Internal problem ID [9319]
Internal file name [OUTPUT/8255_Monday_June_06_2022_02_32_50_AM_2915151/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 986.
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]
\[ \boxed {y^{\prime }-\frac {y^{3}-3 x y^{2} \ln \left (x \right )+3 x^{2} \ln \left (x \right )^{2} y-x^{3} \ln \left (x \right )^{3}+x^{2}+x y}{x^{2}}=0} \]
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 {3 \ln \left (x \right ) y^{2}}{x}-\frac {\left (-3 x^{2} \ln \left (x \right )^{2}-x \right ) y}{x^{2}}-\frac {x^{3} \ln \left (x \right )^{3}-x^{2}}{x^{2}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -x \ln \left (x \right )^{3}+1\\ f_1(x) &= 3 \ln \left (x \right )^{2}+\frac {1}{x}\\ f_2(x) &= -\frac {3 \ln \left (x \right )}{x}\\ f_3(x) &= \frac {1}{x^{2}} \end {align*}
Since \(f_2(x)=-\frac {3 \ln \left (x \right )}{x}\) 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 {3}{x^{2}}} \right ) \\ &= u \left (x \right )+\ln \left (x \right ) x \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 )}{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*} u^{\prime }\left (x \right ) x -u \left (x \right )^{3} x = 0 \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{u^{3}}d u &= x +c_{4}\\ -\frac {1}{2 u^{2}}&=x +c_{4} \end {align*}
Solving for \(u\) gives these solutions \begin {align*} u_1&=\frac {1}{\sqrt {-2 c_{4} -2 x}}\\ u_2&=-\frac {1}{\sqrt {-2 c_{4} -2 x}} \end {align*}
Therefore the solution \(u \left (x \right )\) is \begin {align*} u \left (x \right )&=x u\\ &=\frac {x}{\sqrt {-2 c_{4} -2 x}} \end {align*}
Now we transform the solution \(u \left (x \right ) = \frac {x}{\sqrt {-2 c_{4} -2 x}}\) to \(y\) using \(y = u(x) - \frac {f_2}{3 f_3}\), which gives \begin {align*} y &= \frac {x}{\sqrt {-2 c_{4} -2 x}} - \left (-\ln \left (x \right ) x\right ) \\ &= \frac {x}{\sqrt {-2 c_{4} -2 x}}+3 \ln \left (x \right ) x\\ &= x \left (\frac {1}{\sqrt {-2 c_{4} -2 x}}+3 \ln \left (x \right )\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x \left (\frac {1}{\sqrt {-2 c_{4} -2 x}}+3 \ln \left (x \right )\right ) \\ \end{align*}
Verification of solutions
\[ y = x \left (\frac {1}{\sqrt {-2 c_{4} -2 x}}+3 \ln \left (x \right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \ln \left (x \right )^{3}-3 x^{2} \ln \left (x \right )^{2} y+3 x y^{2} \ln \left (x \right )+y^{\prime } x^{2}-y^{3}-x y-x^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{3}-3 x y^{2} \ln \left (x \right )+3 x^{2} \ln \left (x \right )^{2} y-x^{3} \ln \left (x \right )^{3}+x^{2}+x y}{x^{2}} \end {array} \]
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: 0.015 (sec). Leaf size: 33
dsolve(diff(y(x),x) = (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+x*y(x))/x^2,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= x \left (-\frac {1}{\sqrt {-2 x +c_{1}}}+\ln \left (x \right )\right ) \\ y \left (x \right ) &= x \left (\frac {1}{\sqrt {-2 x +c_{1}}}+\ln \left (x \right )\right ) \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.25 (sec). Leaf size: 49
DSolve[y'[x] == (x^2 - x^3*Log[x]^3 + x*y[x] + 3*x^2*Log[x]^2*y[x] - 3*x*Log[x]*y[x]^2 + y[x]^3)/x^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to x \left (\log (x)-\frac {1}{\sqrt {-2 x+c_1}}\right ) \\ y(x)\to x \left (\log (x)+\frac {1}{\sqrt {-2 x+c_1}}\right ) \\ y(x)\to x \log (x) \\ \end{align*}