Internal problem ID [9317]
Internal file name [OUTPUT/8253_Monday_June_06_2022_02_32_14_AM_17275354/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 984.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(y)]`], _Abel]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\frac {y \left (x^{2} y^{2}+y x \,{\mathrm e}^{x}+{\mathrm e}^{2 x}\right ) {\mathrm e}^{-2 x} \left (x -1\right )}{x}=0} \]
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 {\left (x^{3}-x^{2}\right ) {\mathrm e}^{-2 x} y^{3}}{x}+\frac {\left ({\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x}\right ) {\mathrm e}^{-2 x} y^{2}}{x}+\frac {\left ({\mathrm e}^{2 x} x -{\mathrm e}^{2 x}\right ) {\mathrm e}^{-2 x} y}{x}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= 1-\frac {1}{x}\\ f_2(x) &= x \,{\mathrm e}^{-x}-{\mathrm e}^{-x}\\ f_3(x) &= {\mathrm e}^{-2 x} x^{2}-{\mathrm e}^{-2 x} x \end {align*}
Since \(f_2(x)=x \,{\mathrm e}^{-x}-{\mathrm e}^{-x}\) 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 {x \,{\mathrm e}^{-x}-{\mathrm e}^{-x}}{3 \,{\mathrm e}^{-2 x} x^{2}-3 \,{\mathrm e}^{-2 x} x} \right ) \\ &= \frac {3 u \left (x \right ) x -{\mathrm e}^{x}}{3 x} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = {\mathrm e}^{-2 x} x^{2} u \left (x \right )^{3}-{\mathrm e}^{-2 x} x u \left (x \right )^{3}+\frac {2 u \left (x \right )}{3}-\frac {2 u \left (x \right )}{3 x}+\frac {2 \,{\mathrm e}^{x}}{27 x}-\frac {2 \,{\mathrm e}^{x}}{27 x^{2}}\tag {2} \end {align*}
This is Abel first kind ODE, it has the form \[ u^{\prime }\left (x \right )= f_0(x)+f_1(x) u \left (x \right ) +f_2(x)u \left (x \right )^{2}+f_3(x)u \left (x \right )^{3} \] Comparing the above to given ODE which is \begin {align*} u^{\prime }\left (x \right )&=\frac {\left (27 x^{4}-27 x^{3}\right ) {\mathrm e}^{-2 x} u \left (x \right )^{3}}{27 x^{2}}+\frac {\left (18 \,{\mathrm e}^{2 x} x^{2}-18 \,{\mathrm e}^{2 x} x \right ) {\mathrm e}^{-2 x} u \left (x \right )}{27 x^{2}}+\frac {\left (2 \,{\mathrm e}^{3 x} x -2 \,{\mathrm e}^{3 x}\right ) {\mathrm e}^{-2 x}}{27 x^{2}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {2 \,{\mathrm e}^{x}}{27 x}-\frac {2 \,{\mathrm e}^{x}}{27 x^{2}}\\ f_1(x) &= \frac {2}{3}-\frac {2}{3 x}\\ f_2(x) &= 0\\ f_3(x) &= {\mathrm e}^{-2 x} x^{2}-{\mathrm e}^{-2 x} x \end {align*}
Since \(f_2(x)=0\) then we check the Abel invariant to see if it depends on \(x\) or not. The Abel invariant is given by \begin {align*} -\frac {f_{1}^{3}}{f_{0}^{2} f_{3}} \end {align*}
Which when evaluating gives \begin {align*} -\frac {{\left (-\left (\frac {2 \,{\mathrm e}^{x}}{27 x}-\frac {4 \,{\mathrm e}^{x}}{27 x^{2}}+\frac {4 \,{\mathrm e}^{x}}{27 x^{3}}\right ) \left ({\mathrm e}^{-2 x} x^{2}-{\mathrm e}^{-2 x} x \right )+\left (\frac {2 \,{\mathrm e}^{x}}{27 x}-\frac {2 \,{\mathrm e}^{x}}{27 x^{2}}\right ) \left (-2 \,{\mathrm e}^{-2 x} x^{2}+4 \,{\mathrm e}^{-2 x} x -{\mathrm e}^{-2 x}\right )+3 \left (\frac {2 \,{\mathrm e}^{x}}{27 x}-\frac {2 \,{\mathrm e}^{x}}{27 x^{2}}\right ) \left ({\mathrm e}^{-2 x} x^{2}-{\mathrm e}^{-2 x} x \right ) \left (\frac {2}{3}-\frac {2}{3 x}\right )\right )}^{3}}{27 \left ({\mathrm e}^{-2 x} x^{2}-{\mathrm e}^{-2 x} x \right )^{4} \left (\frac {2 \,{\mathrm e}^{x}}{27 x}-\frac {2 \,{\mathrm e}^{x}}{27 x^{2}}\right )^{5}} \end {align*}
Since the Abel invariant depends on \(x\) then unable to solve this ode at this time.
