problem 26

Internal problem ID [1085]
Internal ﬁle name [OUTPUT/1086_Sunday_June_05_2022_02_02_18_AM_91680699/index.tex]

Book: Elementary diﬀerential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 2, First order equations. Exact equations. Integrating factors. Section 2.6 Page 91
Problem number: 26.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exactWithIntegrationFactor", "ﬁrst_order_ode_lie_symmetry_calculated"

7.25.1 Solving as ﬁrst order ode lie symmetry calculated ode

Writing the ode as \begin {align*} y^{\prime }&=-\frac {6 y \left (y^{2}+2 x \right )}{x \left (10 y^{2}+9 x \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 \left (y^{2}+2 x \right ) \left (b_{3}-a_{2}\right )}{x \left (10 y^{2}+9 x \right )}-\frac {36 y^{2} \left (y^{2}+2 x \right )^{2} a_{3}}{x^{2} \left (10 y^{2}+9 x \right )^{2}}-\left (-\frac {12 y}{x \left (10 y^{2}+9 x \right )}+\frac {6 y \left (y^{2}+2 x \right )}{x^{2} \left (10 y^{2}+9 x \right )}+\frac {54 y \left (y^{2}+2 x \right )}{x \left (10 y^{2}+9 x \right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {6 \left (y^{2}+2 x \right )}{x \left (10 y^{2}+9 x \right )}-\frac {12 y^{2}}{x \left (10 y^{2}+9 x \right )}+\frac {120 y^{2} \left (y^{2}+2 x \right )}{x \left (10 y^{2}+9 x \right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ \frac {160 x^{2} y^{4} b_{2}-96 y^{6} a_{3}+222 x^{3} y^{2} b_{2}+66 x^{2} y^{3} a_{2}-132 x^{2} y^{3} b_{3}-252 x \,y^{4} a_{3}+60 x \,y^{4} b_{1}-60 y^{5} a_{1}+189 x^{4} b_{2}-252 x^{2} y^{2} a_{3}+42 x^{2} y^{2} b_{1}-108 x \,y^{3} a_{1}+108 x^{3} b_{1}-108 x^{2} y a_{1}}{x^{2} \left (10 y^{2}+9 x \right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} 160 x^{2} y^{4} b_{2}-96 y^{6} a_{3}+222 x^{3} y^{2} b_{2}+66 x^{2} y^{3} a_{2}-132 x^{2} y^{3} b_{3}-252 x \,y^{4} a_{3}+60 x \,y^{4} b_{1}-60 y^{5} a_{1}+189 x^{4} b_{2}-252 x^{2} y^{2} a_{3}+42 x^{2} y^{2} b_{1}-108 x \,y^{3} a_{1}+108 x^{3} b_{1}-108 x^{2} y 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} -96 a_{3} v_{2}^{6}+160 b_{2} v_{1}^{2} v_{2}^{4}-60 a_{1} v_{2}^{5}+66 a_{2} v_{1}^{2} v_{2}^{3}-252 a_{3} v_{1} v_{2}^{4}+60 b_{1} v_{1} v_{2}^{4}+222 b_{2} v_{1}^{3} v_{2}^{2}-132 b_{3} v_{1}^{2} v_{2}^{3}-108 a_{1} v_{1} v_{2}^{3}-252 a_{3} v_{1}^{2} v_{2}^{2}+42 b_{1} v_{1}^{2} v_{2}^{2}+189 b_{2} v_{1}^{4}-108 a_{1} v_{1}^{2} v_{2}+108 b_{1} v_{1}^{3} = 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} 189 b_{2} v_{1}^{4}+222 b_{2} v_{1}^{3} v_{2}^{2}+108 b_{1} v_{1}^{3}+160 b_{2} v_{1}^{2} v_{2}^{4}+\left (66 a_{2}-132 b_{3}\right ) v_{1}^{2} v_{2}^{3}+\left (-252 a_{3}+42 b_{1}\right ) v_{1}^{2} v_{2}^{2}-108 a_{1} v_{1}^{2} v_{2}+\left (-252 a_{3}+60 b_{1}\right ) v_{1} v_{2}^{4}-108 a_{1} v_{1} v_{2}^{3}-96 a_{3} v_{2}^{6}-60 a_{1} v_{2}^{5} = 0 \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} -108 a_{1}&=0\\ -60 a_{1}&=0\\ -96 a_{3}&=0\\ 108 b_{1}&=0\\ 160 b_{2}&=0\\ 189 b_{2}&=0\\ 222 b_{2}&=0\\ 66 a_{2}-132 b_{3}&=0\\ -252 a_{3}+42 b_{1}&=0\\ -252 a_{3}+60 b_{1}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=2 b_{3}\\ 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 &= 2 x \\ \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 {6 y \left (y^{2}+2 x \right )}{x \left (10 y^{2}+9 x \right )}\right ) \left (2 x\right ) \\ &= \frac {22 y^{3}+33 y x}{10 y^{2}+9 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 {22 y^{3}+33 y x}{10 y^{2}+9 x}}} dy \end {align*}

