Internal problem ID [9278]
Internal file name [OUTPUT/8214_Monday_June_06_2022_02_19_09_AM_4302220/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 945.
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], _rational, [_Abel, `2nd type`, `class C`]]
\[ \boxed {y^{\prime }-\frac {-32 x y-8 x^{3}-16 a \,x^{2}-32 x +64 y^{3}+48 x^{2} y^{2}+96 y^{2} a x +12 y x^{4}+48 y a \,x^{3}+48 y a^{2} x^{2}+x^{6}+6 x^{5} a +12 a^{2} x^{4}+8 a^{3} x^{3}}{64 y+16 x^{2}+32 a x +64}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {8 a^{3} x^{3}+12 a^{2} x^{4}+6 x^{5} a +x^{6}+48 y \,a^{2} x^{2}+48 y a \,x^{3}+12 y \,x^{4}+96 a x \,y^{2}+48 x^{2} y^{2}-16 a \,x^{2}-8 x^{3}+64 y^{3}-32 x y -32 x}{32 a x +16 x^{2}+64 y +64}\\ 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}+\frac {\left (8 a^{3} x^{3}+12 a^{2} x^{4}+6 x^{5} a +x^{6}+48 y \,a^{2} x^{2}+48 y a \,x^{3}+12 y \,x^{4}+96 a x \,y^{2}+48 x^{2} y^{2}-16 a \,x^{2}-8 x^{3}+64 y^{3}-32 x y -32 x \right ) \left (b_{3}-a_{2}\right )}{32 a x +16 x^{2}+64 y +64}-\frac {\left (8 a^{3} x^{3}+12 a^{2} x^{4}+6 x^{5} a +x^{6}+48 y \,a^{2} x^{2}+48 y a \,x^{3}+12 y \,x^{4}+96 a x \,y^{2}+48 x^{2} y^{2}-16 a \,x^{2}-8 x^{3}+64 y^{3}-32 x y -32 x \right )^{2} a_{3}}{256 \left (2 a x +x^{2}+4 y +4\right )^{2}}-\left (\frac {24 a^{3} x^{2}+48 a^{2} x^{3}+30 a \,x^{4}+6 x^{5}+96 a^{2} x y +144 a \,x^{2} y +48 x^{3} y +96 a \,y^{2}+96 x \,y^{2}-32 a x -24 x^{2}-32 y -32}{32 a x +16 x^{2}+64 y +64}-\frac {\left (8 a^{3} x^{3}+12 a^{2} x^{4}+6 x^{5} a +x^{6}+48 y \,a^{2} x^{2}+48 y a \,x^{3}+12 y \,x^{4}+96 a x \,y^{2}+48 x^{2} y^{2}-16 a \,x^{2}-8 x^{3}+64 y^{3}-32 x y -32 x \right ) \left (2 a +2 x \right )}{16 \left (2 a x +x^{2}+4 y +4\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {48 a^{2} x^{2}+48 a \,x^{3}+12 x^{4}+192 y a x +96 x^{2} y +192 y^{2}-32 x}{32 a x +16 x^{2}+64 y +64}-\frac {8 a^{3} x^{3}+12 a^{2} x^{4}+6 x^{5} a +x^{6}+48 y \,a^{2} x^{2}+48 y a \,x^{3}+12 y \,x^{4}+96 a x \,y^{2}+48 x^{2} y^{2}-16 a \,x^{2}-8 x^{3}+64 y^{3}-32 x y -32 x}{4 \left (2 a x +x^{2}+4 y +4\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ \text {Expression too large to display} \] Setting the numerator to zero gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \{x, y\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} \text {Expression too large to display} \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \text {Expression too large to display} \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -6144 a_{3}&=0\\ -4096 a_{3}&=0\\ -3840 a_{3}&=0\\ -1280 a_{3}&=0\\ -240 a_{3}&=0\\ -24 a_{3}&=0\\ -a_{3}&=0\\ -15360 a a_{3}&=0\\ -12288 a a_{3}&=0\\ -7680 a a_{3}&=0\\ -1920 a a_{3}&=0\\ -240 a a_{3}&=0\\ -12 a a_{3}&=0\\ -15360 a^{2} a_{3}&=0\\ -5760 a^{2} a_{3}&=0\\ -960 a^{2} a_{3}&=0\\ -60 a^{2} a_{3}&=0\\ 2048 a_{1}+4096 b_{2}&=0\\ 4096 a_{1}+2048 a_{3}+8192 b_{2}&=0\\ -10240 a^{3} a_{3}+1024 a_{3}&=0\\ -7680 a^{3} a_{3}+768 a_{3}&=0\\ -1920 a^{3} a_{3}+192 a_{3}&=0\\ -160 a^{3} a_{3}+16 a_{3}&=0\\ -3840 a^{4} a_{3}+2304 a a_{3}-4608 a_{2}&=0\\ -15360 a a_{2}-3072 a_{1}-6144 b_{2}&=0\\ -4096 a a_{3}-4096 a_{2}-4096 b_{3}&=0\\ -1024 a a_{3}-8192 a_{2}-2048 b_{3}&=0\\ -1920 a^{4} a_{3}+1088 a a_{3}-1024 a_{2}+128 b_{3}&=0\\ -240 a^{4} a_{3}+128 a a_{3}-80 a_{2}+16 b_{3}&=0\\ 2048 a a_{1}+4096 a b_{2}+4096 a_{2}-2048 b_{3}&=0\\ -6144 a a_{1}+2048 a_{1}+4096 a_{3}-12288 b_{1}+4096 b_{2}&=0\\ -4096 a a_{1}-6144 a a_{3}-4096 a_{2}+2048 a_{3}-8192 b_{1}-8192 b_{3}&=0\\ -6144 a^{2} a_{3}-12288 a a_{2}-4096 a b_{3}-4096 a_{1}-2048 a_{3}-8192 b_{2}&=0\\ -768 a^{5} a_{3}+1920 a^{2} a_{3}-5376 a a_{2}+768 a b_{3}-768 a_{1}+384 a_{3}-1536 b_{2}&=0\\ -192 a^{5} a_{3}+384 a^{2} a_{3}-576 a a_{2}+128 a b_{3}-64 a_{1}+64 a_{3}-128 b_{2}&=0\\ -6144 a^{2} a_{1}+2048 a a_{1}+2048 a a_{3}-12288 a b_{1}+4096 a b_{2}+8192 a_{2}-4096 b_{3}&=0\\ -3072 a^{3} a_{3}-12288 a^{2} a_{2}-9216 a a_{1}-3072 a a_{3}-12288 a b_{2}-9216 a_{2}-6144 b_{1}-3072 b_{3}&=0\\ 768 a^{3} a_{3}-9216 a^{2} a_{2}+1536 a^{2} b_{3}-3840 a a_{1}+1152 a a_{3}-6144 a b_{2}-3840 a_{2}-384 a_{3}-1536 b_{1}&=0\\ -1536 a^{3} a_{1}+512 a^{2} a_{1}-3072 a^{2} b_{1}+1024 a^{2} b_{2}+4096 a a_{2}-2048 a b_{3}+1024 a_{1}-1024 a_{3}+2048 b_{2}&=0\\ -6144 a^{2} a_{1}-6144 a^{2} a_{3}-12288 a a_{2}+2048 a a_{3}-12288 a b_{1}-6144 a b_{3}-6144 a_{1}+4096 a_{2}-12288 b_{2}-2048 b_{3}&=0\\ -64 a^{6} a_{3}+512 a^{3} a_{3}-1536 a^{2} a_{2}+384 a^{2} b_{3}-448 a a_{1}+384 a a_{3}-768 a b_{2}-448 a_{2}-64 a_{3}-128 b_{1}+64 b_{3}&=0\\ -512 a^{4} a_{3}-5120 a^{3} a_{2}+1024 a^{3} b_{3}-6144 a^{2} a_{1}-6144 a^{2} b_{2}-12288 a a_{2}-512 a a_{3}-6144 a b_{1}-3072 a_{1}+2048 a_{2}-6144 b_{2}-1024 b_{3}&=0\\ -3072 a^{3} a_{1}-1536 a^{3} a_{3}-9216 a^{2} a_{2}+512 a^{2} a_{3}-6144 a^{2} b_{1}-9216 a a_{1}+4096 a a_{2}-12288 a b_{2}-2048 a b_{3}+1024 a_{1}-1024 a_{3}-6144 b_{1}+2048 b_{2}&=0\\ 256 a^{4} a_{3}-1792 a^{3} a_{2}+512 a^{3} b_{3}-1152 a^{2} a_{1}+768 a^{2} a_{3}-1536 a^{2} b_{2}-2304 a a_{2}-256 a a_{3}-768 a b_{1}+384 a b_{3}-384 a_{1}+256 a_{2}-768 b_{2}-128 b_{3}&=0\\ -512 a^{4} a_{1}-2048 a^{3} a_{2}-1024 a^{3} b_{1}+512 a^{3} b_{3}-3072 a^{2} a_{1}+1024 a^{2} a_{2}-3072 a^{2} b_{2}-512 a^{2} b_{3}+512 a a_{1}-1024 a a_{3}-3072 a b_{1}+1024 a b_{2}+2048 a_{2}-1024 b_{3}&=0\\ -768 a^{4} a_{2}+256 a^{4} b_{3}-1280 a^{3} a_{1}+512 a^{3} a_{3}-1024 a^{3} b_{2}-3840 a^{2} a_{2}-256 a^{2} a_{3}-1536 a^{2} b_{1}+768 a^{2} b_{3}-1920 a a_{1}+1024 a a_{2}-3072 a b_{2}-512 a b_{3}+128 a_{1}-512 a_{3}-768 b_{1}+256 b_{2}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=-2 b_{2}\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=a b_{2}\\ b_{2}&=b_{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 &= -2 \\ \eta &= a +x \\ \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 \(S(R)\). Therefore \begin {align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {a +x}{-2}\\ &= -\frac {a}{2}-\frac {x}{2} \end {align*}
