Internal problem ID [3221]
Internal file name [OUTPUT/2713_Sunday_June_05_2022_08_39_07_AM_44701922/index.tex]
Book: Differential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 2. First-Order and Simple Higher-Order Differential Equations. Page
78
Problem number: 79.
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
[[_homogeneous, `class G`], _rational, [_Abel, `2nd type`, `class B`]]
\[ \boxed {6 y^{2}-x \left (2 x^{3}+y\right ) y^{\prime }=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {6 y^{2}}{x \left (2 x^{3}+y \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}+\frac {6 y^{2} \left (b_{3}-a_{2}\right )}{x \left (2 x^{3}+y \right )}-\frac {36 y^{4} a_{3}}{x^{2} \left (2 x^{3}+y \right )^{2}}-\left (-\frac {6 y^{2}}{x^{2} \left (2 x^{3}+y \right )}-\frac {36 y^{2} x}{\left (2 x^{3}+y \right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {12 y}{x \left (2 x^{3}+y \right )}-\frac {6 y^{2}}{x \left (2 x^{3}+y \right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ \frac {4 x^{8} b_{2}-20 x^{5} y b_{2}+36 x^{4} y^{2} a_{2}-12 x^{4} y^{2} b_{3}+48 x^{3} y^{3} a_{3}-24 x^{4} y b_{1}+48 x^{3} y^{2} a_{1}-5 x^{2} y^{2} b_{2}-30 y^{4} a_{3}-6 x \,y^{2} b_{1}+6 y^{3} a_{1}}{x^{2} \left (2 x^{3}+y \right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} 4 x^{8} b_{2}-20 x^{5} y b_{2}+36 x^{4} y^{2} a_{2}-12 x^{4} y^{2} b_{3}+48 x^{3} y^{3} a_{3}-24 x^{4} y b_{1}+48 x^{3} y^{2} a_{1}-5 x^{2} y^{2} b_{2}-30 y^{4} a_{3}-6 x \,y^{2} b_{1}+6 y^{3} a_{1} = 0 \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} 4 b_{2} v_{1}^{8}+36 a_{2} v_{1}^{4} v_{2}^{2}+48 a_{3} v_{1}^{3} v_{2}^{3}-20 b_{2} v_{1}^{5} v_{2}-12 b_{3} v_{1}^{4} v_{2}^{2}+48 a_{1} v_{1}^{3} v_{2}^{2}-24 b_{1} v_{1}^{4} v_{2}-30 a_{3} v_{2}^{4}-5 b_{2} v_{1}^{2} v_{2}^{2}+6 a_{1} v_{2}^{3}-6 b_{1} v_{1} v_{2}^{2} = 0 \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} 4 b_{2} v_{1}^{8}-20 b_{2} v_{1}^{5} v_{2}+\left (36 a_{2}-12 b_{3}\right ) v_{1}^{4} v_{2}^{2}-24 b_{1} v_{1}^{4} v_{2}+48 a_{3} v_{1}^{3} v_{2}^{3}+48 a_{1} v_{1}^{3} v_{2}^{2}-5 b_{2} v_{1}^{2} v_{2}^{2}-6 b_{1} v_{1} v_{2}^{2}-30 a_{3} v_{2}^{4}+6 a_{1} v_{2}^{3} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} 6 a_{1}&=0\\ 48 a_{1}&=0\\ -30 a_{3}&=0\\ 48 a_{3}&=0\\ -24 b_{1}&=0\\ -6 b_{1}&=0\\ -20 b_{2}&=0\\ -5 b_{2}&=0\\ 4 b_{2}&=0\\ 36 a_{2}-12 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=a_{2}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=3 a_{2} \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 &= x \\ \eta &= 3 y \\ \end{align*} Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation \begin {align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= 3 y - \left (\frac {6 y^{2}}{x \left (2 x^{3}+y \right )}\right ) \left (x\right ) \\ &= \frac {6 x^{3} y -3 y^{2}}{2 x^{3}+y}\\ \xi &= 0 \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 \(S(R)\). Since \(\xi =0\) then in this special case \begin {align*} R = x \end {align*}
\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {6 x^{3} y -3 y^{2}}{2 x^{3}+y}}} dy \end {align*}
Which results in \begin {align*} S&= -\frac {2 \ln \left (-2 x^{3}+y \right )}{3}+\frac {\ln \left (y \right )}{3} \end {align*}
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 {6 y^{2}}{x \left (2 x^{3}+y \right )} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= -\frac {4 x^{2}}{2 x^{3}-y}\\ S_{y} &= \frac {2}{6 x^{3}-3 y}+\frac {1}{3 y} \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= -\frac {2}{x}\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 {2}{R} \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 ) = -2 \ln \left (R \right )+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 {2 \ln \left (-2 x^{3}+y\right )}{3}+\frac {\ln \left (y\right )}{3} = -2 \ln \left (x \right )+c_{1} \end {align*}
Which simplifies to \begin {align*} -\frac {2 \ln \left (-2 x^{3}+y\right )}{3}+\frac {\ln \left (y\right )}{3} = -2 \ln \left (x \right )+c_{1} \end {align*}
The following diagram shows solution curves of the original ode and how they transform in the canonical coordinates space using the mapping shown.
