Internal problem ID [9284]
Internal file name [OUTPUT/8220_Monday_June_06_2022_02_20_19_AM_66511834/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 951.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries], _Abel]
\[ \boxed {y^{\prime }-y^{2}-\frac {y x^{2}}{2}-a x y-y^{3}-\frac {3 y^{2} x^{2}}{4}-\frac {3 a y^{2} x}{2}-\frac {3 x^{4} y}{16}-\frac {3 y a \,x^{3}}{4}-\frac {3 y a^{2} x^{2}}{4}=-\frac {1}{2} x +1+\frac {1}{16} x^{4}+\frac {1}{4} x^{3} a +\frac {1}{4} a^{2} x^{2}+\frac {1}{64} x^{6}+\frac {3}{32} a \,x^{5}+\frac {3}{16} a^{2} x^{4}+\frac {1}{8} x^{3} a^{3}} \]
Writing the ode as \begin {align*} y^{\prime }&=-\frac {1}{2} x +1+y^{2}+\frac {1}{2} x^{2} y +a x y +\frac {1}{16} x^{4}+\frac {1}{4} x^{3} a +\frac {1}{4} a^{2} x^{2}+y^{3}+\frac {3}{4} y^{2} x^{2}+\frac {3}{2} a x \,y^{2}+\frac {3}{16} x^{4} y +\frac {3}{4} x^{3} a y +\frac {3}{4} y \,a^{2} x^{2}+\frac {1}{64} x^{6}+\frac {3}{32} a \,x^{5}+\frac {3}{16} a^{2} x^{4}+\frac {1}{8} x^{3} a^{3}\\ 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 {1}{2} x +1+y^{2}+\frac {1}{2} x^{2} y +a x y +\frac {1}{16} x^{4}+\frac {1}{4} x^{3} a +\frac {1}{4} a^{2} x^{2}+y^{3}+\frac {3}{4} y^{2} x^{2}+\frac {3}{2} a x \,y^{2}+\frac {3}{16} x^{4} y +\frac {3}{4} x^{3} a y +\frac {3}{4} y \,a^{2} x^{2}+\frac {1}{64} x^{6}+\frac {3}{32} a \,x^{5}+\frac {3}{16} a^{2} x^{4}+\frac {1}{8} x^{3} a^{3}\right ) \left (b_{3}-a_{2}\right )-\left (-\frac {1}{2} x +1+y^{2}+\frac {1}{2} x^{2} y +a x y +\frac {1}{16} x^{4}+\frac {1}{4} x^{3} a +\frac {1}{4} a^{2} x^{2}+y^{3}+\frac {3}{4} y^{2} x^{2}+\frac {3}{2} a x \,y^{2}+\frac {3}{16} x^{4} y +\frac {3}{4} x^{3} a y +\frac {3}{4} y \,a^{2} x^{2}+\frac {1}{64} x^{6}+\frac {3}{32} a \,x^{5}+\frac {3}{16} a^{2} x^{4}+\frac {1}{8} x^{3} a^{3}\right )^{2} a_{3}-\left (-\frac {1}{2}+y x +a y +\frac {1}{4} x^{3}+\frac {3}{4} a \,x^{2}+\frac {1}{2} a^{2} x +\frac {3}{2} y^{2} x +\frac {3}{2} a \,y^{2}+\frac {3}{4} y \,x^{3}+\frac {9}{4} a \,x^{2} y +\frac {3}{2} a^{2} x y +\frac {3}{32} x^{5}+\frac {15}{32} a \,x^{4}+\frac {3}{4} a^{2} x^{3}+\frac {3}{8} a^{3} x^{2}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (2 y +\frac {1}{2} x^{2}+x a +3 y^{2}+\frac {3}{2} x^{2} y +3 a x y +\frac {3}{16} x^{4}+\frac {3}{4} x^{3} a +\frac {3}{4} a^{2} x^{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*} -8192 a_{3}&=0\\ -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\\ -20480 a 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\\ 2048 a_{1}-4096 a_{2}-4096 a_{3}+4096 b_{2}+4096 b_{3}&=0\\ -8192 