5.19 problem 19

Internal problem ID [7960]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 19
Date solved : Monday, October 21, 2024 at 04:39:37 PM
CAS classification : [[_1st_order, _with_linear_symmetries]]

Solve

\begin{align*} y^{\prime }&=\frac {y}{2 y \ln \left (y\right )+y-x} \end{align*}

5.19.1 Solved as first order Exact ode

Time used: 0.148 (sec)

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \]

Therefore

\begin{align*} \left (2 y \ln \left (y \right )+y -x\right )\mathop {\mathrm {d}y} &= \left (y\right )\mathop {\mathrm {d}x}\\ \left (-y\right )\mathop {\mathrm {d}x} + \left (2 y \ln \left (y \right )+y -x\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(x,y) &= -y\\ N(x,y) &= 2 y \ln \left (y \right )+y -x \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-y\right )\\ &= -1 \end{align*}

And

\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (2 y \ln \left (y \right )+y -x\right )\\ &= -1 \end{align*}

Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)

\begin{align*} \frac {\partial \phi }{\partial x } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= N\tag {2} \end{align*}

Integrating (1) w.r.t. \(x\) gives

\begin{align*} \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int M\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -y\mathop {\mathrm {d}x} \\ \tag{3} \phi &= -x y+ f(y) \\ \end{align*}

Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = -x+f'(y) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial y} = 2 y \ln \left (y \right )+y -x\). Therefore equation (4) becomes

\begin{equation} \tag{5} 2 y \ln \left (y \right )+y -x = -x+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = 2 y \ln \left (y \right )+y \]

Integrating the above w.r.t \(y\) gives

\begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( y \left (2 \ln \left (y \right )+1\right )\right ) \mathop {\mathrm {d}y} \\ f(y) &= y^{2} \ln \left (y \right )+ c_1 \\ \end{align*}

Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = -x y +y^{2} \ln \left (y \right )+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = -x y +y^{2} \ln \left (y \right ) \]

Figure 217: Slope field plot
\(y^{\prime } = \frac {y}{2 y \ln \left (y\right )+y-x}\)

5.19.2 Solved using Lie symmetry for first order ode

Time used: 0.623 (sec)

Writing the ode as

\begin{align*} y^{\prime }&=\frac {y}{2 y \ln \left (y \right )+y -x}\\ 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*}

