Internal problem ID [9310]
Internal file name [OUTPUT/8246_Monday_June_06_2022_02_30_04_AM_65715333/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 977.
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.
\[ \boxed {y^{\prime }-y \left (y^{2}+{\mathrm e}^{-x^{2}} y+{\mathrm e}^{-2 x^{2}}\right ) {\mathrm e}^{2 x^{2}} x=0} \]
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 }&=y^{3} x \,{\mathrm e}^{2 x^{2}}+{\mathrm e}^{x^{2}} y^{2} x +y x\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= x\\ f_2(x) &= x \,{\mathrm e}^{x^{2}}\\ f_3(x) &= x \,{\mathrm e}^{2 x^{2}} \end {align*}
Since \(f_2(x)=x \,{\mathrm e}^{x^{2}}\) 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^{2}}}{3 x \,{\mathrm e}^{2 x^{2}}} \right ) \\ &= u \left (x \right )-\frac {{\mathrm e}^{-x^{2}}}{3} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {2 x u \left (x \right )}{3}+x \,{\mathrm e}^{2 x^{2}} u \left (x \right )^{3}-\frac {25 x \,{\mathrm e}^{-x^{2}}}{27}\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 {2 x u \left (x \right )}{3}+x \,{\mathrm e}^{2 x^{2}} u \left (x \right )^{3}-\frac {25 x \,{\mathrm e}^{-x^{2}}}{27}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -\frac {25 x \,{\mathrm e}^{-x^{2}}}{27}\\ f_1(x) &= \frac {2 x}{3}\\ f_2(x) &= 0\\ f_3(x) &= x \,{\mathrm e}^{2 x^{2}} \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 {531441 {\left (-\left (-\frac {25 \,{\mathrm e}^{-x^{2}}}{27}+\frac {50 x^{2} {\mathrm e}^{-x^{2}}}{27}\right ) x \,{\mathrm e}^{2 x^{2}}-\frac {25 x \,{\mathrm e}^{-x^{2}} \left ({\mathrm e}^{2 x^{2}}+4 x^{2} {\mathrm e}^{2 x^{2}}\right )}{27}-\frac {50 x^{3} {\mathrm e}^{x^{2}}}{27}\right )}^{3} {\mathrm e}^{-3 x^{2}}}{9765625 x^{9}} \end {align*}
Since the Abel invariant depends on \(x\) then unable to solve this ode at this time.
Unable to complete the solution now.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{3}+{\mathrm e}^{-x^{2}} y^{2} x +\left ({\mathrm e}^{-x^{2}}\right )^{2} y x -y^{\prime } \left ({\mathrm e}^{-x^{2}}\right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x y^{3}+{\mathrm e}^{-x^{2}} y^{2} x +\left ({\mathrm e}^{-x^{2}}\right )^{2} y x}{\left ({\mathrm e}^{-x^{2}}\right )^{2}} \end {array} \]
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: 85
dsolve(diff(y(x),x) = y(x)*(y(x)^2+exp(-x^2)*y(x)+exp(-x^2)^2)/exp(-x^2)^2*x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\sqrt {11}\, \tan \left (\operatorname {RootOf}\left (-4 \sqrt {11}\, x^{2}-8 \sqrt {11}\, \ln \left (11\right )-4 \sqrt {11}\, \ln \left (\sec \left (\textit {\_Z} \right )^{2} {\mathrm e}^{2 x^{2}}\right )+8 \sqrt {11}\, \ln \left (5\right )+8 \sqrt {11}\, \ln \left (-\sqrt {11}+11 \tan \left (\textit {\_Z} \right )\right )+9 \sqrt {11}\, c_{1} -8 \textit {\_Z} \right )\right )-1\right ) {\mathrm e}^{-x^{2}}}{2} \]
✓ Solution by Mathematica
Time used: 0.288 (sec). Leaf size: 139
DSolve[y'[x] == E^(2*x^2)*x*y[x]*(E^(-2*x^2) + y[x]/E^x^2 + y[x]^2),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {25}{3} \text {RootSum}\left [-25 \text {$\#$1}^3+24 \sqrt [3]{-1} 5^{2/3} \text {$\#$1}-25\&,\frac {\log \left (\frac {3 e^{2 x^2} x y(x)+e^{x^2} x}{5^{2/3} \sqrt [3]{-e^{3 x^2} x^3}}-\text {$\#$1}\right )}{8 \sqrt [3]{-1} 5^{2/3}-25 \text {$\#$1}^2}\&\right ]=-\frac {5 \sqrt [3]{5} e^{x^2} x^3}{18 \sqrt [3]{-e^{3 x^2} x^3}}+c_1,y(x)\right ] \]