a a_{3}-2048 a_{3}&=0\\ -20480 a^{2} a_{3}-4096 a_{3}&=0\\ -15360 a^{2} a_{3}-10240 a_{3}&=0\\ -15360 a^{2} a_{3}-5120 a_{3}&=0\\ -5760 a^{2} a_{3}-1280 a_{3}&=0\\ -960 a^{2} a_{3}-160 a_{3}&=0\\ -60 a^{2} a_{3}-8 a_{3}&=0\\ -10240 a^{3} a_{3}-20480 a a_{3}&=0\\ -10240 a^{3} a_{3}-6144 a a_{3}&=0\\ -7680 a^{3} a_{3}-7680 a a_{3}&=0\\ -1920 a^{3} a_{3}-1280 a a_{3}&=0\\ -160 a^{3} a_{3}-80 a a_{3}&=0\\ -3840 a^{4} a_{3}-15360 a^{2} a_{3}-1536 a_{3}&=0\\ -1920 a^{4} a_{3}-3840 a^{2} a_{3}-256 a_{3}&=0\\ -240 a^{4} a_{3}-320 a^{2} a_{3}-16 a_{3}&=0\\ -4096 a a_{1}+2048 a_{3}-8192 b_{1}&=0\\ -768 a^{5} a_{3}-5120 a^{3} a_{3}-1536 a a_{3}+384 a_{3}&=0\\ -192 a^{5} a_{3}-640 a^{3} a_{3}-128 a a_{3}+64 a_{3}&=0\\ -6144 a a_{3}-4096 a_{2}-8192 a_{3}-8192 b_{3}&=0\\ -2560 a^{4} a_{3}-3072 a^{2} a_{3}+1152 a a_{3}-3840 a_{2}-1536 a_{3}&=0\\ -6144 a^{2} a_{3}-3072 a a_{3}-9216 a_{2}-6144 a_{3}-3072 b_{3}&=0\\ -2048 a^{2} a_{1}-4096 a b_{1}+4096 a_{2}+4096 a_{3}-2048 b_{3}&=0\\ -2048 a^{3} a_{3}-12288 a a_{2}-6144 a a_{3}-3072 a_{1}+1024 a_{3}-6144 b_{2}&=0\\ -6144 a a_{1}-4096 a a_{3}-4096 a_{2}-8192 a_{3}-12288 b_{1}-4096 b_{3}&=0\\ -6144 a^{2} a_{3}-12288 a a_{2}-12288 a a_{3}-6144 a b_{3}-6144 a_{1}-12288 b_{2}&=0\\ -64 a^{6} a_{3}-640 a^{4} a_{3}-384 a^{2} a_{3}+384 a a_{3}-448 a_{2}-128 a_{3}+64 b_{3}&=0\\ -6144 a^{2} a_{1}-2048 a^{2} a_{3}-8192 a a_{2}-8192 a a_{3}-12288 a b_{1}-4096 a_{1}-8192 b_{2}&=0\\ -1536 a^{3} a_{3}-9216 a^{2} a_{2}-6144 a^{2} a_{3}-9216 a a_{1}+1024 a a_{3}-12288 a b_{2}-6144 a_{2}-4096 a_{3}-6144 b_{1}&=0\\ -256 a^{5} a_{3}-512 a^{3} a_{3}+768 a^{2} a_{3}-2304 a a_{2}-768 a a_{3}+384 a b_{3}-384 a_{1}+256 a_{3}-768 b_{2}&=0\\ -1536 a^{3} a_{1}-3072 a^{2} a_{2}-2048 a^{2} a_{3}-3072 a^{2} b_{1}+1024 a^{2} b_{3}-3072 a a_{1}-4096 a b_{2}-1024 a_{3}-2048 b_{1}&=0\\ -2048 a^{3} a_{2}-1024 a^{3} a_{3}+512 a^{3} b_{3}-3072 a^{2} a_{1}+1024 a^{2} a_{3}-3072 a^{2} b_{2}-4096 a a_{2}-2048 a a_{3}-3072 a b_{1}+1024 a b_{3}-1024 a_{1}-2048 b_{2}&=0\\ -256 a^{4} a_{3}+512 a^{3} a_{3}-3840 a^{2} a_{2}-1536 a^{2} a_{3}+768 a^{2} b_{3}-1920 a a_{1}+1024 a a_{3}-3072 a b_{2}-1280 a_{2}-512 a_{3}-768 b_{1}+256 b_{3}&=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 \(S(R)\). Unable to determine \(R\). Terminating
