Internal problem ID [8380]
Internal file name [OUTPUT/7313_Sunday_June_05_2022_05_44_40_PM_82524767/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 43.
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.
\[ \boxed {y^{\prime }+\left (4 a^{2} x +3 a \,x^{2}+b \right ) y^{3}+3 x y^{2}=0} \]
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 (-4 a^{2} x -3 a \,x^{2}-b \right ) y^{3}-3 x y^{2}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= 0\\ f_2(x) &= -3 x\\ f_3(x) &= -4 a^{2} x -3 a \,x^{2}-b \end {align*}
Since \(f_2(x)=-3 x\) 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 {-3 x}{-12 a^{2} x -9 a \,x^{2}-3 b} \right ) \\ &= u \left (x \right )-\frac {x}{4 a^{2} x +3 a \,x^{2}+b} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = -\frac {64 u \left (x \right )^{3} a^{6} x^{3}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {144 u \left (x \right )^{3} a^{5} x^{4}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {108 u \left (x \right )^{3} a^{4} x^{5}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {27 u \left (x \right )^{3} a^{3} x^{6}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {48 u \left (x \right )^{3} a^{4} b \,x^{2}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {72 u \left (x \right )^{3} a^{3} b \,x^{3}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {27 u \left (x \right )^{3} a^{2} b \,x^{4}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {12 u \left (x \right )^{3} a^{2} b^{2} x}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {9 u \left (x \right )^{3} a \,b^{2} x^{2}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {u \left (x \right )^{3} b^{3}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}+\frac {12 u \left (x \right ) a^{2} x^{3}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}+\frac {9 u \left (x \right ) a \,x^{4}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}+\frac {3 u \left (x \right ) b \,x^{2}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {3 a \,x^{2}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}-\frac {2 x^{3}}{\left (4 a^{2} x +3 a \,x^{2}+b \right )^{2}}+\frac {b}{\left (4 a^{2} x +3 a \,x^{2}+b \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 (64 a^{6} x^{3}+144 a^{5} x^{4}+108 a^{4} x^{5}+27 a^{3} x^{6}+48 a^{4} b \,x^{2}+72 a^{3} b \,x^{3}+27 a^{2} b \,x^{4}+12 a^{2} b^{2} x +9 a \,b^{2} x^{2}+b^{3}\right ) u \left (x \right )^{3}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {\left (-12 a^{2} x^{3}-9 a \,x^{4}-3 b \,x^{2}\right ) u \left (x \right )}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {3 a \,x^{2}+2 x^{3}-b}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -\frac {3 a \,x^{2}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {2 x^{3}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}+\frac {b}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}\\ f_1(x) &= \frac {12 a^{2} x^{3}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}+\frac {9 a \,x^{4}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}+\frac {3 b \,x^{2}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}\\ f_2(x) &= 0\\ f_3(x) &= -\frac {64 a^{6} x^{3}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {144 a^{5} x^{4}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {108 a^{4} x^{5}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {27 a^{3} x^{6}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {48 a^{4} b \,x^{2}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {72 a^{3} b \,x^{3}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {27 a^{2} b \,x^{4}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {12 a^{2} b^{2} x}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {9 a \,b^{2} x^{2}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{2}}-\frac {b^{3}}{16 a^{4} x^{2}+24 a^{3} x^{3}+9 a^{2} x^{4}+8 a^{2} b x +6 a b \,x^{2}+b^{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*} \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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\left (4 a^{2} x +3 a \,x^{2}+b \right ) y^{3}+3 x y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\left (4 a^{2} x +3 a \,x^{2}+b \right ) y^{3}-3 x y^{2} \end {array} \]
