Internal problem ID [12731]
Internal file name [OUTPUT/11384_Friday_November_03_2023_06_31_41_AM_99448235/index.tex
]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 2. The Initial Value Problem. Exercises 2.4.4, page 115
Problem number: 12 (a).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "homogeneousTypeD2", "exactWithIntegrationFactor", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, _dAlembert]
\[ \boxed {y^{\prime }-\frac {x y}{y^{2}+x^{2}}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {x y}{x^{2}+y^{2}} \end {align*}
The \(x\) domain of \(f(x,y)\) when \(y=1\) is \[
\{-\infty The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=1\) is \[
\{-\infty
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} u^{\prime }\left (x \right ) x +u \left (x \right )-\frac {x^{2} u \left (x \right )}{u \left (x \right )^{2} x^{2}+x^{2}} = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u^{3}}{x \left (u^{2}+1\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u^{3}}{u^{2}+1}\). Integrating both sides gives \begin{align*}
\frac {1}{\frac {u^{3}}{u^{2}+1}} \,du &= -\frac {1}{x} \,d x \\
\int { \frac {1}{\frac {u^{3}}{u^{2}+1}} \,du} &= \int {-\frac {1}{x} \,d x} \\
-\frac {1}{2 u^{2}}+\ln \left (u \right )&=-\ln \left (x \right )+c_{2} \\
\end{align*} The solution is \[
-\frac {1}{2 u \left (x \right )^{2}}+\ln \left (u \left (x \right )\right )+\ln \left (x \right )-c_{2} = 0
\] Replacing \(u(x)\) in the above solution
by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \ln \left (\frac {y}{x}\right )-\frac {x^{2}}{2 y^{2}}+\ln \left (x \right )-c_{2} = 0\\ \ln \left (\frac {y}{x}\right )-\frac {x^{2}}{2 y^{2}}+\ln \left (x \right )-c_{2} = 0 \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} \textit {undefined} = 0 \end {align*}
This shows that no solution exist. The solution(s) found are the following \begin{align*}
\tag{1} \ln \left (\frac {y}{x}\right )-\frac {x^{2}}{2 y^{2}}+\ln \left (x \right )-c_{2} &= 0 \\
\end{align*} Verification of solutions
\[
\ln \left (\frac {y}{x}\right )-\frac {x^{2}}{2 y^{2}}+\ln \left (x \right )-c_{2} = 0
\] Warning, solution could not be verified
Writing the ode as \begin {align*} y^{\prime }&=\frac {x y}{x^{2}+y^{2}}\\ 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 {x y \left (b_{3}-a_{2}\right )}{x^{2}+y^{2}}-\frac {x^{2} y^{2} a_{3}}{\left (x^{2}+y^{2}\right )^{2}}-\left (\frac {y}{x^{2}+y^{2}}-\frac {2 x^{2} y}{\left (x^{2}+y^{2}\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {x}{x^{2}+y^{2}}-\frac {2 x \,y^{2}}{\left (x^{2}+y^{2}\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation} Putting the above in
normal form gives \[
-\frac {-3 x^{2} y^{2} b_{2}+2 x \,y^{3} a_{2}-2 x \,y^{3} b_{3}+y^{4} a_{3}-y^{4} b_{2}+x^{3} b_{1}-x^{2} y a_{1}-x \,y^{2} b_{1}+y^{3} a_{1}}{\left (x^{2}+y^{2}\right )^{2}} = 0
\] Setting the numerator to zero gives \begin{equation}
\tag{6E} 3 x^{2} y^{2} b_{2}-2 x \,y^{3} a_{2}+2 x \,y^{3} b_{3}-y^{4} a_{3}+y^{4} b_{2}-x^{3} b_{1}+x^{2} y a_{1}+x \,y^{2} b_{1}-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} -2 a_{2} v_{1} v_{2}^{3}-a_{3} v_{2}^{4}+3 b_{2} v_{1}^{2} v_{2}^{2}+b_{2} v_{2}^{4}+2 b_{3} v_{1} v_{2}^{3}+a_{1} v_{1}^{2} v_{2}-a_{1} v_{2}^{3}-b_{1} v_{1}^{3}+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} -b_{1} v_{1}^{3}+3 b_{2} v_{1}^{2} v_{2}^{2}+a_{1} v_{1}^{2} v_{2}+\left (-2 a_{2}+2 b_{3}\right ) v_{1} v_{2}^{3}+b_{1} v_{1} v_{2}^{2}+\left (-a_{3}+b_{2}\right ) v_{2}^{4}-a_{1} v_{2}^{3} = 0
\end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve
\begin {align*} a_{1}&=0\\ b_{1}&=0\\ -a_{1}&=0\\ -b_{1}&=0\\ 3 b_{2}&=0\\ -2 a_{2}+2 b_{3}&=0\\ -a_{3}+b_{2}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{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 &= 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 {x y}{x^{2}+y^{2}}\right ) \left (x\right ) \\ &= \frac {y^{3}}{x^{2}+y^{2}}\\ \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 {y^{3}}{x^{2}+y^{2}}}} dy \end {align*}
Which results in \begin {align*} S&= \ln \left (y \right )-\frac {x^{2}}{2 y^{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 {x y}{x^{2}+y^{2}} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= -\frac {x}{y^{2}}\\ S_{y} &= \frac {x^{2}+y^{2}}{y^{3}} \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 \(R,S\). Integrating the above gives
\begin {align*} S \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 (y\right ) y^{2}-x^{2}}{2 y^{2}} = c_{1} \end {align*}
Which simplifies to \begin {align*} \frac {2 \ln \left (y\right ) y^{2}-x^{2}}{2 y^{2}} = c_{1} \end {align*}