This is easily solved to give \begin {align*} y = -\frac {1}{4} x^{2}-\frac {1}{2} a x +c_{1} \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {1}{4} x^{2}+\frac {1}{2} a x +y \end {align*}
And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{-2} \end {align*}
Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= -\frac {x}{2} \end {align*}
Where the constant of integration is set to zero as we just need one solution. Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating \begin {align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end {align*}
Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by \begin {align*} \omega (x,y) &= \frac {8 a^{3} x^{3}+12 a^{2} x^{4}+6 x^{5} a +x^{6}+48 y \,a^{2} x^{2}+48 y a \,x^{3}+12 y \,x^{4}+96 a x \,y^{2}+48 x^{2} y^{2}-16 a \,x^{2}-8 x^{3}+64 y^{3}-32 x y -32 x}{32 a x +16 x^{2}+64 y +64} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= \frac {x}{2}+\frac {a}{2}\\ R_{y} &= 1\\ S_{x} &= -{\frac {1}{2}}\\ S_{y} &= 0 \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= \frac {-16 a x -8 x^{2}-32 y -32}{x^{6}+6 x^{5} a +12 \left (a^{2}+y \right ) x^{4}+8 \left (a^{3}+6 a y \right ) x^{3}+8 \left (6 a^{2} y +6 y^{2}+a \right ) x^{2}+16 \left (6 a \,y^{2}+a^{2}\right ) x +32 \left (y +1\right ) a +64 y^{3}}\tag {2A} \end {align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= \frac {-R -1}{2 R^{3}+R a +a} \end {align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \int -\frac {R +1}{2 R^{3}+R a +a}d R +c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,y\) coordinates. This results in \begin {align*} -\frac {x}{2} = \int _{}^{y}-\frac {\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} +1}{2 \left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right )^{3}+\left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right ) a +a}d \textit {\_a} +c_{1} \end {align*}
Which simplifies to \begin {align*} -\frac {x}{2} = \int _{}^{y}-\frac {\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} +1}{2 \left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right )^{3}+\left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right ) a +a}d \textit {\_a} +c_{1} \end {align*}