|
Original ode in \(x,y\) coordinates |
Canonical coordinates transformation |
ODE in canonical coordinates \((R,S)\) |
| \( \frac {dy}{dx} = \frac {6 y^{2}}{x \left (2 x^{3}+y \right )}\) |
|
\( \frac {d S}{d R} = -\frac {2}{R}\) |
| |
\(\!\begin {aligned} R&= x\\ S&= -\frac {2 \ln \left (-2 x^{3}+y \right )}{3}+\frac {\ln \left (y \right )}{3} \end {aligned} \) |
|
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {2 \ln \left (-2 x^{3}+y\right )}{3}+\frac {\ln \left (y\right )}{3} &= -2 \ln \left (x \right )+c_{1} \\ \end{align*}
Verification of solutions
\[ -\frac {2 \ln \left (-2 x^{3}+y\right )}{3}+\frac {\ln \left (y\right )}{3} = -2 \ln \left (x \right )+c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 6 y^{2}-x \left (2 x^{3}+y\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {6 y^{2}}{x \left (2 x^{3}+y\right )} \end {array} \]
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 homogeneous G <- homogeneous successful`
✓ Solution by Maple
Time used: 0.578 (sec). Leaf size: 193
dsolve(6*y(x)^2-(x*(2*x^3+y(x)))*diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -\frac {x^{3} \left (-x^{3}+\sqrt {x^{3} \left (x^{3}+8 c_{1} \right )}-4 c_{1} \right )}{2 c_{1}} \\ y \left (x \right ) &= \frac {x^{3} \left (x^{3}+\sqrt {x^{3} \left (x^{3}+8 c_{1} \right )}+4 c_{1} \right )}{2 c_{1}} \\ y \left (x \right ) &= -\frac {x^{3} \left (-x^{3}+\sqrt {x^{3} \left (x^{3}+8 c_{1} \right )}-4 c_{1} \right )}{2 c_{1}} \\ y \left (x \right ) &= \frac {x^{3} \left (x^{3}+\sqrt {x^{3} \left (x^{3}+8 c_{1} \right )}+4 c_{1} \right )}{2 c_{1}} \\ y \left (x \right ) &= -\frac {x^{3} \left (-x^{3}+\sqrt {x^{3} \left (x^{3}+8 c_{1} \right )}-4 c_{1} \right )}{2 c_{1}} \\ y \left (x \right ) &= \frac {x^{3} \left (x^{3}+\sqrt {x^{3} \left (x^{3}+8 c_{1} \right )}+4 c_{1} \right )}{2 c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.396 (sec). Leaf size: 123
DSolve[6*y[x]^2-(x*(2*x^3+y[x]))*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to 2 x^3 \left (-1+\frac {2}{1-\frac {4 x^{3/2}}{\sqrt {16 x^3+c_1}}}\right ) \\ y(x)\to 2 x^3 \left (-1+\frac {2}{1+\frac {4 x^{3/2}}{\sqrt {16 x^3+c_1}}}\right ) \\ y(x)\to 0 \\ y(x)\to 2 x^3 \\ y(x)\to \frac {2 \left (\left (x^3\right )^{3/2}-x^{9/2}\right )}{x^{3/2}+\sqrt {x^3}} \\ \end{align*}