Internal problem ID [9011]
Internal file name [OUTPUT/7946_Monday_June_06_2022_12_59_34_AM_96518345/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 677.
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]
\[ \boxed {y^{\prime }-\frac {y+x^{3} a \ln \left (x +1\right )+a \,x^{4}+a \,x^{3}-x y^{2} \ln \left (x +1\right )-y^{2} x^{2}-y^{2} x}{x}=0} \]
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*} -x^{3} a \ln \left (x +1\right )-a \,x^{4}+x^{3} u \left (x \right )^{2} \ln \left (x +1\right )+u \left (x \right )^{2} x^{4}-a \,x^{3}+u \left (x \right )^{2} 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)\\ &= x \left (-u^{2}+a \right ) \left (\ln \left (x +1\right )+x +1\right ) \end {align*}
Where \(f(x)=x \left (\ln \left (x +1\right )+x +1\right )\) and \(g(u)=-u^{2}+a\). Integrating both sides gives \begin{align*} \frac {1}{-u^{2}+a} \,du &= x \left (\ln \left (x +1\right )+x +1\right ) \,d x \\ \int { \frac {1}{-u^{2}+a} \,du} &= \int {x \left (\ln \left (x +1\right )+x +1\right ) \,d x} \\ \frac {\operatorname {arctanh}\left (\frac {u}{\sqrt {a}}\right )}{\sqrt {a}}&=\frac {x^{3}}{3}+\frac {x^{2}}{4}+\frac {x}{2}+\frac {7}{12}+\frac {\left (x +1\right )^{2} \ln \left (x +1\right )}{2}-\left (x +1\right ) \ln \left (x +1\right )+c_{2} \\ \end{align*} The solution is \[ \frac {\operatorname {arctanh}\left (\frac {u \left (x \right )}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {x^{3}}{3}-\frac {x^{2}}{4}-\frac {x}{2}-\frac {7}{12}-\frac {\left (x +1\right )^{2} \ln \left (x +1\right )}{2}+\left (x +1\right ) \ln \left (x +1\right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {\operatorname {arctanh}\left (\frac {y}{x \sqrt {a}}\right )}{\sqrt {a}}-\frac {x^{3}}{3}-\frac {x^{2}}{4}-\frac {x}{2}-\frac {7}{12}-\frac {\left (x +1\right )^{2} \ln \left (x +1\right )}{2}+\left (x +1\right ) \ln \left (x +1\right )-c_{2} = 0\\ -\frac {-3 \,\operatorname {arctanh}\left (\frac {y}{x \sqrt {a}}\right )+\sqrt {a}\, \left (\frac {7}{4}+\frac {3 \left (x^{2}-1\right ) \ln \left (x +1\right )}{2}+x^{3}+\frac {3 x^{2}}{4}+\frac {3 x}{2}+3 c_{2} \right )}{3 \sqrt {a}} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {-3 \,\operatorname {arctanh}\left (\frac {y}{x \sqrt {a}}\right )+\sqrt {a}\, \left (\frac {7}{4}+\frac {3 \left (x^{2}-1\right ) \ln \left (x +1\right )}{2}+x^{3}+\frac {3 x^{2}}{4}+\frac {3 x}{2}+3 c_{2} \right )}{3 \sqrt {a}} &= 0 \\ \end{align*}
Verification of solutions
\[ -\frac {-3 \,\operatorname {arctanh}\left (\frac {y}{x \sqrt {a}}\right )+\sqrt {a}\, \left (\frac {7}{4}+\frac {3 \left (x^{2}-1\right ) \ln \left (x +1\right )}{2}+x^{3}+\frac {3 x^{2}}{4}+\frac {3 x}{2}+3 c_{2} \right )}{3 \sqrt {a}} = 0 \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {-y -x^{3} a \ln \left (x +1\right )-a \,x^{4}-a \,x^{3}+x \,y^{2} \ln \left (x +1\right )+x^{2} y^{2}+x \,y^{2}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{2} a \ln \left (x +1\right )+a \,x^{3}-y^{2} \ln \left (x +1\right )+a \,x^{2}-x \,y^{2}-y^{2}+\frac {y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {-x^{3} a \ln \left (x +1\right )-a \,x^{4}-a \,x^{3}}{x}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=-\frac {x \ln \left (x +1\right )+x^{2}+x}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {\left (x \ln \left (x +1\right )+x^{2}+x \right ) 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 {\ln \left (x +1\right )+\frac {x}{x +1}+2 x +1}{x}+\frac {x \ln \left (x +1\right )+x^{2}+x}{x^{2}}\\ f_1 f_2 &=-\frac {x \ln \left (x +1\right )+x^{2}+x}{x^{2}}\\ f_2^2 f_0 &=-\frac {\left (x \ln \left (x +1\right )+x^{2}+x \right )^{2} \left (-x^{3} a \ln \left (x +1\right )-a \,x^{4}-a \,x^{3}\right )}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {\left (x \ln \left (x +1\right )+x^{2}+x \right ) u^{\prime \prime }\left (x \right )}{x}+\frac {\left (\ln \left (x +1\right )+\frac {x}{x +1}+2 x +1\right ) u^{\prime }\left (x \right )}{x}-\frac {\left (x \ln \left (x +1\right )+x^{2}+x \right )^{2} \left (-x^{3} a \ln \left (x +1\right )-a \,x^{4}-a \,x^{3}\right ) u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}+c_{2} {\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = i \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}\, \left (c_{1} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}-c_{2} {\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}\right ) \] Using the above in (1) gives the solution \[ y = \frac {i \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}\, \left (c_{1} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}-c_{2} {\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}\right ) x}{\left (x \ln \left (x +1\right )+x^{2}+x \right ) \left (c_{1} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}+c_{2} {\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {i \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}\, \left (c_{3} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}-{\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}\right )}{\left (\ln \left (x +1\right )+x +1\right ) \left (c_{3} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}+{\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {i \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}\, \left (c_{3} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}-{\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}\right )}{\left (\ln \left (x +1\right )+x +1\right ) \left (c_{3} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}+{\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {i \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}\, \left (c_{3} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}-{\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}\right )}{\left (\ln \left (x +1\right )+x +1\right ) \left (c_{3} {\mathrm e}^{i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}+{\mathrm e}^{-i \left (\int \sqrt {-x^{2} \left (\ln \left (x +1\right )+x +1\right )^{2} a}d x \right )}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -x^{3} a \ln \left (x +1\right )-a \,x^{4}+x y^{2} \ln \left (x +1\right )+y^{2} x^{2}-a \,x^{3}+y^{2} x +y^{\prime } x -y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y+x^{3} a \ln \left (x +1\right )+a \,x^{4}+a \,x^{3}-x y^{2} \ln \left (x +1\right )-y^{2} x^{2}-y^{2} x}{x} \end {array} \]
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: 0.015 (sec). Leaf size: 48
dsolve(diff(y(x),x) = (y(x)+x^3*a*ln(x+1)+a*x^4+x^3*a-x*y(x)^2*ln(x+1)-x^2*y(x)^2-x*y(x)^2)/x,y(x), singsol=all)
\[ y \left (x \right ) = \tanh \left (\frac {\sqrt {a}\, \left (6 \ln \left (x +1\right ) x^{2}+4 x^{3}+3 x^{2}-6 \ln \left (x +1\right )+12 c_{1} +6 x +9\right )}{12}\right ) \sqrt {a}\, x \]
✓ Solution by Mathematica
Time used: 11.983 (sec). Leaf size: 51
DSolve[y'[x] == (a*x^3 + a*x^4 + a*x^3*Log[1 + x] + y[x] - x*y[x]^2 - x^2*y[x]^2 - x*Log[1 + x]*y[x]^2)/x,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {a} x \tanh \left (\frac {1}{12} \sqrt {a} \left (4 x^3+3 x^2+6 \left (x^2-1\right ) \log (x+1)+6 x+12 c_1\right )\right ) \]