2.2.7 Internal
problem
ID
[8542]Book
:
Second
order
enumerated
odes Section
:
section
2 Problem
number
:
8Date
solved
:
Sunday, November 10, 2024 at 04:01:04 AMCAS
classification
:
[[_homogeneous, `class C`], _dAlembert]
Solve
\begin{align*} y^{\prime }&=\left (x +y\right )^{4} \end{align*}
Time used: 0.498 (sec)
Let
\begin{align*} z = x +y\tag {1} \end{align*}
Then
\begin{align*} z^{\prime }\left (x \right )&=1+y^{\prime } \end{align*}
Therefore
\begin{align*} y^{\prime }&=z^{\prime }\left (x \right )-1 \end{align*}
Hence the given ode can now be written as
\begin{align*} z^{\prime }\left (x \right )-1&=z^{4} \end{align*}
This is separable first order ode. Integrating
\begin{align*}
\int d x&=\int \frac {1}{z^{4}+1}d z \\
x +c_1&=\frac {\sqrt {2}\, \left (\ln \left (\frac {z^{2}+z \sqrt {2}+1}{z^{2}-z \sqrt {2}+1}\right )+2 \arctan \left (z \sqrt {2}+1\right )+2 \arctan \left (z \sqrt {2}-1\right )\right )}{8} \\
\end{align*}
Replacing \(z\) back by its value from (1) then the
above gives the solution as
Figure 2.158: Slope field plot\(y^{\prime } = \left (x +y\right )^{4}\) Summary of solutions found
\begin{align*}
\frac {\sqrt {2}\, \left (\ln \left (\frac {\left (x +y\right )^{2}+\left (x +y\right ) \sqrt {2}+1}{\left (x +y\right )^{2}-\left (x +y\right ) \sqrt {2}+1}\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}+1\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}-1\right )\right )}{8} &= x +c_1 \\
\end{align*}
Time used: 1.230 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=\left (x +y \right )^{4}\\ 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}+\left (x +y \right )^{4} \left (b_{3}-a_{2}\right )-\left (x +y \right )^{8} a_{3}-4 \left (x +y \right )^{3} \left (x a_{2}+y a_{3}+a_{1}\right )-4 \left (x +y \right )^{3} \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-x^{8} a_{3}-8 x^{7} y a_{3}-28 x^{6} y^{2} a_{3}-56 x^{5} y^{3} a_{3}-70 x^{4} y^{4} a_{3}-56 x^{3} y^{5} a_{3}-28 x^{2} y^{6} a_{3}-8 x \,y^{7} a_{3}-y^{8} a_{3}-5 x^{4} a_{2}-4 x^{4} b_{2}+x^{4} b_{3}-16 x^{3} y a_{2}-4 x^{3} y a_{3}-12 x^{3} y b_{2}-18 x^{2} y^{2} a_{2}-12 x^{2} y^{2} a_{3}-12 x^{2} y^{2} b_{2}-6 x^{2} y^{2} b_{3}-8 x \,y^{3} a_{2}-12 x \,y^{3} a_{3}-4 x \,y^{3} b_{2}-8 x \,y^{3} b_{3}-y^{4} a_{2}-4 y^{4} a_{3}-3 y^{4} b_{3}-4 x^{3} a_{1}-4 x^{3} b_{1}-12 x^{2} y a_{1}-12 x^{2} y b_{1}-12 x \,y^{2} a_{1}-12 x \,y^{2} b_{1}-4 y^{3} a_{1}-4 y^{3} b_{1}+b_{2} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} -x^{8} a_{3}-8 x^{7} y a_{3}-28 x^{6} y^{2} a_{3}-56 x^{5} y^{3} a_{3}-70 x^{4} y^{4} a_{3}-56 x^{3} y^{5} a_{3}-28 x^{2} y^{6} a_{3}-8 x \,y^{7} a_{3}-y^{8} a_{3}-5 x^{4} a_{2}-4 x^{4} b_{2}+x^{4} b_{3}-16 x^{3} y a_{2}-4 x^{3} y a_{3}-12 x^{3} y b_{2}-18 x^{2} y^{2} a_{2}-12 x^{2} y^{2} a_{3}-12 x^{2} y^{2} b_{2}-6 x^{2} y^{2} b_{3}-8 x \,y^{3} a_{2}-12 x \,y^{3} a_{3}-4 x \,y^{3} b_{2}-8 x \,y^{3} b_{3}-y^{4} a_{2}-4 y^{4} a_{3}-3 y^{4} b_{3}-4 x^{3} a_{1}-4 x^{3} b_{1}-12 x^{2} y a_{1}-12 x^{2} y b_{1}-12 x \,y^{2} a_{1}-12 x \,y^{2} b_{1}-4 y^{3} a_{1}-4 y^{3} b_{1}+b_{2} = 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} -a_{3} v_{1}^{8}-8 a_{3} v_{1}^{7} v_{2}-28 a_{3} v_{1}^{6} v_{2}^{2}-56 a_{3} v_{1}^{5} v_{2}^{3}-70 a_{3} v_{1}^{4} v_{2}^{4}-56 a_{3} v_{1}^{3} v_{2}^{5}-28 a_{3} v_{1}^{2} v_{2}^{6}-8 a_{3} v_{1} v_{2}^{7}-a_{3} v_{2}^{8}-5 a_{2} v_{1}^{4}-16 a_{2} v_{1}^{3} v_{2}-18 a_{2} v_{1}^{2} v_{2}^{2}-8 a_{2} v_{1} v_{2}^{3}-a_{2} v_{2}^{4}-4 a_{3} v_{1}^{3} v_{2}-12 a_{3} v_{1}^{2} v_{2}^{2}-12 a_{3} v_{1} v_{2}^{3}-4 a_{3} v_{2}^{4}-4 b_{2} v_{1}^{4}-12 b_{2} v_{1}^{3} v_{2}-12 b_{2} v_{1}^{2} v_{2}^{2}-4 b_{2} v_{1} v_{2}^{3}+b_{3} v_{1}^{4}-6 