Internal problem ID [4680]
Internal file name [OUTPUT/4173_Sunday_June_05_2022_12_36_26_PM_78426666/index.tex]
Book: Differential Gleichungen, Kamke, 3rd ed, Abel ODEs
Section: Abel ODE’s with constant invariant
Problem number: problem 169.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind"
Maple gives the following as the ode type
[_rational, _Abel]
Unable to solve or complete the solution.
\[ \boxed {\left (a x +b \right )^{2} y^{\prime }+\left (a x +b \right ) y^{3}+c y^{2}=0} \]
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 }&=-\frac {y^{3}}{a x +b}-\frac {c y^{2}}{\left (a x +b \right )^{2}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= 0\\ f_2(x) &= -\frac {c}{\left (a x +b \right )^{2}}\\ f_3(x) &= -\frac {1}{a x +b} \end {align*}
Since \(f_2(x)=-\frac {c}{\left (a x +b \right )^{2}}\) 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 {-\frac {c}{\left (a x +b \right )^{2}}}{-\frac {3}{a x +b}} \right ) \\ &= u \left (x \right )-\frac {c}{3 a x +3 b} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = -\frac {u \left (x \right )^{3} a^{3} x^{3}}{\left (a x +b \right )^{4}}-\frac {3 u \left (x \right )^{3} a^{2} b \,x^{2}}{\left (a x +b \right )^{4}}-\frac {3 u \left (x \right )^{3} a \,b^{2} x}{\left (a x +b \right )^{4}}-\frac {u \left (x \right )^{3} b^{3}}{\left (a x +b \right )^{4}}-\frac {a^{3} c \,x^{2}}{3 \left (a x +b \right )^{4}}+\frac {u \left (x \right ) a \,c^{2} x}{3 \left (a x +b \right )^{4}}-\frac {2 a^{2} b c x}{3 \left (a x +b \right )^{4}}+\frac {u \left (x \right ) b \,c^{2}}{3 \left (a x +b \right )^{4}}-\frac {a \,b^{2} c}{3 \left (a x +b \right )^{4}}-\frac {2 c^{3}}{27 \left (a x +b \right )^{4}}\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 a^{3} x^{3}+81 a^{2} b \,x^{2}+81 a \,b^{2} x +27 b^{3}\right ) u \left (x \right )^{3}}{27 \left (a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}\right )}-\frac {\left (-9 a \,c^{2} x -9 b \,c^{2}\right ) u \left (x \right )}{27 \left (a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}\right )}-\frac {9 a^{3} c \,x^{2}+18 a^{2} b c x +9 a \,b^{2} c +2 c^{3}}{27 \left (a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}\right )}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -\frac {a^{3} c \,x^{2}}{3 \left (a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}\right )}-\frac {2 a^{2} b c x}{3 \left (a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}\right )}-\frac {a \,b^{2} c}{3 \left (a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}\right )}-\frac {2 c^{3}}{27 \left (a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}\right )}\\ f_1(x) &= \frac {a \,c^{2} x}{3 a^{4} x^{4}+12 a^{3} b \,x^{3}+18 a^{2} b^{2} x^{2}+12 a \,b^{3} x +3 b^{4}}+\frac {b \,c^{2}}{3 a^{4} x^{4}+12 a^{3} b \,x^{3}+18 a^{2} b^{2} x^{2}+12 a \,b^{3} x +3 b^{4}}\\ f_2(x) &= 0\\ f_3(x) &= -\frac {a^{3} x^{3}}{a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}}-\frac {3 a^{2} b \,x^{2}}{a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}}-\frac {3 a \,b^{2} x}{a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{4}}-\frac {b^{3}}{a^{4} x^{4}+4 a^{3} b \,x^{3}+6 a^{2} b^{2} x^{2}+4 a \,b^{3} x +b^{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*} \text {Expression too large to display} \end {align*}
Since the Abel invariant depends on \(x\) then unable to solve this ode at this time.
Unable to complete the solution now.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a x +b \right )^{2} y^{\prime }+\left (a x +b \right ) y^{3}+c y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\left (a x +b \right ) y^{3}+c y^{2}}{\left (a x +b \right )^{2}} \end {array} \]
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: 126
dsolve((a*x+b)^2*diff(y(x),x)+(a*x+b)*y(x)^3+c*y(x)^2 = 0,y(x), singsol=all)
\[ \frac {\left (\sqrt {a}\, b +a^{\frac {3}{2}} x \right ) {\mathrm e}^{-\frac {\left (\left (a x +b +c \right ) y \left (x \right )+a \left (a x +b \right )\right ) \left (\left (-a x -b +c \right ) y \left (x \right )+a \left (a x +b \right )\right )}{2 y \left (x \right )^{2} \left (a x +b \right )^{2} a}}+\frac {c \sqrt {2}\, \sqrt {\pi }\, {\mathrm e}^{\frac {1}{2 a}} \operatorname {erf}\left (\frac {\left (c y \left (x \right )+a \left (a x +b \right )\right ) \sqrt {2}}{2 \sqrt {a}\, y \left (x \right ) \left (a x +b \right )}\right )}{2}+c_{1} a^{\frac {3}{2}}}{a^{\frac {3}{2}}} = 0 \]
✓ Solution by Mathematica
Time used: 1.43 (sec). Leaf size: 149
DSolve[(a*x+b)^2*y'[x]+(a*x+b)*y[x]^3+c*y[x]^2 == 0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {c}{\sqrt {-a (a x+b)^2}}=\frac {2 \exp \left (\frac {1}{2} \left (-\frac {c}{\sqrt {-a (a x+b)^2}}-\frac {\left (-a (a x+b)^2\right )^{3/2}}{a y(x) (a x+b)^3}\right )^2\right )}{\sqrt {2 \pi } \text {erfi}\left (\frac {-\frac {c}{\sqrt {-a (a x+b)^2}}-\frac {\left (-a (a x+b)^2\right )^{3/2}}{a y(x) (a x+b)^3}}{\sqrt {2}}\right )+2 c_1},y(x)\right ] \]