This results in \begin {align*} -\frac {x}{2} = \int _{}^{y}-\frac {\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} +1}{2 \left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right )^{3}+\left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right ) a +a}d \textit {\_a} +c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {x}{2} &= \int _{}^{y}-\frac {\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} +1}{2 \left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right )^{3}+\left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right ) a +a}d \textit {\_a} +c_{1} \\ \end{align*}
Verification of solutions
\[ -\frac {x}{2} = \int _{}^{y}-\frac {\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} +1}{2 \left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right )^{3}+\left (\frac {1}{4} x^{2}+\frac {1}{2} a x +\textit {\_a} \right ) a +a}d \textit {\_a} +c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -8 a^{3} x^{3}-12 a^{2} x^{4}-6 x^{5} a -x^{6}-48 y a^{2} x^{2}-48 y a \,x^{3}-12 y x^{4}-96 y^{2} a x -48 x^{2} y^{2}+32 y^{\prime } a x +16 y^{\prime } x^{2}-64 y^{3}+16 a \,x^{2}+8 x^{3}+64 y^{\prime } y+32 x y+64 y^{\prime }+32 x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-32 x y-8 x^{3}-16 a \,x^{2}-32 x +64 y^{3}+48 x^{2} y^{2}+96 y^{2} a x +12 y x^{4}+48 y a \,x^{3}+48 y a^{2} x^{2}+x^{6}+6 x^{5} a +12 a^{2} x^{4}+8 a^{3} x^{3}}{64 y+16 x^{2}+32 a x +64} \end {array} \]
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 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: 42
dsolve(diff(y(x),x) = (-32*x*y(x)-8*x^3-16*a*x^2-32*x+64*y(x)^3+48*x^2*y(x)^2+96*a*x*y(x)^2+12*y(x)*x^4+48*y(x)*a*x^3+48*a^2*x^2*y(x)+x^6+6*x^5*a+12*a^2*x^4+8*a^3*x^3)/(64*y(x)+16*x^2+32*a*x+64),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x^{2}}{4}-\frac {a x}{2}+\operatorname {RootOf}\left (-x +2 \left (\int _{}^{\textit {\_Z}}\frac {\textit {\_a} +1}{2 \textit {\_a}^{3}+\textit {\_a} a +a}d \textit {\_a} \right )+c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 18.129 (sec). Leaf size: 213
DSolve[y'[x] == (-32*x - 16*a*x^2 - 8*x^3 + 8*a^3*x^3 + 12*a^2*x^4 + 6*a*x^5 + x^6 - 32*x*y[x] + 48*a^2*x^2*y[x] + 48*a*x^3*y[x] + 12*x^4*y[x] + 96*a*x*y[x]^2 + 48*x^2*y[x]^2 + 64*y[x]^3)/(64 + 32*a*x + 16*x^2 + 64*y[x]),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [x-4 \text {RootSum}\left [\text {$\#$1}^6+6 \text {$\#$1}^5 a+12 \text {$\#$1}^4 a^2+12 \text {$\#$1}^4 y(x)+8 \text {$\#$1}^3 a^3+48 \text {$\#$1}^3 a y(x)+48 \text {$\#$1}^2 a^2 y(x)+8 \text {$\#$1}^2 a+48 \text {$\#$1}^2 y(x)^2+16 \text {$\#$1} a^2+96 \text {$\#$1} a y(x)^2+32 a y(x)+32 a+64 y(x)^3\&,\frac {\text {$\#$1}^2 \log (x-\text {$\#$1})+2 \text {$\#$1} a \log (x-\text {$\#$1})+4 y(x) \log (x-\text {$\#$1})+4 \log (x-\text {$\#$1})}{3 \text {$\#$1}^4+12 \text {$\#$1}^3 a+12 \text {$\#$1}^2 a^2+24 \text {$\#$1}^2 y(x)+48 \text {$\#$1} a y(x)+8 a+48 y(x)^2}\&\right ]=c_1,y(x)\right ] \]