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 Looking for potential symmetries Looking for potential symmetries <- Abel successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 470
dsolve(diff(y(x),x) + (3*a*x^2 + 4*a^2*x + b)*y(x)^3 + 3*x*y(x)^2=0,y(x), singsol=all)
\[ \frac {a \sqrt {3}\, \left (\operatorname {BesselI}\left (1+\frac {\sqrt {\frac {4 a^{3}-3 b}{a^{3}}}}{2}, -\frac {\sqrt {3}\, \sqrt {\frac {\left (4 a^{2} x +3 a \,x^{2}+b \right ) y \left (x \right )-2 a}{y \left (x \right ) a^{3}}}}{2}\right ) c_{1} -\operatorname {BesselK}\left (1+\frac {\sqrt {\frac {4 a^{3}-3 b}{a^{3}}}}{2}, -\frac {\sqrt {3}\, \sqrt {\frac {\left (4 a^{2} x +3 a \,x^{2}+b \right ) y \left (x \right )-2 a}{y \left (x \right ) a^{3}}}}{2}\right )\right ) \sqrt {\frac {\left (4 a^{2} x +3 a \,x^{2}+b \right ) y \left (x \right )-2 a}{y \left (x \right ) a^{3}}}-\left (c_{1} \operatorname {BesselI}\left (\frac {\sqrt {\frac {4 a^{3}-3 b}{a^{3}}}}{2}, -\frac {\sqrt {3}\, \sqrt {\frac {\left (4 a^{2} x +3 a \,x^{2}+b \right ) y \left (x \right )-2 a}{y \left (x \right ) a^{3}}}}{2}\right )+\operatorname {BesselK}\left (\frac {\sqrt {\frac {4 a^{3}-3 b}{a^{3}}}}{2}, -\frac {\sqrt {3}\, \sqrt {\frac {\left (4 a^{2} x +3 a \,x^{2}+b \right ) y \left (x \right )-2 a}{y \left (x \right ) a^{3}}}}{2}\right )\right ) \left (a \sqrt {\frac {4 a^{3}-3 b}{a^{3}}}-2 a -3 x \right )}{\operatorname {BesselI}\left (1+\frac {\sqrt {\frac {4 a^{3}-3 b}{a^{3}}}}{2}, -\frac {\sqrt {3}\, \sqrt {\frac {\left (4 a^{2} x +3 a \,x^{2}+b \right ) y \left (x \right )-2 a}{y \left (x \right ) a^{3}}}}{2}\right ) \sqrt {3}\, \sqrt {\frac {\left (4 a^{2} x +3 a \,x^{2}+b \right ) y \left (x \right )-2 a}{y \left (x \right ) a^{3}}}\, a -\operatorname {BesselI}\left (\frac {\sqrt {\frac {4 a^{3}-3 b}{a^{3}}}}{2}, -\frac {\sqrt {3}\, \sqrt {\frac {\left (4 a^{2} x +3 a \,x^{2}+b \right ) y \left (x \right )-2 a}{y \left (x \right ) a^{3}}}}{2}\right ) \left (a \sqrt {\frac {4 a^{3}-3 b}{a^{3}}}-2 a -3 x \right )} = 0 \]
✓ Solution by Mathematica
Time used: 4.252 (sec). Leaf size: 490
DSolve[y'[x] + (3*a*x^2 + 4*a^2*x + b)*y[x]^3 + 3*x*y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [c_1=-\frac {i \sqrt {-\frac {4 a^3-3 b}{4 a^3}-\frac {3}{2 a^2 y(x)}+\frac {(-2 a-3 x)^2}{4 a^2}} \operatorname {BesselJ}\left (\frac {1}{2} \sqrt {\frac {4 a^3-3 b}{a^3}}+1,-i \sqrt {\frac {(-2 a-3 x)^2}{4 a^2}-\frac {4 a^3-3 b}{4 a^3}-\frac {3}{2 a^2 y(x)}}\right )+\left (\frac {1}{2} \sqrt {\frac {4 a^3-3 b}{a^3}}+\frac {-2 a-3 x}{2 a}\right ) \operatorname {BesselJ}\left (\frac {1}{2} \sqrt {\frac {4 a^3-3 b}{a^3}},-i \sqrt {\frac {(-2 a-3 x)^2}{4 a^2}-\frac {4 a^3-3 b}{4 a^3}-\frac {3}{2 a^2 y(x)}}\right )}{i \sqrt {-\frac {4 a^3-3 b}{4 a^3}-\frac {3}{2 a^2 y(x)}+\frac {(-2 a-3 x)^2}{4 a^2}} \operatorname {BesselY}\left (\frac {1}{2} \sqrt {\frac {4 a^3-3 b}{a^3}}+1,-i \sqrt {\frac {(-2 a-3 x)^2}{4 a^2}-\frac {4 a^3-3 b}{4 a^3}-\frac {3}{2 a^2 y(x)}}\right )+\left (\frac {1}{2} \sqrt {\frac {4 a^3-3 b}{a^3}}+\frac {-2 a-3 x}{2 a}\right ) \operatorname {BesselY}\left (\frac {1}{2} \sqrt {\frac {4 a^3-3 b}{a^3}},-i \sqrt {\frac {(-2 a-3 x)^2}{4 a^2}-\frac {4 a^3-3 b}{4 a^3}-\frac {3}{2 a^2 y(x)}}\right )},y(x)\right ] \]