Internal problem ID [8884]
Internal file name [OUTPUT/7819_Monday_June_06_2022_12_36_05_AM_41370797/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 550.
ODE order: 1.
ODE degree: 0.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_homogeneous, `class G`]]
Unable to solve or complete the solution.
\[ \boxed {{y^{\prime }}^{r}-a y^{s}=b \,x^{\frac {r s}{r -s}}} \] Solving the given ode for \(y^{\prime }\) results in \(1\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\left (a y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} \tag {1} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Writing the ode as \begin {align*} y^{\prime }&=\left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}}\\ 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 (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} \left (b_{3}-a_{2}\right )-\left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {2}{r}} a_{3}-\frac {\left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b \,x^{\frac {r s}{r -s}} s \left (x a_{2}+y a_{3}+a_{1}\right )}{\left (r -s \right ) x \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )}-\frac {\left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,y^{s} s \left (x b_{2}+y b_{3}+b_{1}\right )}{r y \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )} = 0 \end{equation} Putting the above in normal form gives \[ -\frac {x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {2}{r}} b \,r^{2} x y a_{3}-x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {2}{r}} b r s x y a_{3}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {2}{r}} a \,r^{2} x y a_{3}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {2}{r}} a r s x y a_{3}+x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b \,r^{2} x y a_{2}-x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b \,r^{2} x y b_{3}+x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b r s x y b_{3}+x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b r s \,y^{2} a_{3}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,r^{2} x y a_{2}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,r^{2} x y b_{3}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a r s \,x^{2} b_{2}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a r s x y a_{2}+2 y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a r s x y b_{3}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,s^{2} x^{2} b_{2}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,s^{2} x y b_{3}+x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b r s y a_{1}-x^{\frac {r s}{r -s}} b \,r^{2} x y b_{2}+x^{\frac {r s}{r -s}} b r s x y b_{2}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a r s x b_{1}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,s^{2} x b_{1}-y^{s} a \,r^{2} x y b_{2}+y^{s} a r s x y b_{2}}{\left (r -s \right ) x \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right ) r y} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {2}{r}} b \,r^{2} x y a_{3}+x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {2}{r}} b r s x y a_{3}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {2}{r}} a \,r^{2} x y a_{3}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {2}{r}} a r s x y a_{3}-x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b \,r^{2} x y a_{2}+x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b \,r^{2} x y b_{3}-x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b r s x y b_{3}-x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b r s \,y^{2} a_{3}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,r^{2} x y a_{2}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,r^{2} x y b_{3}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a r s \,x^{2} b_{2}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a r s x y a_{2}-2 y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a r s x y b_{3}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,s^{2} x^{2} b_{2}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,s^{2} x y b_{3}-x^{\frac {r s}{r -s}} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} b r s y a_{1}+x^{\frac {r s}{r -s}} b \,r^{2} x y b_{2}-x^{\frac {r s}{r -s}} b r s x y b_{2}-y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a r s x b_{1}+y^{s} \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} a \,s^{2} x b_{1}+y^{s} a \,r^{2} x y b_{2}-y^{s} a r s x y b_{2} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, x^{\frac {r s}{r -s}}, y^{s}, \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, x^{\frac {r s}{r -s}} = v_{3}, y^{s} = v_{4}, \left (a \,y^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -v_{4} v_{5}^{2} a \,r^{2} v_{1} v_{2} a_{3}+v_{4} v_{5}^{2} a r s v_{1} v_{2} a_{3}-v_{3} v_{5}^{2} b \,r^{2} v_{1} v_{2} a_{3}+v_{3} v_{5}^{2} b r s v_{1} v_{2} a_{3}-v_{4} v_{5} a \,r^{2} v_{1} v_{2} a_{2}+v_{4} v_{5} a \,r^{2} v_{1} v_{2} b_{3}+v_{4} v_{5} a r s v_{1} v_{2} a_{2}-v_{4} v_{5} a r s v_{1}^{2} b_{2}-2 v_{4} v_{5} a r s v_{1} v_{2} b_{3}+v_{4} v_{5} a \,s^{2} v_{1}^{2} b_{2}+v_{4} v_{5} a \,s^{2} v_{1} v_{2} b_{3}-v_{3} v_{5} b \,r^{2} v_{1} v_{2} a_{2}+v_{3} v_{5} b \,r^{2} v_{1} v_{2} b_{3}-v_{3} v_{5} b r s v_{2}^{2} a_{3}-v_{3} v_{5} b r s v_{1} v_{2} b_{3}+v_{4} a \,r^{2} v_{1} v_{2} b_{2}-v_{4} v_{5} a r s v_{1} b_{1}-v_{4} a r s v_{1} v_{2} b_{2}+v_{4} v_{5} a \,s^{2} v_{1} b_{1}+v_{3} b \,r^{2} v_{1} v_{2} b_{2}-v_{3} v_{5} b r s v_{2} a_{1}-v_{3} b r s v_{1} v_{2} b_{2} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (-a r s b_{2}+a \,s^{2} b_{2}\right ) v_{1}^{2} v_{4} v_{5}+\left (-b \,r^{2} a_{3}+b r s a_{3}\right ) v_{1} v_{2} v_{3} v_{5}^{2}+\left (-b \,r^{2} a_{2}+b \,r^{2} b_{3}-b r s b_{3}\right ) v_{1} v_{2} v_{3} v_{5}+\left (b \,r^{2} b_{2}-b r s b_{2}\right ) v_{1} v_{2} v_{3}+\left (-a \,r^{2} a_{3}+a r s a_{3}\right ) v_{1} v_{2} v_{4} v_{5}^{2}+\left (-a \,r^{2} a_{2}+a \,r^{2} b_{3}+a r s a_{2}-2 a r s b_{3}+a \,s^{2} b_{3}\right ) v_{1} v_{2} v_{4} v_{5}+\left (a \,r^{2} b_{2}-a r s b_{2}\right ) v_{1} v_{2} v_{4}+\left (-a r s b_{1}+a \,s^{2} b_{1}\right ) v_{1} v_{4} v_{5}-v_{3} v_{5} b r s v_{2}^{2} a_{3}-v_{3} v_{5} b r s v_{2} a_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -b r s a_{1}&=0\\ -b r s a_{3}&=0\\ -a \,r^{2} a_{3}+a r s a_{3}&=0\\ -a r s b_{1}+a \,s^{2} b_{1}&=0\\ -a r s b_{2}+a \,s^{2} b_{2}&=0\\ a \,r^{2} b_{2}-a r s b_{2}&=0\\ -b \,r^{2} a_{3}+b r s a_{3}&=0\\ b \,r^{2} b_{2}-b r s b_{2}&=0\\ -b \,r^{2} a_{2}+b \,r^{2} b_{3}-b r s b_{3}&=0\\ -a \,r^{2} a_{2}+a \,r^{2} b_{3}+a r s a_{2}-2 a r s b_{3}+a \,s^{2} b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=\frac {\left (r -s \right ) b_{3}}{r}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \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 &= \frac {x \left (r -s \right )}{r} \\ \eta &= y \\ \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
Unable to determine ODE type.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{r}-a y^{s}=b \,x^{\frac {r s}{r -s}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{\frac {\ln \left (a y^{s}+b \,x^{\frac {r s}{r -s}}\right )}{r}} \end {array} \]
Maple trace
`Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables trying simple symmetries for implicit equations --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous G 1st order, trying the canonical coordinates of the invariance group -> Calling odsolve with the ODE`, diff(y(x), x) = r*y(x)/((r-s)*x), y(x)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful <- 1st order, canonical coordinates successful <- homogeneous successful`
✓ Solution by Maple
Time used: 0.125 (sec). Leaf size: 64
dsolve(diff(y(x),x)^r-a*y(x)^s-b*x^(r*s/(r-s))=0,y(x), singsol=all)
\[ -\left (\int _{\textit {\_b}}^{y \left (x \right )}\frac {1}{x \left (r -s \right ) \left (a \,\textit {\_a}^{s}+b \,x^{\frac {r s}{r -s}}\right )^{\frac {1}{r}}-r \textit {\_a}}d \textit {\_a} \right )+\frac {\ln \left (x \right )}{r -s}-c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 0.932 (sec). Leaf size: 488
DSolve[-(b*x^((r*s)/(r - s))) - a*y[x]^s + y'[x]^r==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (\frac {r}{-r x \left (a K[2]^s+b x^{\frac {r s}{r-s}}\right )^{\frac {1}{r}}+s x \left (a K[2]^s+b x^{\frac {r s}{r-s}}\right )^{\frac {1}{r}}+r K[2]}-\int _1^x\left (\frac {a s K[2]^{s-1} \left (a K[2]^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}-1}}{r K[1] \left (a K[2]^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}}-s K[1] \left (a K[2]^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}}-r K[2]}-\frac {r \left (a K[2]^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}} \left (-\frac {a s^2 K[1] K[2]^{s-1} \left (a K[2]^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}-1}}{r}+a s K[1] K[2]^{s-1} \left (a K[2]^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}-1}-r\right )}{\left (r K[1] \left (a K[2]^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}}-s K[1] \left (a K[2]^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}}-r K[2]\right )^2}\right )dK[1]\right )dK[2]+\int _1^x\frac {r \left (a y(x)^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}}}{r K[1] \left (a y(x)^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}}-s K[1] \left (a y(x)^s+b K[1]^{\frac {r s}{r-s}}\right )^{\frac {1}{r}}-r y(x)}dK[1]=c_1,y(x)\right ] \]