Internal problem ID [9260]
Internal file name [OUTPUT/8196_Monday_June_06_2022_02_14_52_AM_79041309/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 927.
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(x)]`], _Abel]
\[ \boxed {y^{\prime }+\frac {x \left ({\mathrm e}^{-3 x^{2}} x^{6}-6 \,{\mathrm e}^{-2 x^{2}} x^{4} y-2 \,{\mathrm e}^{-2 x^{2}} x^{4}+12 x^{2} {\mathrm e}^{-x^{2}} y^{2}+8 x^{2} {\mathrm e}^{-x^{2}} y+8 x^{2} {\mathrm e}^{-x^{2}}-8 y^{3}-8 y^{2}-8 \,{\mathrm e}^{-x^{2}}-8\right )}{8}=0} \]
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 }&=x y^{3}-\frac {\left (12 x^{2} {\mathrm e}^{-x^{2}}-8\right ) x y^{2}}{8}-\frac {\left (-6 \,{\mathrm e}^{-2 x^{2}} x^{4}+8 x^{2} {\mathrm e}^{-x^{2}}\right ) x y}{8}-\frac {\left ({\mathrm e}^{-3 x^{2}} x^{6}-2 \,{\mathrm e}^{-2 x^{2}} x^{4}+8 x^{2} {\mathrm e}^{-x^{2}}-8 \,{\mathrm e}^{-x^{2}}-8\right ) x}{8}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -\frac {{\mathrm e}^{-3 x^{2}} x^{7}}{8}+\frac {{\mathrm e}^{-2 x^{2}} x^{5}}{4}-{\mathrm e}^{-x^{2}} x^{3}+{\mathrm e}^{-x^{2}} x +x\\ f_1(x) &= \frac {3 \,{\mathrm e}^{-2 x^{2}} x^{5}}{4}-{\mathrm e}^{-x^{2}} x^{3}\\ f_2(x) &= -\frac {3 \,{\mathrm e}^{-x^{2}} x^{3}}{2}+x\\ f_3(x) &= x \end {align*}
Since \(f_2(x)=-\frac {3 \,{\mathrm e}^{-x^{2}} x^{3}}{2}+x\) 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 \,{\mathrm e}^{-x^{2}} x^{3}}{2}+x}{3 x} \right ) \\ &= u \left (x \right )+\frac {x^{2} {\mathrm e}^{-x^{2}}}{2}-\frac {1}{3} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = x u \left (x \right )^{3}-\frac {x u \left (x \right )}{3}+\frac {29 x}{27}\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= x \left (u^{3}-\frac {1}{3} u +\frac {29}{27}\right ) \end {align*}
Where \(f(x)=x\) and \(g(u)=u^{3}-\frac {1}{3} u +\frac {29}{27}\). Integrating both sides gives \begin{align*} \frac {1}{u^{3}-\frac {1}{3} u +\frac {29}{27}} \,du &= x \,d x \\ \int { \frac {1}{u^{3}-\frac {1}{3} u +\frac {29}{27}} \,du} &= \int {x \,d x} \\ \int _{}^{u}\frac {1}{\textit {\_a}^{3}-\frac {1}{3} \textit {\_a} +\frac {29}{27}}d \textit {\_a}&=\frac {x^{2}}{2}+c_{2} \\ \end{align*} Which results in \[ \int _{}^{u}\frac {1}{\textit {\_a}^{3}-\frac {1}{3} \textit {\_a} +\frac {29}{27}}d \textit {\_a}=\frac {x^{2}}{2}+c_{2} \] The solution is \[ \int _{}^{u \left (x \right )}\frac {1}{\textit {\_a}^{3}-\frac {1}{3} \textit {\_a} +\frac {29}{27}}d \textit {\_a} -\frac {x^{2}}{2}-c_{2} = 0 \] Substituting \(u=y-\frac {-\frac {3 \,{\mathrm e}^{-x^{2}} x^{3}}{2}+x}{3 x}\) in the