Internal problem ID [9324]
Internal file name [OUTPUT/8260_Monday_June_06_2022_02_34_18_AM_35500280/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 991.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati", "homogeneousTypeD2"
Maple gives the following as the ode type
[[_homogeneous, `class D`], _Riccati]
\[ \boxed {y^{\prime }+F \left (x \right ) \left (-y^{2}+2 y x +x^{2}\right )-\frac {y}{x}=0} \]
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} F \left (x \right ) u \left (x \right )^{2} x^{3}-2 F \left (x \right ) u \left (x \right ) x^{3}-F \left (x \right ) x^{3}-\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x +u \left (x \right ) x = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= F \left (x \right ) x \left (u^{2}-2 u -1\right ) \end {align*}
Where \(f(x)=F \left (x \right ) x\) and \(g(u)=u^{2}-2 u -1\). Integrating both sides gives \begin{align*} \frac {1}{u^{2}-2 u -1} \,du &= F \left (x \right ) x \,d x \\ \int { \frac {1}{u^{2}-2 u -1} \,du} &= \int {F \left (x \right ) x \,d x} \\ -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2 u -2\right ) \sqrt {2}}{4}\right )}{2}&=\int F \left (x \right ) x d x +c_{2} \\ \end{align*} The solution is \[ -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2 u \left (x \right )-2\right ) \sqrt {2}}{4}\right )}{2}-\left (\int F \left (x \right ) x d x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (\frac {2 y}{x}-2\right ) \sqrt {2}}{4}\right )}{2}-\left (\int F \left (x \right ) x d x \right )-c_{2} = 0\\ -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (y-x \right ) \sqrt {2}}{2 x}\right )}{2}-\left (\int F \left (x \right ) x d x \right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (y-x \right ) \sqrt {2}}{2 x}\right )}{2}-\left (\int F \left (x \right ) x d x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (y-x \right ) \sqrt {2}}{2 x}\right )}{2}-\left (\int F \left (x \right ) x d x \right )-c_{2} = 0 \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-F \left (x \right ) x^{3}-2 F \left (x \right ) x^{2} y +F \left (x \right ) y^{2} x +y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -F \left (x \right ) x^{2}-2 F \left (x \right ) y x +F \left (x \right ) y^{2}+\frac {y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-F \left (x \right ) x^{2}\), \(f_1(x)=\frac {-2 F \left (x \right ) x^{2}+1}{x}\) and \(f_2(x)=F \left (x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{F \left (x \right ) 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' &=F^{\prime }\left (x \right )\\ f_1 f_2 &=\frac {\left (-2 F \left (x \right ) x^{2}+1\right ) F \left (x \right )}{x}\\ f_2^2 f_0 &=-F \left (x \right )^{3} x^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} F \left (x \right ) u^{\prime \prime }\left (x \right )-\left (F^{\prime }\left (x \right )+\frac {\left (-2 F \left (x \right ) x^{2}+1\right ) F \left (x \right )}{x}\right ) u^{\prime }\left (x \right )-F \left (x \right )^{3} x^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\left (\int F \left (x \right ) x d x \right ) \left (1+\sqrt {2}\right )} \left (c_{2} {\mathrm e}^{2 \left (\int F \left (x \right ) x d x \right ) \sqrt {2}}+c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -x \left (c_{1} \left (1+\sqrt {2}\right ) {\mathrm e}^{-\left (\int F \left (x \right ) x d x \right ) \left (1+\sqrt {2}\right )}-c_{2} {\mathrm e}^{\left (\int F \left (x \right ) x d x \right ) \left (\sqrt {2}-1\right )} \left (\sqrt {2}-1\right )\right ) F \left (x \right ) \] Using the above in (1) gives the solution \[ y = \frac {x \left (c_{1} \left (1+\sqrt {2}\right ) {\mathrm e}^{-\left (\int F \left (x \right ) x d x \right ) \left (1+\sqrt {2}\right )}-c_{2} {\mathrm e}^{\left (\int F \left (x \right ) x d x \right ) \left (\sqrt {2}-1\right )} \left (\sqrt {2}-1\right )\right ) {\mathrm e}^{\int F \left (x \right ) x \left (1+\sqrt {2}\right )d x}}{c_{2} {\mathrm e}^{2 \left (\int F \left (x \right ) x d x \right ) \sqrt {2}}+c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x \left (-c_{2} {\mathrm e}^{2 \left (\int F \left (x \right ) x d x \right ) \sqrt {2}} \left (\sqrt {2}-1\right )+c_{1} \left (1+\sqrt {2}\right )\right )}{c_{2} {\mathrm e}^{2 \left (\int F \left (x \right ) x d x \right ) \sqrt {2}}+c_{1}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x \left (-c_{2} {\mathrm e}^{2 \left (\int F \left (x \right ) x d x \right ) \sqrt {2}} \left (\sqrt {2}-1\right )+c_{1} \left (1+\sqrt {2}\right )\right )}{c_{2} {\mathrm e}^{2 \left (\int F \left (x \right ) x d x \right ) \sqrt {2}}+c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \frac {x \left (-c_{2} {\mathrm e}^{2 \left (\int F \left (x \right ) x d x \right ) \sqrt {2}} \left (\sqrt {2}-1\right )+c_{1} \left (1+\sqrt {2}\right )\right )}{c_{2} {\mathrm e}^{2 \left (\int F \left (x \right ) x d x \right ) \sqrt {2}}+c_{1}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & F \left (x \right ) y^{2} x -2 F \left (x \right ) y x^{2}-F \left (x \right ) x^{3}-y^{\prime } x +y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {F \left (x \right ) x^{3}+2 F \left (x \right ) y x^{2}-F \left (x \right ) y^{2} x -y}{x} \end {array} \]
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 homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 29
dsolve(diff(y(x),x) = -F(x)*(x^2+2*x*y(x)-y(x)^2)+y(x)/x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x \left (\sqrt {2}-2 \tanh \left (\left (\int F \left (x \right ) x d x +c_{1} \right ) \sqrt {2}\right )\right ) \sqrt {2}}{2} \]
✓ Solution by Mathematica
Time used: 0.56 (sec). Leaf size: 104
DSolve[y'[x] == y[x]/x - F[x]*(x^2 + 2*x*y[x] - y[x]^2),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x \left (-\left (\sqrt {2}-1\right ) \exp \left (2 \sqrt {2} \left (\int _1^xF(K[1]) K[1]dK[1]+c_1\right )\right )+1+\sqrt {2}\right )}{1+\exp \left (2 \sqrt {2} \left (\int _1^xF(K[1]) K[1]dK[1]+c_1\right )\right )} \\ y(x)\to \left (1+\sqrt {2}\right ) x \\ y(x)\to x-\sqrt {2} x \\ \end{align*}