Internal problem ID [9322]
Internal file name [OUTPUT/8258_Monday_June_06_2022_02_33_53_AM_30547686/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 989.
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]
\[ \boxed {y^{\prime }+F \left (x \right ) \left (-a y^{2}-b \,x^{2}\right )-\frac {y}{x}=0} \]
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} a +F \left (x \right ) b \,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} a +b \right ) \end {align*}
Where \(f(x)=F \left (x \right ) x\) and \(g(u)=u^{2} a +b\). Integrating both sides gives \begin{align*} \frac {1}{u^{2} a +b} \,du &= F \left (x \right ) x \,d x \\ \int { \frac {1}{u^{2} a +b} \,du} &= \int {F \left (x \right ) x \,d x} \\ \frac {\arctan \left (\frac {a u}{\sqrt {a b}}\right )}{\sqrt {a b}}&=\int F \left (x \right ) x d x +c_{2} \\ \end{align*} The solution is \[ \frac {\arctan \left (\frac {a u \left (x \right )}{\sqrt {a b}}\right )}{\sqrt {a b}}-\left (\int F \left (x \right ) x d x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {\arctan \left (\frac {a y}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\left (\int F \left (x \right ) x d x \right )-c_{2} = 0\\ \frac {\arctan \left (\frac {a y}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\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 {\arctan \left (\frac {a y}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\left (\int F \left (x \right ) x d x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {\arctan \left (\frac {a y}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\left (\int F \left (x \right ) x d x \right )-c_{2} = 0 \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {F \left (x \right ) a \,y^{2} x +F \left (x \right ) b \,x^{3}+y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = F \left (x \right ) a \,y^{2}+F \left (x \right ) b \,x^{2}+\frac {y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=F \left (x \right ) b \,x^{2}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=a F \left (x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a 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' &=a F^{\prime }\left (x \right )\\ f_1 f_2 &=\frac {a F \left (x \right )}{x}\\ f_2^2 f_0 &=a^{2} F \left (x \right )^{3} b \,x^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a F \left (x \right ) u^{\prime \prime }\left (x \right )-\left (a F^{\prime }\left (x \right )+\frac {a F \left (x \right )}{x}\right ) u^{\prime }\left (x \right )+a^{2} F \left (x \right )^{3} b \,x^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}+c_{2} {\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = i F \left (x \right ) x \sqrt {a b}\, \left (c_{1} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}-c_{2} {\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {i x \sqrt {a b}\, \left (c_{1} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}-c_{2} {\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}\right )}{a \left (c_{1} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}+c_{2} {\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {i x \sqrt {a b}\, \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}-{\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}\right )}{a \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}+{\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {i x \sqrt {a b}\, \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}-{\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}\right )}{a \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}+{\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {i x \sqrt {a b}\, \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}-{\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}\right )}{a \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}+{\mathrm e}^{-i \sqrt {a b}\, \left (\int F \left (x \right ) x d x \right )}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & F \left (x \right ) y^{2} a x +F \left (x \right ) b \,x^{3}-y^{\prime } x +y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {F \left (x \right ) y^{2} a x +F \left (x \right ) b \,x^{3}+y}{x} \end {array} \]
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)*(-a*y(x)^2-b*x^2)+y(x)/x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\tan \left (\left (\int F \left (x \right ) x d x +c_{1} \right ) \sqrt {a b}\right ) x \sqrt {a b}}{a} \]
✓ Solution by Mathematica
Time used: 0.289 (sec). Leaf size: 45
DSolve[y'[x] == y[x]/x - F[x]*(-(b*x^2) - a*y[x]^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\sqrt {b} x \tan \left (\sqrt {a} \sqrt {b} \left (\int _1^xF(K[1]) K[1]dK[1]+c_1\right )\right )}{\sqrt {a}} \]