Internal problem ID [8382]
Internal file name [OUTPUT/7315_Sunday_June_05_2022_05_44_51_PM_13889413/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 45.
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.
\[ \boxed {y^{\prime }+2 \left (a^{2} x^{3}-x \,b^{2}\right ) y^{3}+3 b y^{2}=0} \]
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 (-2 a^{2} x^{3}+2 x \,b^{2}\right ) y^{3}-3 b y^{2}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= 0\\ f_2(x) &= -3 b\\ f_3(x) &= -2 a^{2} x^{3}+2 x \,b^{2} \end {align*}
Since \(f_2(x)=-3 b\) 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 b}{-6 a^{2} x^{3}+6 x \,b^{2}} \right ) \\ &= u \left (x \right )-\frac {b}{2 a^{2} x^{3}-2 x \,b^{2}} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = -\frac {2 x^{7} u \left (x \right )^{3} a^{6}}{\left (a^{2} x^{2}-b^{2}\right )^{2}}+\frac {6 x^{5} u \left (x \right )^{3} a^{4} b^{2}}{\left (a^{2} x^{2}-b^{2}\right )^{2}}-\frac {6 x^{3} u \left (x \right )^{3} a^{2} b^{4}}{\left (a^{2} x^{2}-b^{2}\right )^{2}}+\frac {2 x u \left (x \right )^{3} b^{6}}{\left (a^{2} x^{2}-b^{2}\right )^{2}}+\frac {3 x u \left (x \right ) a^{2} b^{2}}{2 \left (a^{2} x^{2}-b^{2}\right )^{2}}-\frac {3 u \left (x \right ) b^{4}}{2 x \left (a^{2} x^{2}-b^{2}\right )^{2}}-\frac {3 b \,a^{2}}{2 \left (a^{2} x^{2}-b^{2}\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 (4 a^{6} x^{8}-12 a^{4} b^{2} x^{6}+12 a^{2} b^{4} x^{4}-4 x^{2} b^{6}\right ) u \left (x \right )^{3}}{2 x \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}-\frac {\left (-3 a^{2} b^{2} x^{2}+3 b^{4}\right ) u \left (x \right )}{2 x \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}-\frac {3 a^{2} b}{2 \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -\frac {3 a^{2} b}{2 \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}\\ f_1(x) &= \frac {3 x \,a^{2} b^{2}}{2 \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}-\frac {3 b^{4}}{2 x \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}\\ f_2(x) &= 0\\ f_3(x) &= -\frac {2 x^{7} a^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {6 x^{5} a^{4} b^{2}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}-\frac {6 x^{3} a^{2} b^{4}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {2 x \,b^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+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*} \frac {32 \left (-\frac {3 a^{2} b \left (4 a^{4} x^{3}-4 a^{2} b^{2} x \right ) \left (-\frac {2 x^{7} a^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {6 x^{5} a^{4} b^{2}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}-\frac {6 x^{3} a^{2} b^{4}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {2 x \,b^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}\right )}{2 \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )^{2}}-\frac {3 a^{2} b \left (-\frac {14 x^{6} a^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {2 x^{7} a^{6} \left (4 a^{4} x^{3}-4 a^{2} b^{2} x \right )}{\left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )^{2}}+\frac {30 x^{4} a^{4} b^{2}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}-\frac {6 x^{5} a^{4} b^{2} \left (4 a^{4} x^{3}-4 a^{2} b^{2} x \right )}{\left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )^{2}}-\frac {18 x^{2} a^{2} b^{4}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {6 x^{3} a^{2} b^{4} \left (4 a^{4} x^{3}-4 a^{2} b^{2} x \right )}{\left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )^{2}}+\frac {2 b^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}-\frac {2 x \,b^{6} \left (4 a^{4} x^{3}-4 a^{2} b^{2} x \right )}{\left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )^{2}}\right )}{2 \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}-\frac {9 a^{2} b \left (-\frac {2 x^{7} a^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {6 x^{5} a^{4} b^{2}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}-\frac {6 x^{3} a^{2} b^{4}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {2 x \,b^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}\right ) \left (\frac {3 x \,a^{2} b^{2}}{2 \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}-\frac {3 b^{4}}{2 x \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}\right )}{2 \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )}\right )^{3} \left (a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}\right )^{5}}{6561 \left (-\frac {2 x^{7} a^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {6 x^{5} a^{4} b^{2}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}-\frac {6 x^{3} a^{2} b^{4}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}+\frac {2 x \,b^{6}}{a^{4} x^{4}-2 a^{2} b^{2} x^{2}+b^{4}}\right )^{4} a^{10} b^{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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+2 \left (a^{2} x^{3}-x \,b^{2}\right ) y^{3}+3 b y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 \left (a^{2} x^{3}-x \,b^{2}\right ) y^{3}-3 b y^{2} \end {array} \]
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: 113
dsolve(diff(y(x),x) + 2*(a^2*x^3 - b^2*x)*y(x)^3 + 3*b*y(x)^2=0,y(x), singsol=all)
\[ c_{1} +\frac {\left (\frac {a^{2} y \left (x \right )^{2} x^{4}-y \left (x \right )^{2} b^{2} x^{2}+2 b x y \left (x \right )-1}{\left (b x y \left (x \right )-1\right )^{2}}\right )^{\frac {1}{4}} a x}{\sqrt {\frac {a \,x^{2} y \left (x \right )}{b x y \left (x \right )-1}}\, b \left (b x y \left (x \right )-1\right )}-\left (\int _{}^{\frac {a \,x^{2} y \left (x \right )}{b x y \left (x \right )-1}}\frac {\left (\textit {\_a}^{2}-1\right )^{\frac {1}{4}}}{\sqrt {\textit {\_a}}}d \textit {\_a} \right ) = 0 \]
✓ Solution by Mathematica
Time used: 0.442 (sec). Leaf size: 133
DSolve[y'[x] + 2*(a^2*x^3 - b^2*x)*y[x]^3 + 3*b*y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [c_1=\sqrt [4]{\left (\frac {b}{a x}-\frac {1}{a x^2 y(x)}\right )^2-1} \left (-\frac {\left (\frac {b}{a x}-\frac {1}{a x^2 y(x)}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\left (\frac {b}{a x}-\frac {1}{a x^2 y(x)}\right )^2\right )}{2 \sqrt [4]{1-\left (\frac {b}{a x}-\frac {1}{a x^2 y(x)}\right )^2}}-\frac {a x}{b}\right ),y(x)\right ] \]