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 {y \left (b_{3}-a_{2}\right )}{2 y \ln \left (y \right )+y -x}-\frac {y^{2} a_{3}}{\left (2 y \ln \left (y \right )+y -x \right )^{2}}-\frac {y \left (x a_{2}+y a_{3}+a_{1}\right )}{\left (2 y \ln \left (y \right )+y -x \right )^{2}}-\left (\frac {1}{2 y \ln \left (y \right )+y -x}-\frac {y \left (2 \ln \left (y \right )+3\right )}{\left (2 y \ln \left (y \right )+y -x \right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \frac {4 \ln \left (y \right )^{2} y^{2} b_{2}-4 \ln \left (y \right ) x y b_{2}-2 \ln \left (y \right ) y^{2} a_{2}+4 \ln \left (y \right ) y^{2} b_{2}+2 \ln \left (y \right ) y^{2} b_{3}+2 x^{2} b_{2}-y^{2} a_{2}-2 y^{2} a_{3}+y^{2} b_{2}+3 y^{2} b_{3}+x b_{1}-y a_{1}+2 y b_{1}}{\left (2 y \ln \left (y \right )+y -x \right )^{2}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} 4 \ln \left (y \right )^{2} y^{2} b_{2}-4 \ln \left (y \right ) x y b_{2}-2 \ln \left (y \right ) y^{2} a_{2}+4 \ln \left (y \right ) y^{2} b_{2}+2 \ln \left (y \right ) y^{2} b_{3}+2 x^{2} b_{2}-y^{2} a_{2}-2 y^{2} a_{3}+y^{2} b_{2}+3 y^{2} b_{3}+x b_{1}-y a_{1}+2 y b_{1} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \{x, y, \ln \left (y \right )\} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \{x = v_{1}, y = v_{2}, \ln \left (y \right ) = v_{3}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} 4 v_{3}^{2} v_{2}^{2} b_{2}-2 v_{3} v_{2}^{2} a_{2}-4 v_{3} v_{1} v_{2} b_{2}+4 v_{3} v_{2}^{2} b_{2}+2 v_{3} v_{2}^{2} b_{3}-v_{2}^{2} a_{2}-2 v_{2}^{2} a_{3}+2 v_{1}^{2} b_{2}+v_{2}^{2} b_{2}+3 v_{2}^{2} b_{3}-v_{2} a_{1}+v_{1} b_{1}+2 v_{2} b_{1} = 0 \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}, v_{3}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} 2 v_{1}^{2} b_{2}-4 v_{3} v_{1} v_{2} b_{2}+v_{1} b_{1}+4 v_{3}^{2} v_{2}^{2} b_{2}+\left (-2 a_{2}+4 b_{2}+2 b_{3}\right ) v_{2}^{2} v_{3}+\left (-a_{2}-2 a_{3}+b_{2}+3 b_{3}\right ) v_{2}^{2}+\left (-a_{1}+2 b_{1}\right ) v_{2} = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} b_{1}&=0\\ -4 b_{2}&=0\\ 2 b_{2}&=0\\ 4 b_{2}&=0\\ -a_{1}+2 b_{1}&=0\\ -2 a_{2}+4 b_{2}+2 b_{3}&=0\\ -a_{2}-2 a_{3}+b_{2}+3 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=b_{3}\\ 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 &= x +y \\ \eta &= 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 \\ &= y - \left (\frac {y}{2 y \ln \left (y \right )+y -x}\right ) \left (x +y\right ) \\ &= \frac {-2 x y +2 y^{2} \ln \left (y \right )}{2 y \ln \left (y \right )+y -x}\\ \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 {-2 x y +2 y^{2} \ln \left (y \right )}{2 y \ln \left (y \right )+y -x}}} dy \end{align*}

Which results in

\begin{align*} S&= \frac {\ln \left (-x y +y^{2} \ln \left (y \right )\right )}{2} \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 {y}{2 y \ln \left (y \right )+y -x} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {1}{-2 y \ln \left (y \right )+2 x}\\ S_{y} &= \frac {2 y \ln \left (y \right )+y -x}{2 y \left (y \ln \left (y \right )-x \right )} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= 0\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} &= 0 \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 \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {0\, dR} + c_2 \\ S \left (R \right ) &= c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} \frac {\ln \left (y\right )}{2}+\frac {\ln \left (y \ln \left (y\right )-x \right )}{2} = c_2 \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 \((R,S)\)
\( \frac {dy}{dx} = \frac {y}{2 y \ln \left (y \right )+y -x}\)		\( \frac {d S}{d R} = 0\)
	\(\!\begin {aligned} R&= x\\ S&= \frac {\ln \left (y \right )}{2}+\frac {\ln \left (y \ln \left (y \right )-x \right )}{2} \end {aligned} \)

Figure 218: Slope field plot
\(y^{\prime } = \frac {y}{2 y \ln \left (y\right )+y-x}\)

5.19.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{2 y \ln \left (y\right )+y-x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{2 y \ln \left (y\right )+y-x} \end {array} \]

5.19.4 Maple trace

Methods for first order ODEs:
 

5.19.5 Maple dsolve solution

Solving time : 0.014 (sec)
Leaf size : 19

dsolve(diff(y(x),x) = y(x)/(2*y(x)*ln(y(x))+y(x)-x), 
       y(x),singsol=all)
 

\[ y = {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z} \,{\mathrm e}^{2 \textit {\_Z}}-x \,{\mathrm e}^{\textit {\_Z}}+c_1 \right )} \]

5.19.6 Mathematica DSolve solution

Solving time : 0.188 (sec)
Leaf size : 19

DSolve[{D[y[x],x]==y[x]/(2*y[x]*Log[y[x]]+y[x]-x),{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [x=y(x) \log (y(x))+\frac {c_1}{y(x)},y(x)\right ] \]