1.50 problem 50

1.50.1 Solving as abelFirstKind ode
1.50.2 Maple step by step solution

Internal problem ID [8387]
Internal file name [OUTPUT/7320_Sunday_June_05_2022_05_47_49_PM_33136285/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 50.
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.

\[ \boxed {y^{\prime }-f_{3} \left (x \right ) y^{3}-f_{2} \left (x \right ) y^{2}-f_{1} \left (x \right ) y=f_{0} \left (x \right )} \]

1.50.1 Solving as abelFirstKind ode

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 }&=f_{3} \left (x \right ) y^{3}+f_{2} \left (x \right ) y^{2}+f_{1} \left (x \right ) y+f_{0} \left (x \right )\tag {1} \end {align*}

Therefore \begin {align*} f_0(x) &= f_{0} \left (x \right )\\ f_1(x) &= f_{1} \left (x \right )\\ f_2(x) &= f_{2} \left (x \right )\\ f_3(x) &= f_{3} \left (x \right ) \end {align*}

Since \(f_2(x)=f_{2} \left (x \right )\) 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 {f_{2} \left (x \right )}{3 f_{3} \left (x \right )} \right ) \\ &= u \left (x \right )-\frac {f_{2} \left (x \right )}{3 f_{3} \left (x \right )} \end {align*}

The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = f_{3} \left (x \right ) u \left (x \right )^{3}+u \left (x \right ) f_{1} \left (x \right )-\frac {f_{2} \left (x \right )^{2} u \left (x \right )}{3 f_{3} \left (x \right )}+f_{0} \left (x \right )-\frac {f_{2} \left (x \right ) f_{1} \left (x \right )}{3 f_{3} \left (x \right )}+\frac {2 f_{2} \left (x \right )^{3}}{27 f_{3} \left (x \right )^{2}}+\frac {f_{2}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )}-\frac {f_{2} \left (x \right ) f_{3}^{\prime }\left (x \right )}{3 f_{3} \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 )&=f_{3} \left (x \right ) u \left (x \right )^{3}+\frac {\left (27 f_{1} \left (x \right ) f_{3} \left (x \right )^{2}-9 f_{3} \left (x \right ) f_{2} \left (x \right )^{2}\right ) u \left (x \right )}{27 f_{3} \left (x \right )^{2}}+\frac {27 f_{3} \left (x \right )^{2} f_{0} \left (x \right )-9 f_{3} \left (x \right ) f_{2} \left (x \right ) f_{1} \left (x \right )+2 f_{2} \left (x \right )^{3}+9 f_{2}^{\prime }\left (x \right ) f_{3} \left (x \right )-9 f_{2} \left (x \right ) f_{3}^{\prime }\left (x \right )}{27 f_{3} \left (x \right )^{2}}\tag {1} \end {align*}

Therefore \begin {align*} f_0(x) &= f_{0} \left (x \right )-\frac {f_{2} \left (x \right ) f_{1} \left (x \right )}{3 f_{3} \left (x \right )}+\frac {2 f_{2} \left (x \right )^{3}}{27 f_{3} \left (x \right )^{2}}+\frac {f_{2}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )}-\frac {f_{2} \left (x \right ) f_{3}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )^{2}}\\ f_1(x) &= f_{1} \left (x \right )-\frac {f_{2} \left (x \right )^{2}}{3 f_{3} \left (x \right )}\\ f_2(x) &= 0\\ f_3(x) &= f_{3} \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 (f_{0}^{\prime }\left (x \right )+\frac {f_{2} \left (x \right ) f_{1} \left (x \right ) f_{3}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )^{2}}-\frac {f_{2}^{\prime }\left (x \right ) f_{1} \left (x \right )}{3 f_{3} \left (x \right )}-\frac {f_{2} \left (x \right ) f_{1}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )}-\frac {4 f_{2} \left (x \right )^{3} f_{3}^{\prime }\left (x \right )}{27 f_{3} \left (x \right )^{3}}+\frac {2 f_{2} \left (x \right )^{2} f_{2}^{\prime }\left (x \right )}{9 f_{3} \left (x \right )^{2}}+\frac {f_{2}^{\prime \prime }\left (x \right )}{3 f_{3} \left (x \right )}-\frac {2 f_{2}^{\prime }\left (x \right ) f_{3}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )^{2}}+\frac {2 f_{2} \left (x \right ) f_{3}^{\prime }\left (x \right )^{2}}{3 f_{3} \left (x \right )^{3}}-\frac {f_{2} \left (x \right ) f_{3}^{\prime \prime }\left (x \right )}{3 f_{3} \left (x \right )^{2}}\right ) f_{3} \left (x \right )+\left (f_{0} \left (x \right )-\frac {f_{2} \left (x \right ) f_{1} \left (x \right )}{3 f_{3} \left (x \right )}+\frac {2 f_{2} \left (x \right )^{3}}{27 f_{3} \left (x \right )^{2}}+\frac {f_{2}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )}-\frac {f_{2} \left (x \right ) f_{3}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )^{2}}\right ) f_{3}^{\prime }\left (x \right )+3 \left (f_{0} \left (x \right )-\frac {f_{2} \left (x \right ) f_{1} \left (x \right )}{3 f_{3} \left (x \right )}+\frac {2 f_{2} \left (x \right )^{3}}{27 f_{3} \left (x \right )^{2}}+\frac {f_{2}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )}-\frac {f_{2} \left (x \right ) f_{3}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )^{2}}\right ) f_{3} \left (x \right ) \left (f_{1} \left (x \right )-\frac {f_{2} \left (x \right )^{2}}{3 f_{3} \left (x \right )}\right )\right )}^{3}}{27 f_{3} \left (x \right )^{4} \left (f_{0} \left (x \right )-\frac {f_{2} \left (x \right ) f_{1} \left (x \right )}{3 f_{3} \left (x \right )}+\frac {2 f_{2} \left (x \right )^{3}}{27 f_{3} \left (x \right )^{2}}+\frac {f_{2}^{\prime }\left (x \right )}{3 f_{3} \left (x \right )}-\frac {f_{2} \left (x \right ) f_{3}^{\prime }\left (x \right )}{3 f_{3} \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.

1.50.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-f_{3} \left (x \right ) y^{3}-f_{2} \left (x \right ) y^{2}-f_{1} \left (x \right ) y=f_{0} \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=f_{3} \left (x \right ) y^{3}+f_{2} \left (x \right ) y^{2}+f_{1} \left (x \right ) y+f_{0} \left (x \right ) \end {array} \]

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) - f__3(x)*y(x)^3 - f__2(x)*y(x)^2 - f__1(x)*y(x) - f__0(x)=0,y(x), singsol=all)
 

\[ \text {No solution found} \]

Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

DSolve[y'[x] - f3[x]*y[x]^3 - f2[x]*y[x]^2 - f1[x]*y[x] - f0[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

Not solved