Internal problem ID [9266]
Internal file name [OUTPUT/8202_Monday_June_06_2022_02_15_53_AM_43814739/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 933.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind"
Maple gives the following as the ode type
[_Abel]
\[ \boxed {y^{\prime }+\frac {-x^{2}-x y-x^{3}-x y^{2}+2 y x^{2} \ln \left (x \right )-x^{3} \ln \left (x \right )^{2}-y^{3}+3 x y^{2} \ln \left (x \right )-3 x^{2} \ln \left (x \right )^{2} y+x^{3} \ln \left (x \right )^{3}}{x^{2}}=0} \]
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=\frac {y^{3}}{x^{2}}-\frac {\left (3 \ln \left (x \right ) x -x \right ) y^{2}}{x^{2}}-\frac {\left (-3 x^{2} \ln \left (x \right )^{2}+2 x^{2} \ln \left (x \right )-x \right ) y}{x^{2}}-\frac {x^{3} \ln \left (x \right )^{3}-x^{3} \ln \left (x \right )^{2}-x^{3}-x^{2}}{x^{2}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -x \ln \left (x \right )^{3}+x \ln \left (x \right )^{2}+x +1\\ f_1(x) &= 3 \ln \left (x \right )^{2}-2 \ln \left (x \right )+\frac {1}{x}\\ f_2(x) &= -\frac {3 \ln \left (x \right )}{x}+\frac {1}{x}\\ f_3(x) &= \frac {1}{x^{2}} \end {align*}
Since \(f_2(x)=-\frac {3 \ln \left (x \right )}{x}+\frac {1}{x}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {-\frac {3 \ln \left (x \right )}{x}+\frac {1}{x}}{\frac {3}{x^{2}}} \right ) \\ &= \ln \left (x \right ) x +u \left (x \right )-\frac {x}{3} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{3}}{x^{2}}-\frac {u \left (x \right )}{3}+\frac {29 x}{27}+\frac {u \left (x \right )}{x}\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
Using the change of variables \(u \left (x \right ) = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} 27 u \left (x \right )^{3} x^{3}-9 u \left (x \right ) x^{3}-27 \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2}+29 x^{3}+27 u \left (x \right ) x^{2} = 0 \end {align*}
Integrating both sides gives \begin {align*} \int _{}^{u \left (x \right )}\frac {1}{\textit {\_a}^{3}-\frac {1}{3} \textit {\_a} +\frac {29}{27}}d \textit {\_a} = x +c_{4} \end {align*}
Replacing \(u(x)\) in the above solution by \(\frac {u \left (x \right )}{x}\) results in the solution for \(u \left (x \right )\) in implicit form \begin {align*} \int _{}^{\frac {u \left (x \right )}{x}}\frac {1}{\textit {\_a}^{3}-\frac {1}{3} \textit {\_a} +\frac {29}{27}}d \textit {\_a} = x +c_{4}\\ 27 \left (\int _{}^{\frac {u \left (x \right )}{x}}\frac {1}{27 \textit {\_a}^{3}-9 \textit {\_a} +29}d \textit {\_a} \right ) = x +c_{4} \end {align*}
Substituting \(u=y-\frac {\left (-\frac {3 \ln \left (x \right )}{x}+\frac {1}{x}\right ) x^{2}}{3}\) in the above solution gives \begin {align*} 27 \left (\int _{}^{\frac {y-\frac {\left (-\frac {3 \ln \left (x \right )}{x}+\frac {1}{x}\right ) x^{2}}{3}}{x}}\frac {1}{27 \textit {\_a}^{3}-9 \textit {\_a} +29}d \textit {\_a} \right ) = x +c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} 27 \left (\int _{}^{\frac {y-\frac {\left (-\frac {3 \ln \left (x \right )}{x}+\frac {1}{x}\right ) x^{2}}{3}}{x}}\frac {1}{27 \textit {\_a}^{3}-9 \textit {\_a} +29}d \textit {\_a} \right ) &= x +c_{4} \\ \end{align*}
Verification of solutions
\[ 27 \left (\int _{}^{\frac {y-\frac {\left (-\frac {3 \ln \left (x \right )}{x}+\frac {1}{x}\right ) x^{2}}{3}}{x}}\frac {1}{27 \textit {\_a}^{3}-9 \textit {\_a} +29}d \textit {\_a} \right ) = x +c_{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \ln \left (x \right )^{3}-3 x^{2} \ln \left (x \right )^{2} y-x^{3} \ln \left (x \right )^{2}+3 x y^{2} \ln \left (x \right )+2 y x^{2} \ln \left (x \right )+y^{\prime } x^{2}-y^{3}-x y^{2}-x^{3}-x y-x^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-x^{3} \ln \left (x \right )^{3}+3 x^{2} \ln \left (x \right )^{2} y+x^{3} \ln \left (x \right )^{2}-3 x y^{2} \ln \left (x \right )-2 y x^{2} \ln \left (x \right )+y^{3}+x y^{2}+x^{3}+x y+x^{2}}{x^{2}} \end {array} \]
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 trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 39
dsolve(diff(y(x),x) = -(-x^2-x*y(x)-x^3-x*y(x)^2+2*y(x)*x^2*ln(x)-x^3*ln(x)^2-y(x)^3+3*x*y(x)^2*ln(x)-3*x^2*ln(x)^2*y(x)+x^3*ln(x)^3)/x^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x \left (9 \ln \left (x \right )-3+29 \operatorname {RootOf}\left (-81 \left (\int _{}^{\textit {\_Z}}\frac {1}{841 \textit {\_a}^{3}-27 \textit {\_a} +27}d \textit {\_a} \right )+x +3 c_{1} \right )\right )}{9} \]
✓ Solution by Mathematica
Time used: 1.197 (sec). Leaf size: 99
DSolve[y'[x] == (x^2 + x^3 + x^3*Log[x]^2 - x^3*Log[x]^3 + x*y[x] - 2*x^2*Log[x]*y[x] + 3*x^2*Log[x]^2*y[x] + x*y[x]^2 - 3*x*Log[x]*y[x]^2 + y[x]^3)/x^2,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {29}{3} \text {RootSum}\left [-29 \text {$\#$1}^3+3 \sqrt [3]{29} \text {$\#$1}-29\&,\frac {\log \left (\frac {\frac {3 y(x)}{x^2}+\frac {1-3 \log (x)}{x}}{\sqrt [3]{29} \sqrt [3]{\frac {1}{x^3}}}-\text {$\#$1}\right )}{\sqrt [3]{29}-29 \text {$\#$1}^2}\&\right ]=\frac {29^{2/3}}{9 \sqrt [3]{\frac {1}{x^3}}}+c_1,y(x)\right ] \]