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 }&=y^{3}+\left (1+\frac {3}{4} x^{2}+\frac {3}{2} x a \right ) y^{2}+\left (\frac {1}{2} x^{2}+x a +\frac {3}{16} x^{4}+\frac {3}{4} x^{3} a +\frac {3}{4} a^{2} x^{2}\right ) y-\frac {x}{2}+1+\frac {x^{4}}{16}+\frac {x^{3} a}{4}+\frac {a^{2} x^{2}}{4}+\frac {x^{6}}{64}+\frac {3 a \,x^{5}}{32}+\frac {3 a^{2} x^{4}}{16}+\frac {x^{3} a^{3}}{8}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -\frac {1}{2} x +1+\frac {1}{16} x^{4}+\frac {1}{4} x^{3} a +\frac {1}{4} a^{2} x^{2}+\frac {1}{64} x^{6}+\frac {3}{32} a \,x^{5}+\frac {3}{16} a^{2} x^{4}+\frac {1}{8} x^{3} a^{3}\\ f_1(x) &= \frac {1}{2} x^{2}+x a +\frac {3}{16} x^{4}+\frac {3}{4} x^{3} a +\frac {3}{4} a^{2} x^{2}\\ f_2(x) &= 1+\frac {3}{4} x^{2}+\frac {3}{2} x a\\ f_3(x) &= 1 \end {align*}
Since \(f_2(x)=1+\frac {3}{4} x^{2}+\frac {3}{2} x a\) 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 {1+\frac {3}{4} x^{2}+\frac {3}{2} x a}{3} \right ) \\ &= u \left (x \right )-\frac {1}{3}-\frac {x^{2}}{4}-\frac {x a}{2} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {a}{2}+\frac {29}{27}+u \left (x \right )^{3}-\frac {u \left (x \right )}{3}\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
Integrating both sides gives \begin {align*} \int _{}^{u \left (x \right )}\frac {1}{\frac {1}{2} a +\frac {29}{27}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} = x +c_{2} \end {align*}
Substituting \(u=y+\frac {1}{3}+\frac {x^{2}}{4}+\frac {x a}{2}\) in the above solution gives \begin {align*} \int _{}^{y+\frac {1}{3}+\frac {x^{2}}{4}+\frac {x a}{2}}\frac {1}{\frac {1}{2} a +\frac {29}{27}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} = x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y+\frac {1}{3}+\frac {x^{2}}{4}+\frac {x a}{2}}\frac {1}{\frac {1}{2} a +\frac {29}{27}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} &= x +c_{2} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y+\frac {1}{3}+\frac {x^{2}}{4}+\frac {x a}{2}}\frac {1}{\frac {1}{2} a +\frac {29}{27}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-\frac {y x^{2}}{2}-a x y-y^{3}-\frac {3 y^{2} x^{2}}{4}-\frac {3 a y^{2} x}{2}-\frac {3 x^{4} y}{16}-\frac {3 y a \,x^{3}}{4}-\frac {3 y a^{2} x^{2}}{4}=-\frac {1}{2} x +1+\frac {1}{16} x^{4}+\frac {1}{4} x^{3} a +\frac {1}{4} a^{2} x^{2}+\frac {1}{64} x^{6}+\frac {3}{32} a \,x^{5}+\frac {3}{16} a^{2} x^{4}+\frac {1}{8} x^{3} a^{3} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {x}{2}+1+y^{2}+\frac {y x^{2}}{2}+a x y+\frac {x^{4}}{16}+\frac {x^{3} a}{4}+\frac {a^{2} x^{2}}{4}+y^{3}+\frac {3 y^{2} x^{2}}{4}+\frac {3 a y^{2} x}{2}+\frac {3 x^{4} y}{16}+\frac {3 y a \,x^{3}}{4}+\frac {3 y a^{2} x^{2}}{4}+\frac {x^{6}}{64}+\frac {3 a \,x^{5}}{32}+\frac {3 a^{2} x^{4}}{16}+\frac {x^{3} a^{3}}{8} \end {array} \]
