Internal problem ID [9156]
Internal file name [OUTPUT/8091_Monday_June_06_2022_01_43_54_AM_84553523/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 822.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime }-\frac {x \left ({\mathrm e}^{-2 x^{2}} x^{4}-4 x^{2} {\mathrm e}^{-x^{2}} y-4 x^{2} {\mathrm e}^{-x^{2}}+4 y^{2}+4 \,{\mathrm e}^{-x^{2}}\right )}{4}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\left ({\mathrm e}^{-2 x^{2}} x^{4}-4 x^{2} {\mathrm e}^{-x^{2}} y -4 x^{2} {\mathrm e}^{-x^{2}}+4 y^{2}+4 \,{\mathrm e}^{-x^{2}}\right ) x}{4} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {{\mathrm e}^{-2 x^{2}} x^{5}}{4}-{\mathrm e}^{-x^{2}} x^{3} y -{\mathrm e}^{-x^{2}} x^{3}+x \,y^{2}+{\mathrm e}^{-x^{2}} x \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {\left ({\mathrm e}^{-2 x^{2}} x^{4}-4 x^{2} {\mathrm e}^{-x^{2}}+4 \,{\mathrm e}^{-x^{2}}\right ) x}{4}\), \(f_1(x)=-{\mathrm e}^{-x^{2}} x^{3}\) and \(f_2(x)=x\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{x u} \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' &=1\\ f_1 f_2 &=-{\mathrm e}^{-x^{2}} x^{4}\\ f_2^2 f_0 &=\frac {x^{3} \left ({\mathrm e}^{-2 x^{2}} x^{4}-4 x^{2} {\mathrm e}^{-x^{2}}+4 \,{\mathrm e}^{-x^{2}}\right )}{4} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} x u^{\prime \prime }\left (x \right )-\left (1-{\mathrm e}^{-x^{2}} x^{4}\right ) u^{\prime }\left (x \right )+\frac {x^{3} \left ({\mathrm e}^{-2 x^{2}} x^{4}-4 x^{2} {\mathrm e}^{-x^{2}}+4 \,{\mathrm e}^{-x^{2}}\right ) u \left (x \right )}{4} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {{\mathrm e}^{-x^{2}} \left (x^{2}+1\right )}{4}} \left (c_{2} x^{2}+c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {x \left (x^{2} \left (c_{2} x^{2}+c_{1} \right ) {\mathrm e}^{-x^{2}}-4 c_{2} \right ) {\mathrm e}^{\frac {{\mathrm e}^{-x^{2}} \left (x^{2}+1\right )}{4}}}{2} \] Using the above in (1) gives the solution \[ y = \frac {x^{2} \left (c_{2} x^{2}+c_{1} \right ) {\mathrm e}^{-x^{2}}-4 c_{2}}{2 c_{2} x^{2}+2 c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x^{2} \left (x^{2}+c_{3} \right ) {\mathrm e}^{-x^{2}}-4}{2 x^{2}+2 c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2} \left (x^{2}+c_{3} \right ) {\mathrm e}^{-x^{2}}-4}{2 x^{2}+2 c_{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2} \left (x^{2}+c_{3} \right ) {\mathrm e}^{-x^{2}}-4}{2 x^{2}+2 c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {\left (4 \,{\mathrm e}^{-x^{2}}-4 x^{2} {\mathrm e}^{-x^{2}}+4 y^{2}-4 x^{2} {\mathrm e}^{-x^{2}} y+x^{4} \left ({\mathrm e}^{-x^{2}}\right )^{2}\right ) x}{4}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (4 \,{\mathrm e}^{-x^{2}}-4 x^{2} {\mathrm e}^{-x^{2}}+4 y^{2}-4 x^{2} {\mathrm e}^{-x^{2}} y+x^{4} \left ({\mathrm e}^{-x^{2}}\right )^{2}\right ) x}{4} \end {array} \]
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 Looking for potential symmetries trying Riccati trying Riccati sub-methods: <- Riccati particular case Kamke (d) successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 35
dsolve(diff(y(x),x) = 1/4*(4*exp(-x^2)-4*x^2*exp(-x^2)+4*y(x)^2-4*x^2*exp(-x^2)*y(x)+x^4*exp(-x^2)^2)*x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-4+x^{2} \left (x^{2}-2 c_{1} \right ) {\mathrm e}^{-x^{2}}}{2 x^{2}-4 c_{1}} \]
✓ Solution by Mathematica
Time used: 0.502 (sec). Leaf size: 50
DSolve[y'[x] == (x*(4/E^x^2 - (4*x^2)/E^x^2 + x^4/E^(2*x^2) - (4*x^2*y[x])/E^x^2 + 4*y[x]^2))/4,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{2} e^{-x^2} x^2+\frac {1}{-\frac {x^2}{2}+c_1} \\ y(x)\to \frac {1}{2} e^{-x^2} x^2 \\ \end{align*}