Which results in \begin {align*} S&= \frac {3 \ln \left (y \right )}{11}+\frac {\ln \left (2 y^{2}+3 x \right )}{11} \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 \left (y^{2}+2 x \right )}{x \left (10 y^{2}+9 x \right )} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {3}{22 y^{2}+33 x}\\ S_{y} &= \frac {10 y^{2}+9 x}{22 y^{3}+33 y x} \end {align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= -\frac {3}{11 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 {3}{11 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 ) = -\frac {3 \ln \left (R \right )}{11}+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 {3 \ln \left (y\right )}{11}+\frac {\ln \left (2 y^{2}+3 x \right )}{11} = -\frac {3 \ln \left (x \right )}{11}+c_{1} \end {align*}

Which simpliﬁes to \begin {align*} \frac {3 \ln \left (y\right )}{11}+\frac {\ln \left (2 y^{2}+3 x \right )}{11} = -\frac {3 \ln \left (x \right )}{11}+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 \left (y^{2}+2 x \right )}{x \left (10 y^{2}+9 x \right )}$		$ \frac {d S}{d R} = -\frac {3}{11 R}$
	$\!\begin {aligned} R&= x\\ S&= \frac {3 \ln \left (y \right )}{11}+\frac {\ln \left (2 y^{2}+3 x \right )}{11} \end {aligned} $

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {3 \ln \left (y\right )}{11}+\frac {\ln \left (2 y^{2}+3 x \right )}{11} &= -\frac {3 \ln \left (x \right )}{11}+c_{1} \\ \end{align*}

Veriﬁcation of solutions

\[ \frac {3 \ln \left (y\right )}{11}+\frac {\ln \left (2 y^{2}+3 x \right )}{11} = -\frac {3 \ln \left (x \right )}{11}+c_{1} \] Veriﬁed OK.

7.25.2 Solving as exact ode

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 (10 x \,y^{2}+9 x^{2}\right )\mathop {\mathrm {d}y} &= \left (-6 y^{3}-12 y x\right )\mathop {\mathrm {d}x}\\ \left (6 y^{3}+12 y x\right )\mathop {\mathrm {d}x} + \left (10 x \,y^{2}+9 x^{2}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}

Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= 6 y^{3}+12 y x\\ N(x,y) &= 10 x \,y^{2}+9 x^{2} \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 (6 y^{3}+12 y x\right )\\ &= 18 y^{2}+12 x \end {align*}

And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (10 x \,y^{2}+9 x^{2}\right )\\ &= 10 y^{2}+18 x \end {align*}

Since $\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}$, then the ODE is not exact. Since the ODE is not exact, we will try to ﬁnd an integrating factor to make it exact. Let \begin {align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=\frac {1}{10 x \,y^{2}+9 x^{2}}\left ( \left ( 18 y^{2}+12 x\right ) - \left (10 y^{2}+18 x \right ) \right ) \\ &=\frac {8 y^{2}-6 x}{10 x \,y^{2}+9 x^{2}} \end {align*}

Since $A$ depends on $y$, it can not be used to obtain an integrating factor. We will now try a second method to ﬁnd an integrating factor. Let \begin {align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} \right ) \\ &=\frac {1}{6 y^{3}+12 y x}\left ( \left ( 10 y^{2}+18 x\right ) - \left (18 y^{2}+12 x \right ) \right ) \\ &=\frac {-4 y^{2}+3 x}{3 y^{3}+6 y x} \end {align*}

Since $B$ depends on $x$, it can not be used to obtain an integrating factor.We will now try a third method to ﬁnd an integrating factor. Let \[ R = \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \] $R$ is now checked to see if it is a function of only $t=xy$. Therefore \begin {align*} R &= \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \\ &= \frac {\left (10 y^{2}+18 x\right )-\left (18 y^{2}+12 x\right )} {x\left (6 y^{3}+12 y x\right ) - y\left (10 x \,y^{2}+9 x^{2}\right )} \\ &= \frac {2}{y x} \end {align*}

Replacing all powers of terms $xy$ by $t$ gives \[ R = \frac {2}{t} \] Since $R$ depends on $t$ only, then it can be used to ﬁnd an integrating factor. Let the integrating factor be $\mu $ then \begin {align*} \mu &= e^{\int R \mathop {\mathrm {d}t}} \\ &= e^{\int \left (\frac {2}{t}\right )\mathop {\mathrm {d}t} } \end {align*}

The result of integrating gives \begin {align*} \mu &= e^{2 \ln \left (t \right ) } \\ &= t^{2} \end {align*}

