Internal problem ID [8913]
Internal file name [OUTPUT/7848_Monday_June_06_2022_12_46_46_AM_94272223/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 579.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-F \left (y+\frac {a \,x^{2}}{4}+\frac {x b}{2}\right )=-\frac {x a}{2}} \]
Writing the ode as \begin {align*} y^{\prime }&=-\frac {x a}{2}+F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right )\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}+\left (-\frac {x a}{2}+F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right )\right ) \left (b_{3}-a_{2}\right )-{\left (-\frac {x a}{2}+F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right )\right )}^{2} a_{3}-\left (-\frac {a}{2}+D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) \left (\frac {x a}{2}+\frac {b}{2}\right )\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ b_{2}-\frac {x a b_{3}}{2}+x a a_{2}+F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b_{3}-F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) a_{2}-\frac {a_{3} a^{2} x^{2}}{4}+a_{3} x a F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right )-a_{3} F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right )^{2}+\frac {a y a_{3}}{2}+\frac {a a_{1}}{2}-\frac {D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) a \,x^{2} a_{2}}{2}-\frac {D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) a x y a_{3}}{2}-\frac {D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) a x a_{1}}{2}-\frac {D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b x a_{2}}{2}-\frac {D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b y a_{3}}{2}-\frac {D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b a_{1}}{2}-D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) x b_{2}-D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) y b_{3}-D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b_{1} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -2 D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) a \,x^{2} a_{2}-2 D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) a x y a_{3}-a_{3} a^{2} x^{2}+4 a_{3} x a F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right )-2 D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) a x a_{1}-2 D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b x a_{2}-2 D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b y a_{3}-4 a_{3} F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right )^{2}-2 D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b a_{1}-4 D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) x b_{2}-4 D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) y b_{3}+4 x a a_{2}-2 x a b_{3}+2 a y a_{3}-4 F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) a_{2}+4 F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b_{3}-4 D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) b_{1}+2 a a_{1}+4 b_{2} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ), D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right )\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, F \left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) = v_{3}, D\left (F \right )\left (y +\frac {1}{4} a \,x^{2}+\frac {1}{2} x b \right ) = v_{4}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -a_{3} a^{2} v_{1}^{2}-2 v_{4} a v_{1}^{2} a_{2}-2 v_{4} a v_{1} v_{2} a_{3}-2 v_{4} a v_{1} a_{1}+4 a_{3} v_{1} a v_{3}-2 v_{4} b v_{1} a_{2}-2 v_{4} b v_{2} a_{3}+4 v_{1} a a_{2}+2 a v_{2} a_{3}-2 v_{1} a b_{3}-2 v_{4} b a_{1}-4 a_{3} v_{3}^{2}-4 v_{4} v_{1} b_{2}-4 v_{4} v_{2} b_{3}+2 a a_{1}-4 v_{3} a_{2}-4 v_{4} b_{1}+4 v_{3} b_{3}+4 b_{2} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} -2 v_{4} a v_{1}^{2} a_{2}-a_{3} a^{2} v_{1}^{2}-2 v_{4} a v_{1} v_{2} a_{3}+4 a_{3} v_{1} a v_{3}+\left (-2 a a_{1}-2 b a_{2}-4 b_{2}\right ) v_{1} v_{4}+\left (4 a a_{2}-2 a b_{3}\right ) v_{1}+\left (-2 b a_{3}-4 b_{3}\right ) v_{2} v_{4}+2 a v_{2} a_{3}-4 a_{3} v_{3}^{2}+\left (-4 a_{2}+4 b_{3}\right ) v_{3}+\left (-2 b a_{1}-4 b_{1}\right ) v_{4}+2 a a_{1}+4 b_{2} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -4 a_{3}&=0\\ -2 a a_{2}&=0\\ -2 a a_{3}&=0\\ 2 a a_{3}&=0\\ 4 a a_{3}&=0\\ -a^{2} a_{3}&=0\\ -4 a_{2}+4 b_{3}&=0\\ 2 a a_{1}+4 b_{2}&=0\\ 4 a a_{2}-2 a b_{3}&=0\\ -2 b a_{1}-4 b_{1}&=0\\ -2 b a_{3}-4 b_{3}&=0\\ -2 a a_{1}-2 b a_{2}-4 b_{2}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=a_{1}\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=-\frac {b a_{1}}{2}\\ b_{2}&=-\frac {a a_{1}}{2}\\ b_{3}&=0 \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= 1 \\ \eta &= -\frac {x a}{2}-\frac {b}{2} \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-F \left (y+\frac {a \,x^{2}}{4}+\frac {x b}{2}\right )=-\frac {x a}{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {x a}{2}+F \left (y+\frac {a \,x^{2}}{4}+\frac {x b}{2}\right ) \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: differential order: 1; looking for linear symmetries differential order: 1; found: 1 linear symmetries. Trying reduction of order 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 35
dsolve(diff(y(x),x) = -1/2*a*x+F(y(x)+1/4*a*x^2+1/2*b*x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {a \,x^{2}}{4}-\frac {b x}{2}+\operatorname {RootOf}\left (-x +2 \left (\int _{}^{\textit {\_Z}}\frac {1}{2 F \left (\textit {\_a} \right )+b}d \textit {\_a} \right )+c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.227 (sec). Leaf size: 514
DSolve[y'[x] == -1/2*(a*x) + F[(b*x)/2 + (a*x^2)/4 + y[x]],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}-\frac {b \int _1^x\left (\frac {2 a K[1] F'\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )}{\left (b+2 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )\right )^2}+\frac {2 F'\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )}{b+2 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )}-\frac {4 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right ) F'\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )}{\left (b+2 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )\right )^2}\right )dK[1]+2 F\left (\frac {a x^2}{4}+\frac {b x}{2}+K[2]\right ) \int _1^x\left (\frac {2 a K[1] F'\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )}{\left (b+2 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )\right )^2}+\frac {2 F'\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )}{b+2 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )}-\frac {4 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right ) F'\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )}{\left (b+2 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+K[2]\right )\right )^2}\right )dK[1]+2}{b+2 F\left (\frac {a x^2}{4}+\frac {b x}{2}+K[2]\right )}dK[2]+\int _1^x\left (\frac {2 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+y(x)\right )}{b+2 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+y(x)\right )}-\frac {a K[1]}{b+2 F\left (\frac {1}{4} a K[1]^2+\frac {1}{2} b K[1]+y(x)\right )}\right )dK[1]=c_1,y(x)\right ] \]