Internal problem ID [8379]
Internal file name [OUTPUT/7312_Sunday_June_05_2022_05_44_32_PM_88067671/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 42.
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]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-x \left (2+x \right ) y^{3}-\left (x +3\right ) y^{2}=0} \]
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 }&=\left (x^{2}+2 x \right ) y^{3}+\left (x +3\right ) y^{2}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= 0\\ f_2(x) &= x +3\\ f_3(x) &= x^{2}+2 x \end {align*}
Since \(f_2(x)=x +3\) 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 +3}{3 x^{2}+6 x} \right ) \\ &= u \left (x \right )-\frac {x +3}{3 x \left (2+x \right )} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {x^{4} u \left (x \right )^{3}}{\left (2+x \right )^{2}}+\frac {6 x^{3} u \left (x \right )^{3}}{\left (2+x \right )^{2}}+\frac {12 x^{2} u \left (x \right )^{3}}{\left (2+x \right )^{2}}+\frac {8 x u \left (x \right )^{3}}{\left (2+x \right )^{2}}-\frac {x^{2} u \left (x \right )}{3 \left (2+x \right )^{2}}-\frac {8 x u \left (x \right )}{3 \left (2+x \right )^{2}}-\frac {7 u \left (x \right )}{\left (2+x \right )^{2}}+\frac {2 x}{27 \left (2+x \right )^{2}}-\frac {6 u \left (x \right )}{x \left (2+x \right )^{2}}+\frac {1}{3 \left (2+x \right )^{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^{5}+162 x^{4}+324 x^{3}+216 x^{2}\right ) u \left (x \right )^{3}}{27 x \left (x^{2}+4 x +4\right )}+\frac {\left (-9 x^{3}-72 x^{2}-189 x -162\right ) u \left (x \right )}{27 x \left (x^{2}+4 x +4\right )}+\frac {2 x^{2}+9 x}{27 x \left (x^{2}+4 x +4\right )}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {2 x}{27 \left (x^{2}+4 x +4\right )}+\frac {1}{3 x^{2}+12 x +12}\\ f_1(x) &= -\frac {x^{2}}{3 \left (x^{2}+4 x +4\right )}-\frac {8 x}{3 \left (x^{2}+4 x +4\right )}-\frac {7}{x^{2}+4 x +4}-\frac {6}{x \left (x^{2}+4 x +4\right )}\\ f_2(x) &= 0\\ f_3(x) &= \frac {x^{4}}{x^{2}+4 x +4}+\frac {6 x^{3}}{x^{2}+4 x +4}+\frac {12 x^{2}}{x^{2}+4 x +4}+\frac {8 x}{x^{2}+4 x +4} \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}{27 \left (x^{2}+4 x +4\right )}-\frac {2 x \left (4+2 x \right )}{27 \left (x^{2}+4 x +4\right )^{2}}-\frac {4+2 x}{3 \left (x^{2}+4 x +4\right )^{2}}\right ) \left (\frac {x^{4}}{x^{2}+4 x +4}+\frac {6 x^{3}}{x^{2}+4 x +4}+\frac {12 x^{2}}{x^{2}+4 x +4}+\frac {8 x}{x^{2}+4 x +4}\right )+\left (\frac {2 x}{27 \left (x^{2}+4 x +4\right )}+\frac {1}{3 x^{2}+12 x +12}\right ) \left (\frac {4 x^{3}}{x^{2}+4 x +4}-\frac {x^{4} \left (4+2 x \right )}{\left (x^{2}+4 x +4\right )^{2}}+\frac {18 x^{2}}{x^{2}+4 x +4}-\frac {6 x^{3} \left (4+2 x \right )}{\left (x^{2}+4 x +4\right )^{2}}+\frac {24 x}{x^{2}+4 x +4}-\frac {12 x^{2} \left (4+2 x \right )}{\left (x^{2}+4 x +4\right )^{2}}+\frac {8}{x^{2}+4 x +4}-\frac {8 x \left (4+2 x \right )}{\left (x^{2}+4 x +4\right )^{2}}\right )+3 \left (\frac {2 x}{27 \left (x^{2}+4 x +4\right )}+\frac {1}{3 x^{2}+12 x +12}\right ) \left (\frac {x^{4}}{x^{2}+4 x +4}+\frac {6 x^{3}}{x^{2}+4 x +4}+\frac {12 x^{2}}{x^{2}+4 x +4}+\frac {8 x}{x^{2}+4 x +4}\right ) \left (-\frac {x^{2}}{3 \left (x^{2}+4 x +4\right )}-\frac {8 x}{3 \left (x^{2}+4 x +4\right )}-\frac {7}{x^{2}+4 x +4}-\frac {6}{x \left (x^{2}+4 x +4\right )}\right )\right )}^{3}}{27 \left (\frac {x^{4}}{x^{2}+4 x +4}+\frac {6 x^{3}}{x^{2}+4 x +4}+\frac {12 x^{2}}{x^{2}+4 x +4}+\frac {8 x}{x^{2}+4 x +4}\right )^{4} {\left (\frac {2 x}{27 \left (x^{2}+4 x +4\right )}+\frac {1}{3 x^{2}+12 x +12}\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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-x \left (2+x \right ) y^{3}-\left (x +3\right ) y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=x \left (2+x \right ) y^{3}+\left (x +3\right ) y^{2} \end {array} \]
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: 53
dsolve(diff(y(x),x) - x*(x+2)*y(x)^3 - (x+3)*y(x)^2=0,y(x), singsol=all)
\[ \frac {\frac {\sqrt {2+\left (x^{2}+2 x \right ) y \left (x \right )}}{2}+\left (\operatorname {arctanh}\left (\frac {\sqrt {y \left (x \right )}\, x}{\sqrt {2+\left (x^{2}+2 x \right ) y \left (x \right )}}\right )+c_{1} \right ) \sqrt {y \left (x \right )}}{\sqrt {y \left (x \right )}} = 0 \]
✓ Solution by Mathematica
Time used: 0.764 (sec). Leaf size: 485
DSolve[y'[x] - x*(x+2)*y[x]^3 - (x+3)*y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [c_1=-\frac {\frac {i \sqrt {\frac {2}{\pi }} \sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}} \left (\frac {\sinh \left (\sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}\right )}{\sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}}-\cosh \left (\sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}\right )\right )}{\sqrt {-i \sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}}}-\frac {i \sqrt {\frac {2}{\pi }} \left (\frac {x+1}{2}+\frac {1}{2}\right ) \sinh \left (\sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}\right )}{\sqrt {-i \sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}}}}{\frac {i \sqrt {\frac {2}{\pi }} \sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}} \left (i \sinh \left (\sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}\right )-\frac {i \cosh \left (\sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}\right )}{\sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}}\right )}{\sqrt {-i \sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}}}-\frac {\sqrt {\frac {2}{\pi }} \left (\frac {x+1}{2}+\frac {1}{2}\right ) \cosh \left (\sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}\right )}{\sqrt {-i \sqrt {\frac {1}{2 y(x)}+\frac {1}{4} (x+1)^2-\frac {1}{4}}}}},y(x)\right ] \]