above solution gives \begin {align*} \int _{}^{y-\frac {-\frac {3 \,{\mathrm e}^{-x^{2}} x^{3}}{2}+x}{3 x}}\frac {1}{\textit {\_a}^{3}-\frac {1}{3} \textit {\_a} +\frac {29}{27}}d \textit {\_a} -\frac {x^{2}}{2}-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y-\frac {-\frac {3 \,{\mathrm e}^{-x^{2}} x^{3}}{2}+x}{3 x}}\frac {1}{\textit {\_a}^{3}-\frac {1}{3} \textit {\_a} +\frac {29}{27}}d \textit {\_a} -\frac {x^{2}}{2}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \int _{}^{y-\frac {-\frac {3 \,{\mathrm e}^{-x^{2}} x^{3}}{2}+x}{3 x}}\frac {1}{\textit {\_a}^{3}-\frac {1}{3} \textit {\_a} +\frac {29}{27}}d \textit {\_a} -\frac {x^{2}}{2}-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\frac {\left (-8 \,{\mathrm e}^{-x^{2}}+8 x^{2} {\mathrm e}^{-x^{2}}-8-8 y^{2}+8 x^{2} {\mathrm e}^{-x^{2}} y-2 x^{4} \left ({\mathrm e}^{-x^{2}}\right )^{2}-8 y^{3}+12 x^{2} {\mathrm e}^{-x^{2}} y^{2}-6 y x^{4} \left ({\mathrm e}^{-x^{2}}\right )^{2}+x^{6} \left ({\mathrm e}^{-x^{2}}\right )^{3}\right ) x}{8}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\left (-8 \,{\mathrm e}^{-x^{2}}+8 x^{2} {\mathrm e}^{-x^{2}}-8-8 y^{2}+8 x^{2} {\mathrm e}^{-x^{2}} y-2 x^{4} \left ({\mathrm e}^{-x^{2}}\right )^{2}-8 y^{3}+12 x^{2} {\mathrm e}^{-x^{2}} y^{2}-6 y x^{4} \left ({\mathrm e}^{-x^{2}}\right )^{2}+x^{6} \left ({\mathrm e}^{-x^{2}}\right )^{3}\right ) x}{8} \end {array} \]
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: 45
dsolve(diff(y(x),x) = -1/8*(-8*exp(-x^2)+8*x^2*exp(-x^2)-8-8*y(x)^2+8*x^2*exp(-x^2)*y(x)-2*x^4*exp(-x^2)^2-8*y(x)^3+12*x^2*exp(-x^2)*y(x)^2-6*y(x)*x^4*exp(-x^2)^2+x^6*exp(-x^2)^3)*x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-x^{2}} x^{2}}{2}-\frac {1}{3}+\frac {29 \operatorname {RootOf}\left (x^{2}-162 \left (\int _{}^{\textit {\_Z}}\frac {1}{841 \textit {\_a}^{3}-27 \textit {\_a} +27}d \textit {\_a} \right )+6 c_{1} \right )}{9} \]
✓ Solution by Mathematica
Time used: 0.417 (sec). Leaf size: 112
DSolve[y'[x] == -1/8*(x*(-8 - 8/E^x^2 + (8*x^2)/E^x^2 - (2*x^4)/E^(2*x^2) + x^6/E^(3*x^2) + (8*x^2*y[x])/E^x^2 - (6*x^4*y[x])/E^(2*x^2) - 8*y[x]^2 + (12*x^2*y[x]^2)/E^x^2 - 8*y[x]^3)),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {29}{3} \text {RootSum}\left [-29 \text {$\#$1}^3+3 \sqrt [3]{29} \text {$\#$1}-29\&,\frac {\log \left (\frac {\frac {1}{2} e^{-x^2} x \left (2 e^{x^2}-3 x^2\right )+3 x y(x)}{\sqrt [3]{29} \sqrt [3]{x^3}}-\text {$\#$1}\right )}{\sqrt [3]{29}-29 \text {$\#$1}^2}\&\right ]=\frac {1}{18} 29^{2/3} \left (x^3\right )^{2/3}+c_1,y(x)\right ] \]