Now $t$ is replaced back with $xy$ giving \[ \mu =x^{2} y^{2} \] Multiplying $M$ and $N$ by this integrating factor gives new $M$ and new $N$ which are called $ \overline {M}$ and $ \overline {N}$ so not to confuse them with the original $M$ and $N$ \begin {align*} \overline {M} &=\mu M \\ &= x^{2} y^{2}\left (6 y^{3}+12 y x\right ) \\ &= 6 x^{2} y^{5}+12 y^{3} x^{3} \end {align*}

And \begin {align*} \overline {N} &=\mu N \\ &= x^{2} y^{2}\left (10 x \,y^{2}+9 x^{2}\right ) \\ &= x^{3} \left (10 y^{2}+9 x \right ) y^{2} \end {align*}

A modiﬁed ODE is now obtained from the original ODE, which is exact and can solved. The modiﬁed ODE is \begin {align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (6 x^{2} y^{5}+12 y^{3} x^{3}\right ) + \left (x^{3} \left (10 y^{2}+9 x \right ) y^{2}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end {align*}

The following equations are now set up to solve for the function $\phi \left (x,y\right )$ \begin {align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {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 \overline {M}\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int 6 x^{2} y^{5}+12 y^{3} x^{3}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= y^{3} x^{3} \left (2 y^{2}+3 x \right )+ 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{align*} \tag{4} \frac {\partial \phi }{\partial y} &= 3 y^{2} x^{3} \left (2 y^{2}+3 x \right )+4 y^{4} x^{3}+f'(y) \\ &=x^{3} \left (10 y^{2}+9 x \right ) y^{2}+f'(y) \\ \end{align*} But equation (2) says that $\frac {\partial \phi }{\partial y} = x^{3} \left (10 y^{2}+9 x \right ) y^{2}$. Therefore equation (4) becomes \begin{equation} \tag{5} x^{3} \left (10 y^{2}+9 x \right ) y^{2} = x^{3} \left (10 y^{2}+9 x \right ) y^{2}+f'(y) \end{equation} Solving equation (5) for $ f'(y)$ gives \[ f'(y) = 0 \] Therefore \[ f(y) = c_{1} \] Where $c_{1}$ is constant of integration. Substituting this result for $f(y)$ into equation (3) gives $\phi $ \[ \phi = y^{3} x^{3} \left (2 y^{2}+3 x \right )+ c_{1} \] 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 new constant $c_{1}$ gives the solution as \[ c_{1} = y^{3} x^{3} \left (2 y^{2}+3 x \right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y^{3} x^{3} \left (2 y^{2}+3 x \right ) &= c_{1} \\ \end{align*}

Veriﬁcation of solutions

\[ y^{3} x^{3} \left (2 y^{2}+3 x \right ) = c_{1} \] Veriﬁed OK.

7.25.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 12 y x +6 y^{3}+\left (9 x^{2}+10 x y^{2}\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-12 y x -6 y^{3}}{9 x^{2}+10 x y^{2}} \end {array} \]

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`

\[ \ln \left (x \right )-c_{1} +\frac {6 \ln \left (\frac {y \left (x \right )}{\sqrt {x}}\right )}{11}+\frac {2 \ln \left (\frac {2 y \left (x \right )^{2}+3 x}{x}\right )}{11} = 0 \]

\begin{align*} y(x)\to \text {Root}\left [2 \text {$\#$1}^5 x^3+3 \text {$\#$1}^3 x^4-c_1\&,1\right ] \\ y(x)\to \text {Root}\left [2 \text {$\#$1}^5 x^3+3 \text {$\#$1}^3 x^4-c_1\&,2\right ] \\ y(x)\to \text {Root}\left [2 \text {$\#$1}^5 x^3+3 \text {$\#$1}^3 x^4-c_1\&,3\right ] \\ y(x)\to \text {Root}\left [2 \text {$\#$1}^5 x^3+3 \text {$\#$1}^3 x^4-c_1\&,4\right ] \\ y(x)\to \text {Root}\left [2 \text {$\#$1}^5 x^3+3 \text {$\#$1}^3 x^4-c_1\&,5\right ] \\ \end{align*}


Original ode in \(x,y\) coordinates	Canonical coordinates transformation	ODE in canonical coordinates \((R,S)\)

\( \frac {dy}{dx} = -\frac {6 y \left (y^{2}+2 x \right )}{x \left (10 y^{2}+9 x \right )}\)		\( \frac {d S}{d R} = -\frac {3}{11 R}\)
	\(\!\begin {aligned} R&= x\\ S&= \frac {3 \ln \left (y \right )}{11}+\frac {\ln \left (2 y^{2}+3 x \right )}{11} \end {aligned} \)

7.25 problem 26

7.25.1 Solving as ﬁrst order ode lie symmetry calculated ode

7.25.2 Solving as exact ode

7.25.3 Maple step by step solution