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-x \cos \left (x \right )+\sin \left (x \right )-1\right ) \ln \left (x \right )-\cos \left (x \right ) x c_{3} +\sin \left (x \right ) c_{3} -x -c_{3}}{\left (\ln \left (x \right )+c_{3} \right ) x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{2} x^{2}+4 y \cos \left (x \right ) x^{2}-2 y^{\prime } x^{3}+2 x^{3} \sin \left (x \right )-4 y \sin \left (x \right ) x +\cos \left (2 x \right ) x^{2}+2 x^{2} \cos \left (x \right )+4 x y-2 \sin \left (2 x \right ) x -2 x \sin \left (x \right )+4 x \cos \left (x \right )+x^{2}-\cos \left (2 x \right )-4 \sin \left (x \right )+2 x +3=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-2 x^{3} \sin \left (x \right )-4 y \cos \left (x \right ) x^{2}-2 y^{2} x^{2}-\cos \left (2 x \right ) x^{2}+4 y \sin \left (x \right ) x -2 x^{2} \cos \left (x \right )+2 \sin \left (2 x \right ) x +2 x \sin \left (x \right )-4 x \cos \left (x \right )-x^{2}-4 x y+\cos \left (2 x \right )+4 \sin \left (x \right )-2 x -3}{2 x^{3}} \end {array} \]
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: 44
dsolve(diff(y(x),x) = 1/2*(2*x^2*cos(x)+2*sin(x)*x^3-2*x*sin(x)+2*x+2*x^2*y(x)^2-4*y(x)*sin(x)*x+4*y(x)*cos(x)*x^2+4*x*y(x)+3-cos(2*x)-2*sin(2*x)*x-4*sin(x)+x^2*cos(2*x)+x^2+4*cos(x)*x)/x^3,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\cos \left (x \right ) x -\sin \left (x \right )+1\right ) \ln \left (x \right )-\cos \left (x \right ) c_{1} x +c_{1} \sin \left (x \right )+x -c_{1}}{x \left (-\ln \left (x \right )+c_{1} \right )} \]
✓ Solution by Mathematica
Time used: 0.631 (sec). Leaf size: 45
DSolve[y'[x] == (3/2 + x + x^2/2 + 2*x*Cos[x] + x^2*Cos[x] - Cos[2*x]/2 + (x^2*Cos[2*x])/2 - 2*Sin[x] - x*Sin[x] + x^3*Sin[x] - x*Sin[2*x] + 2*x*y[x] + 2*x^2*Cos[x]*y[x] - 2*x*Sin[x]*y[x] + x^2*y[x]^2)/x^3,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sin (x)-x \cos (x)-1}{x}+\frac {1}{-\log (x)+c_1} \\ y(x)\to \frac {\sin (x)-x \cos (x)-1}{x} \\ \end{align*}