Unable to complete the solution now.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -x^{3} y^{3}+y^{3} x^{2}-y^{2} {\mathrm e}^{x} x^{2}+y^{\prime } x \left ({\mathrm e}^{x}\right )^{2}+y^{2} {\mathrm e}^{x} x -y \left ({\mathrm e}^{x}\right )^{2} x +y \left ({\mathrm e}^{x}\right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{3} y^{3}-y^{3} x^{2}+y^{2} {\mathrm e}^{x} x^{2}-y^{2} {\mathrm e}^{x} x +y \left ({\mathrm e}^{x}\right )^{2} x -y \left ({\mathrm e}^{x}\right )^{2}}{\left ({\mathrm e}^{x}\right )^{2} x} \end {array} \]
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 Chini differential order: 1; looking for linear symmetries trying exact trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 44
dsolve(diff(y(x),x) = y(x)/x*(x^2*y(x)^2+y(x)*x*exp(x)+exp(x)^2)/exp(x)^2*(x-1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{\operatorname {RootOf}\left ({\mathrm e}^{\textit {\_Z}} \ln \left (2\right )-{\mathrm e}^{\textit {\_Z}} \ln \left (\left ({\mathrm e}^{\textit {\_Z}}+9\right ) x \right )+3 c_{1} {\mathrm e}^{\textit {\_Z}}+{\mathrm e}^{\textit {\_Z}} \textit {\_Z} +x \,{\mathrm e}^{\textit {\_Z}}+9\right )+x}}{9 x} \]
✓ Solution by Mathematica
Time used: 7.806 (sec). Leaf size: 428
DSolve[y'[x] == ((-1 + x)*y[x]*(E^(2*x) + E^x*x*y[x] + x^2*y[x]^2))/(E^(2*x)*x),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\frac {\sqrt [3]{2} \left (\frac {3 e^{-2 x} x (x-1) y(x)+e^{-x} (x-1)}{\sqrt [3]{2} \sqrt [3]{e^{-3 x} (x-1)^3}}+2^{2/3}\right ) \left (2^{2/3}-\frac {2^{2/3} \left (3 e^{-2 x} x (x-1) y(x)+e^{-x} (x-1)\right )}{\sqrt [3]{e^{-3 x} (x-1)^3}}\right ) \left (\left (1-\frac {3 e^{-2 x} x (x-1) y(x)+e^{-x} (x-1)}{\sqrt [3]{e^{-3 x} (x-1)^3}}\right ) \log \left (2^{2/3}-\frac {2^{2/3} \left (3 e^{-2 x} x (x-1) y(x)+e^{-x} (x-1)\right )}{\sqrt [3]{e^{-3 x} (x-1)^3}}\right )+\left (\frac {3 e^{-2 x} x (x-1) y(x)+e^{-x} (x-1)}{\sqrt [3]{e^{-3 x} (x-1)^3}}-1\right ) \log \left (2 \left (\frac {3 e^{-2 x} x (x-1) y(x)+e^{-x} (x-1)}{\sqrt [3]{2} \sqrt [3]{e^{-3 x} (x-1)^3}}+2^{2/3}\right )\right )-3\right )}{9 \left (-\frac {e^{3 x} \left (3 e^{-2 x} x (x-1) y(x)+e^{-x} (x-1)\right )^3}{(x-1)^3}+\frac {3 \left (3 e^{-2 x} x (x-1) y(x)+e^{-x} (x-1)\right )}{\sqrt [3]{e^{-3 x} (x-1)^3}}-2\right )}=\frac {2^{2/3} e^{-x} (x-1) (x-\log (x))}{9 \sqrt [3]{e^{-3 x} (x-1)^3}}+c_1,y(x)\right ] \]