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: 41
dsolve(diff(y(x),x) = -1/2*x+1+y(x)^2+1/2*x^2*y(x)+y(x)*a*x+1/16*x^4+1/4*x^3*a+1/4*a^2*x^2+y(x)^3+3/4*x^2*y(x)^2+3/2*a*x*y(x)^2+3/16*y(x)*x^4+3/4*y(x)*a*x^3+3/4*a^2*x^2*y(x)+1/64*x^6+3/32*x^5*a+3/16*a^2*x^4+1/8*a^3*x^3,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x^{2}}{4}-\frac {a x}{2}+\operatorname {RootOf}\left (-x +2 \left (\int _{}^{\textit {\_Z}}\frac {1}{2 \textit {\_a}^{3}+2 \textit {\_a}^{2}+a +2}d \textit {\_a} \right )+c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 6.585 (sec). Leaf size: 906
DSolve[y'[x] == 1 - x/2 + (a^2*x^2)/4 + (a*x^3)/4 + (a^3*x^3)/8 + x^4/16 + (3*a^2*x^4)/16 + (3*a*x^5)/32 + x^6/64 + a*x*y[x] + (x^2*y[x])/2 + (3*a^2*x^2*y[x])/4 + (3*a*x^3*y[x])/4 + (3*x^4*y[x])/16 + y[x]^2 + (3*a*x*y[x]^2)/2 + (3*x^2*y[x]^2)/4 + y[x]^3,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\frac {1}{9} \text {RootSum}\left [729 a^2 \text {$\#$1}^9+3132 a \text {$\#$1}^9+3364 \text {$\#$1}^9+2187 a^2 \text {$\#$1}^6+9396 a \text {$\#$1}^6+10092 \text {$\#$1}^6+2187 a^2 \text {$\#$1}^3+9396 a \text {$\#$1}^3+9984 \text {$\#$1}^3+729 a^2+3132 a+3364\&,\frac {729 a^2 \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}^6+3132 a \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}^6+3364 \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}^6+81\ 2^{2/3} a \sqrt [3]{27 a+58} \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}^4+174\ 2^{2/3} \sqrt [3]{27 a+58} \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}^4+1458 a^2 \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}^3+6264 a \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}^3+6728 \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}^3+18 \sqrt [3]{2} (27 a+58)^{2/3} \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}^2+81\ 2^{2/3} a \sqrt [3]{27 a+58} \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}+174\ 2^{2/3} \sqrt [3]{27 a+58} \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right ) \text {$\#$1}+729 a^2 \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right )+3132 a \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right )+3364 \log \left (\frac {\sqrt [3]{2} \left (\frac {1}{4} \left (3 x^2+6 a x+4\right )+3 y(x)\right )}{\sqrt [3]{27 a+58}}-\text {$\#$1}\right )}{729 a^2 \text {$\#$1}^8+3132 a \text {$\#$1}^8+3364 \text {$\#$1}^8+1458 a^2 \text {$\#$1}^5+6264 a \text {$\#$1}^5+6728 \text {$\#$1}^5+729 a^2 \text {$\#$1}^2+3132 a \text {$\#$1}^2+3328 \text {$\#$1}^2}\&\right ]=\frac {(27 a+58)^{2/3} x}{9\ 2^{2/3}}+c_1,y(x)\right ] \]