Internal problem ID [3636]
Internal file name [OUTPUT/3129_Sunday_June_05_2022_08_53_02_AM_1337703/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 14
Problem number: 380.
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.
\[ \boxed {x^{7} y^{\prime }+5 y^{2} x^{3}+2 \left (x^{2}+1\right ) y^{3}=0} \]
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 {\left (2 x^{2}+2\right ) y^{3}}{x^{7}}-\frac {5 y^{2}}{x^{4}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= 0\\ f_2(x) &= -\frac {5}{x^{4}}\\ f_3(x) &= -\frac {2}{x^{5}}-\frac {2}{x^{7}} \end {align*}
Since \(f_2(x)=-\frac {5}{x^{4}}\) 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 {5}{x^{4}}}{-\frac {6}{x^{5}}-\frac {6}{x^{7}}} \right ) \\ &= u \left (x \right )-\frac {5 x^{3}}{6 x^{2}+6} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {5 x^{4}}{6 \left (x^{2}+1\right )^{2}}-\frac {2 u \left (x \right )^{3}}{\left (x^{2}+1\right )^{2} x}+\frac {25 x u \left (x \right )}{6 \left (x^{2}+1\right )^{2}}+\frac {5 x^{2}}{27 \left (x^{2}+1\right )^{2}}-\frac {6 u \left (x \right )^{3}}{\left (x^{2}+1\right )^{2} x^{3}}+\frac {25 u \left (x \right )}{6 \left (x^{2}+1\right )^{2} x}-\frac {6 u \left (x \right )^{3}}{\left (x^{2}+1\right )^{2} x^{5}}-\frac {2 u \left (x \right )^{3}}{\left (x^{2}+1\right )^{2} x^{7}}\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 (108 x^{6}+324 x^{4}+324 x^{2}+108\right ) u \left (x \right )^{3}}{54 x^{7} \left (x^{4}+2 x^{2}+1\right )}-\frac {\left (-225 x^{8}-225 x^{6}\right ) u \left (x \right )}{54 x^{7} \left (x^{4}+2 x^{2}+1\right )}-\frac {-45 x^{11}-10 x^{9}}{54 x^{7} \left (x^{4}+2 x^{2}+1\right )}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {5 x^{4}}{6 \left (x^{4}+2 x^{2}+1\right )}+\frac {5 x^{2}}{27 \left (x^{4}+2 x^{2}+1\right )}\\ f_1(x) &= \frac {25 x}{6 \left (x^{4}+2 x^{2}+1\right )}+\frac {25}{6 x \left (x^{4}+2 x^{2}+1\right )}\\ f_2(x) &= 0\\ f_3(x) &= -\frac {2}{x \left (x^{4}+2 x^{2}+1\right )}-\frac {6}{x^{3} \left (x^{4}+2 x^{2}+1\right )}-\frac {6}{x^{5} \left (x^{4}+2 x^{2}+1\right )}-\frac {2}{x^{7} \left (x^{4}+2 x^{2}+1\right )} \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 {10 x^{3}}{3 \left (x^{4}+2 x^{2}+1\right )}-\frac {5 x^{4} \left (4 x^{3}+4 x \right )}{6 \left (x^{4}+2 x^{2}+1\right )^{2}}+\frac {10 x}{27 \left (x^{4}+2 x^{2}+1\right )}-\frac {5 x^{2} \left (4 x^{3}+4 x \right )}{27 \left (x^{4}+2 x^{2}+1\right )^{2}}\right ) \left (-\frac {2}{x \left (x^{4}+2 x^{2}+1\right )}-\frac {6}{x^{3} \left (x^{4}+2 x^{2}+1\right )}-\frac {6}{x^{5} \left (x^{4}+2 x^{2}+1\right )}-\frac {2}{x^{7} \left (x^{4}+2 x^{2}+1\right )}\right )+\left (\frac {5 x^{4}}{6 \left (x^{4}+2 x^{2}+1\right )}+\frac {5 x^{2}}{27 \left (x^{4}+2 x^{2}+1\right )}\right ) \left (\frac {2}{x^{2} \left (x^{4}+2 x^{2}+1\right )}+\frac {8 x^{3}+8 x}{x \left (x^{4}+2 x^{2}+1\right )^{2}}+\frac {18}{x^{4} \left (x^{4}+2 x^{2}+1\right )}+\frac {24 x^{3}+24 x}{x^{3} \left (x^{4}+2 x^{2}+1\right )^{2}}+\frac {30}{x^{6} \left (x^{4}+2 x^{2}+1\right )}+\frac {24 x^{3}+24 x}{x^{5} \left (x^{4}+2 x^{2}+1\right )^{2}}+\frac {14}{x^{8} \left (x^{4}+2 x^{2}+1\right )}+\frac {8 x^{3}+8 x}{x^{7} \left (x^{4}+2 x^{2}+1\right )^{2}}\right )+3 \left (\frac {5 x^{4}}{6 \left (x^{4}+2 x^{2}+1\right )}+\frac {5 x^{2}}{27 \left (x^{4}+2 x^{2}+1\right )}\right ) \left (-\frac {2}{x \left (x^{4}+2 x^{2}+1\right )}-\frac {6}{x^{3} \left (x^{4}+2 x^{2}+1\right )}-\frac {6}{x^{5} \left (x^{4}+2 x^{2}+1\right )}-\frac {2}{x^{7} \left (x^{4}+2 x^{2}+1\right )}\right ) \left (\frac {25 x}{6 \left (x^{4}+2 x^{2}+1\right )}+\frac {25}{6 x \left (x^{4}+2 x^{2}+1\right )}\right )\right )}^{3}}{27 {\left (-\frac {2}{x \left (x^{4}+2 x^{2}+1\right )}-\frac {6}{x^{3} \left (x^{4}+2 x^{2}+1\right )}-\frac {6}{x^{5} \left (x^{4}+2 x^{2}+1\right )}-\frac {2}{x^{7} \left (x^{4}+2 x^{2}+1\right )}\right )}^{4} \left (\frac {5 x^{4}}{6 \left (x^{4}+2 x^{2}+1\right )}+\frac {5 x^{2}}{27 \left (x^{4}+2 x^{2}+1\right )}\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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{7} y^{\prime }+5 y^{2} x^{3}+2 \left (x^{2}+1\right ) y^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-5 y^{2} x^{3}-2 \left (x^{2}+1\right ) y^{3}}{x^{7}} \end {array} \]
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: 0.015 (sec). Leaf size: 78
dsolve(x^7*diff(y(x),x)+5*x^3*y(x)^2+2*(x^2+1)*y(x)^3 = 0,y(x), singsol=all)
\[ c_{1} +\frac {x}{\left (\frac {x^{6}+x^{2} y \left (x \right )^{2}+2 x^{3} y \left (x \right )+y \left (x \right )^{2}}{x^{2} y \left (x \right )^{2}}\right )^{\frac {1}{4}}}+\frac {\left (x^{3}+y \left (x \right )\right ) \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {3}{2}\right ], -\frac {\left (x^{3}+y \left (x \right )\right )^{2}}{x^{2} y \left (x \right )^{2}}\right )}{2 y \left (x \right ) x} = 0 \]
✓ Solution by Mathematica
Time used: 0.313 (sec). Leaf size: 123
DSolve[x^7 y'[x]+5 x^3 y[x]^2+2(1+x^2)y[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [c_1=\frac {\frac {1}{2} \sqrt [4]{1-\left (\frac {i x^2}{y(x)}+\frac {i}{x}\right )^2} \left (\frac {i x^2}{y(x)}+\frac {i}{x}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {3}{2},\left (\frac {i x^2}{y(x)}+\frac {i}{x}\right )^2\right )+i x}{\sqrt [4]{-1+\left (\frac {i x^2}{y(x)}+\frac {i}{x}\right )^2}},y(x)\right ] \]