Internal problem ID [3352]
Internal file name [OUTPUT/2844_Sunday_June_05_2022_08_41_37_AM_31688915/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 4
Problem number: 92.
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.
\[ \boxed {y^{\prime }-\operatorname {f1} \left (x \right ) y-\operatorname {f2} \left (x \right ) y^{2}-\operatorname {f3} \left (x \right ) y^{3}=\operatorname {f0} \left (x \right )} \]
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 }&=\operatorname {f0} \left (x \right )+\operatorname {f1} \left (x \right ) y+\operatorname {f2} \left (x \right ) y^{2}+\operatorname {f3} \left (x \right ) y^{3}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \operatorname {f0} \left (x \right )\\ f_1(x) &= \operatorname {f1} \left (x \right )\\ f_2(x) &= \operatorname {f2} \left (x \right )\\ f_3(x) &= \operatorname {f3} \left (x \right ) \end {align*}
Since \(f_2(x)=\operatorname {f2} \left (x \right )\) 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 {\operatorname {f2} \left (x \right )}{3 \operatorname {f3} \left (x \right )} \right ) \\ &= u \left (x \right )-\frac {\operatorname {f2} \left (x \right )}{3 \operatorname {f3} \left (x \right )} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \operatorname {f3} \left (x \right ) u \left (x \right )^{3}+\operatorname {f1} \left (x \right ) u \left (x \right )-\frac {\operatorname {f2} \left (x \right )^{2} u \left (x \right )}{3 \operatorname {f3} \left (x \right )}+\operatorname {f0} \left (x \right )-\frac {\operatorname {f1} \left (x \right ) \operatorname {f2} \left (x \right )}{3 \operatorname {f3} \left (x \right )}+\frac {2 \operatorname {f2} \left (x \right )^{3}}{27 \operatorname {f3} \left (x \right )^{2}}+\frac {\operatorname {f2}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )}-\frac {\operatorname {f2} \left (x \right ) \operatorname {f3}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (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 )&=\operatorname {f3} \left (x \right ) u \left (x \right )^{3}+\frac {\left (27 \operatorname {f1} \left (x \right ) \operatorname {f3} \left (x \right )^{2}-9 \operatorname {f2} \left (x \right )^{2} \operatorname {f3} \left (x \right )\right ) u \left (x \right )}{27 \operatorname {f3} \left (x \right )^{2}}+\frac {27 \operatorname {f3} \left (x \right )^{2} \operatorname {f0} \left (x \right )-9 \operatorname {f1} \left (x \right ) \operatorname {f2} \left (x \right ) \operatorname {f3} \left (x \right )+2 \operatorname {f2} \left (x \right )^{3}-9 \operatorname {f3}^{\prime }\left (x \right ) \operatorname {f2} \left (x \right )+9 \operatorname {f2}^{\prime }\left (x \right ) \operatorname {f3} \left (x \right )}{27 \operatorname {f3} \left (x \right )^{2}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \operatorname {f0} \left (x \right )-\frac {\operatorname {f1} \left (x \right ) \operatorname {f2} \left (x \right )}{3 \operatorname {f3} \left (x \right )}+\frac {2 \operatorname {f2} \left (x \right )^{3}}{27 \operatorname {f3} \left (x \right )^{2}}+\frac {\operatorname {f2}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )}-\frac {\operatorname {f2} \left (x \right ) \operatorname {f3}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )^{2}}\\ f_1(x) &= \operatorname {f1} \left (x \right )-\frac {\operatorname {f2} \left (x \right )^{2}}{3 \operatorname {f3} \left (x \right )}\\ f_2(x) &= 0\\ f_3(x) &= \operatorname {f3} \left (x \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 (\operatorname {f0}^{\prime }\left (x \right )+\frac {\operatorname {f1} \left (x \right ) \operatorname {f2} \left (x \right ) \operatorname {f3}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )^{2}}-\frac {\operatorname {f1}^{\prime }\left (x \right ) \operatorname {f2} \left (x \right )}{3 \operatorname {f3} \left (x \right )}-\frac {\operatorname {f1} \left (x \right ) \operatorname {f2}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )}-\frac {4 \operatorname {f2} \left (x \right )^{3} \operatorname {f3}^{\prime }\left (x \right )}{27 \operatorname {f3} \left (x \right )^{3}}+\frac {2 \operatorname {f2} \left (x \right )^{2} \operatorname {f2}^{\prime }\left (x \right )}{9 \operatorname {f3} \left (x \right )^{2}}+\frac {\operatorname {f2}^{\prime \prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )}-\frac {2 \operatorname {f2}^{\prime }\left (x \right ) \operatorname {f3}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )^{2}}+\frac {2 \operatorname {f2} \left (x \right ) \operatorname {f3}^{\prime }\left (x \right )^{2}}{3 \operatorname {f3} \left (x \right )^{3}}-\frac {\operatorname {f2} \left (x \right ) \operatorname {f3}^{\prime \prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )^{2}}\right ) \operatorname {f3} \left (x \right )+\left (\operatorname {f0} \left (x \right )-\frac {\operatorname {f1} \left (x \right ) \operatorname {f2} \left (x \right )}{3 \operatorname {f3} \left (x \right )}+\frac {2 \operatorname {f2} \left (x \right )^{3}}{27 \operatorname {f3} \left (x \right )^{2}}+\frac {\operatorname {f2}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )}-\frac {\operatorname {f2} \left (x \right ) \operatorname {f3}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )^{2}}\right ) \operatorname {f3}^{\prime }\left (x \right )+3 \left (\operatorname {f0} \left (x \right )-\frac {\operatorname {f1} \left (x \right ) \operatorname {f2} \left (x \right )}{3 \operatorname {f3} \left (x \right )}+\frac {2 \operatorname {f2} \left (x \right )^{3}}{27 \operatorname {f3} \left (x \right )^{2}}+\frac {\operatorname {f2}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )}-\frac {\operatorname {f2} \left (x \right ) \operatorname {f3}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )^{2}}\right ) \operatorname {f3} \left (x \right ) \left (\operatorname {f1} \left (x \right )-\frac {\operatorname {f2} \left (x \right )^{2}}{3 \operatorname {f3} \left (x \right )}\right )\right )}^{3}}{27 \operatorname {f3} \left (x \right )^{4} \left (\operatorname {f0} \left (x \right )-\frac {\operatorname {f1} \left (x \right ) \operatorname {f2} \left (x \right )}{3 \operatorname {f3} \left (x \right )}+\frac {2 \operatorname {f2} \left (x \right )^{3}}{27 \operatorname {f3} \left (x \right )^{2}}+\frac {\operatorname {f2}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )}-\frac {\operatorname {f2} \left (x \right ) \operatorname {f3}^{\prime }\left (x \right )}{3 \operatorname {f3} \left (x \right )^{2}}\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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\mathit {f1} \left (x \right ) y-\mathit {f2} \left (x \right ) y^{2}-\mathit {f3} \left (x \right ) y^{3}=\mathit {f0} \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\mathit {f0} \left (x \right )+\mathit {f1} \left (x \right ) y+\mathit {f2} \left (x \right ) y^{2}+\mathit {f3} \left (x \right ) y^{3} \end {array} \]
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 Looking for potential symmetries trying inverse_Riccati trying an equivalence to an Abel ODE differential order: 1; trying a linearization to 2nd order --- trying a change of variables {x -> y(x), y(x) -> x} differential order: 1; trying a linearization to 2nd order trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 2 trying symmetry patterns for 1st order ODEs -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying symmetry patterns of the forms [F(x),G(y)] and [G(y),F(x)] -> trying a symmetry pattern of the form [F(x),G(x)] -> trying a symmetry pattern of the form [F(y),G(y)] -> trying a symmetry pattern of the form [F(x)+G(y), 0] -> trying a symmetry pattern of the form [0, F(x)+G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] -> trying a symmetry pattern of conformal type`
✗ Solution by Maple
dsolve(diff(y(x),x) = f0(x)+f1(x)*y(x)+f2(x)*y(x)^2+f3(x)*y(x)^3,y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==f0[x]+f1[x]y[x]+f2[x] y[x]^2+f3[x]y[x]^3,y[x],x,IncludeSingularSolutions -> True]
Not solved