Which gives \begin {align*} y = {\mathrm e}^{\frac {\operatorname {LambertW}\left (x^{2} {\mathrm e}^{-2 c_{1}}\right )}{2}+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 \((R,S)\) \( \frac {dy}{dx} = \frac {x y}{x^{2}+y^{2}}\) \( \frac {d S}{d R} = 0\) \(\!\begin {aligned} R&= x\\ S&= \frac {2 \ln \left (y \right ) y^{2}-x^{2}}{2 y^{2}} \end {aligned} \) Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 1 = {\mathrm e}^{c_{1}} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\sqrt {\frac {x^{2}}{\operatorname {LambertW}\left (x^{2}\right )}} \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \sqrt {\frac {x^{2}}{\operatorname {LambertW}\left (x^{2}\right )}} \\
\end{align*} Verification of solutions
\[
y = \sqrt {\frac {x^{2}}{\operatorname {LambertW}\left (x^{2}\right )}}
\] Verified OK. Entering Exact first order ODE solver. (Form one type)
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 (x^{2}+y^{2}\right )\mathop {\mathrm {d}y} &= \left (y x\right )\mathop {\mathrm {d}x}\\ \left (-y x\right )\mathop {\mathrm {d}x} + \left (x^{2}+y^{2}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -y x\\ N(x,y) &= x^{2}+y^{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 (-y x\right )\\ &= -x \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (x^{2}+y^{2}\right )\\ &= 2 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}{x^{2}+y^{2}}\left ( \left ( -x\right ) - \left (2 x \right ) \right ) \\ &=-\frac {3 x}{x^{2}+y^{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}{y x}\left ( \left ( 2 x\right ) - \left (-x \right ) \right ) \\ &=-\frac {3}{y} \end {align*}
Since \(B\) does not depend on \(x\), it can be used to obtain an integrating factor. Let the
integrating factor be \(\mu \). Then \begin {align*} \mu &= e^{\int B \mathop {\mathrm {d}y}} \\ &= e^{\int -\frac {3}{y}\mathop {\mathrm {d}y} } \end {align*}
The result of integrating gives \begin {align*} \mu &= e^{-3 \ln \left (y \right ) } \\ &= \frac {1}{y^{3}} \end {align*}
\(M\) and \(N\) are now multiplied by this integrating factor, giving 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 \\ &= \frac {1}{y^{3}}\left (-y x\right ) \\ &= -\frac {x}{y^{2}} \end {align*}
And \begin {align*} \overline {N} &=\mu N \\ &= \frac {1}{y^{3}}\left (x^{2}+y^{2}\right ) \\ &= \frac {x^{2}+y^{2}}{y^{3}} \end {align*}
So now a modified ODE is obtained from the original ODE which will be exact and can be
solved using the standard method. The modified ODE is \begin {align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (-\frac {x}{y^{2}}\right ) + \left (\frac {x^{2}+y^{2}}{y^{3}}\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 -\frac {x}{y^{2}}\mathop {\mathrm {d}x} \\
\tag{3} \phi &= -\frac {x^{2}}{2 y^{2}}+ 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} = \frac {x^{2}}{y^{3}}+f'(y)
\end{equation} But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {x^{2}+y^{2}}{y^{3}}\).
Therefore equation (4) becomes \begin{equation}
\tag{5} \frac {x^{2}+y^{2}}{y^{3}} = \frac {x^{2}}{y^{3}}+f'(y)
\end{equation} Solving equation (5) for \( f'(y)\) gives \[
f'(y) = \frac {1}{y}
\] Integrating the above w.r.t \(y\)
gives \begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {1}{y}\right ) \mathop {\mathrm {d}y} \\
f(y) &= \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 \(\phi \) \[
\phi = \ln \left (y \right )-\frac {x^{2}}{2 y^{2}}+ c_{1}
\] 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 \[
c_{1} = \ln \left (y \right )-\frac {x^{2}}{2 y^{2}}
\]
The solution becomes\[
y = {\mathrm e}^{\frac {\operatorname {LambertW}\left (x^{2} {\mathrm e}^{-2 c_{1}}\right )}{2}+c_{1}}
\] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=1\)
in the above solution gives an equation to solve for the constant of integration.
\begin {align*} 1 = {\mathrm e}^{c_{1}} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\sqrt {\frac {x^{2}}{\operatorname {LambertW}\left (x^{2}\right )}} \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \sqrt {\frac {x^{2}}{\operatorname {LambertW}\left (x^{2}\right )}} \\
\end{align*} Verification of solutions
\[
y = \sqrt {\frac {x^{2}}{\operatorname {LambertW}\left (x^{2}\right )}}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {x y}{y^{2}+x^{2}}=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x y}{y^{2}+x^{2}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 1.75 (sec). Leaf size: 11
\[
y \left (x \right ) = \sqrt {\frac {x^{2}}{\operatorname {LambertW}\left (x^{2}\right )}}
\]
✓ Solution by Mathematica
Time used: 10.851 (sec). Leaf size: 15
\[
y(x)\to \frac {x}{\sqrt {W\left (x^2\right )}}
\]
8.34.2 Solving as homogeneousTypeD2 ode
8.34.3 Solving as first order ode lie symmetry calculated ode
8.34.4 Solving as exact ode
8.34.5 Maple step by step solution
`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 D
<- homogeneous successful`
dsolve([diff(y(x),x)=x*y(x)/(x^2+y(x)^2),y(0) = 1],y(x), singsol=all)
DSolve[{y'[x]==x*y[x]/(x^2+y[x]^2),{y[0]==1}},y[x],x,IncludeSingularSolutions -> True]