b_{3} v_{1}^{2} v_{2}^{2}-8 b_{3} v_{1} v_{2}^{3}-3 b_{3} v_{2}^{4}-4 a_{1} v_{1}^{3}-12 a_{1} v_{1}^{2} v_{2}-12 a_{1} v_{1} v_{2}^{2}-4 a_{1} v_{2}^{3}-4 b_{1} v_{1}^{3}-12 b_{1} v_{1}^{2} v_{2}-12 b_{1} v_{1} v_{2}^{2}-4 b_{1} v_{2}^{3}+b_{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} -a_{3} v_{1}^{8}-8 a_{3} v_{1}^{7} v_{2}-28 a_{3} v_{1}^{6} v_{2}^{2}-56 a_{3} v_{1}^{5} v_{2}^{3}-70 a_{3} v_{1}^{4} v_{2}^{4}+\left (-5 a_{2}-4 b_{2}+b_{3}\right ) v_{1}^{4}-56 a_{3} v_{1}^{3} v_{2}^{5}+\left (-16 a_{2}-4 a_{3}-12 b_{2}\right ) v_{1}^{3} v_{2}+\left (-4 a_{1}-4 b_{1}\right ) v_{1}^{3}-28 a_{3} v_{1}^{2} v_{2}^{6}+\left (-18 a_{2}-12 a_{3}-12 b_{2}-6 b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (-12 a_{1}-12 b_{1}\right ) v_{1}^{2} v_{2}-8 a_{3} v_{1} v_{2}^{7}+\left (-8 a_{2}-12 a_{3}-4 b_{2}-8 b_{3}\right ) v_{1} v_{2}^{3}+\left (-12 a_{1}-12 b_{1}\right ) v_{1} v_{2}^{2}-a_{3} v_{2}^{8}+\left (-a_{2}-4 a_{3}-3 b_{3}\right ) v_{2}^{4}+\left (-4 a_{1}-4 b_{1}\right ) v_{2}^{3}+b_{2} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} b_{2}&=0\\ -70 a_{3}&=0\\ -56 a_{3}&=0\\ -28 a_{3}&=0\\ -8 a_{3}&=0\\ -a_{3}&=0\\ -12 a_{1}-12 b_{1}&=0\\ -4 a_{1}-4 b_{1}&=0\\ -16 a_{2}-4 a_{3}-12 b_{2}&=0\\ -5 a_{2}-4 b_{2}+b_{3}&=0\\ -a_{2}-4 a_{3}-3 b_{3}&=0\\ -18 a_{2}-12 a_{3}-12 b_{2}-6 b_{3}&=0\\ -8 a_{2}-12 a_{3}-4 b_{2}-8 b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=-b_{1}\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=b_{1}\\ b_{2}&=0\\ 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 &= -1 \\
\eta &= 1 \\
\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 \(S(R)\) .
Therefore
\begin{align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {1}{-1}\\ &= -1 \end{align*}
This is easily solved to give
\begin{align*} y = -x +c_1 \end{align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence
\begin{align*} R &= x +y \end{align*}
And \(S\) is found from
\begin{align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{-1} \end{align*}
Integrating gives
\begin{align*} S &= \int { \frac {dx}{T}}\\ &= -x \end{align*}
Where the constant of integration is set to zero as we just need one solution. 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) &= \left (x +y \right )^{4} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 1\\ S_{x} &= -1\\ S_{y} &= 0 \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= -\frac {1}{1+\left (x +y \right )^{4}}\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 {1}{R^{4}+1} \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 {-\frac {1}{R^{4}+1}\, dR}\\ S \left (R \right ) &= -\frac {\sqrt {2}\, \left (\ln \left (\frac {R^{2}+R \sqrt {2}+1}{R^{2}-R \sqrt {2}+1}\right )+2 \arctan \left (R \sqrt {2}+1\right )+2 \arctan \left (R \sqrt {2}-1\right )\right )}{8} + 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*} -x = -\frac {\sqrt {2}\, \left (\ln \left (\frac {\left (x +y\right )^{2}+\left (x +y\right ) \sqrt {2}+1}{\left (x +y\right )^{2}-\left (x +y\right ) \sqrt {2}+1}\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}+1\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}-1\right )\right )}{8}+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 \((R,S)\)
\( \frac {dy}{dx} = \left (x +y \right )^{4}\)
\( \frac {d S}{d R} = -\frac {1}{R^{4}+1}\)
\(\!\begin {aligned} R&= x +y\\ S&= -x \end {aligned} \)
Figure 2.159: Slope field plot\(y^{\prime } = \left (x +y\right )^{4}\) Summary of solutions found
\begin{align*}
-x &= -\frac {\sqrt {2}\, \left (\ln \left (\frac {\left (x +y\right )^{2}+\left (x +y\right ) \sqrt {2}+1}{\left (x +y\right )^{2}-\left (x +y\right ) \sqrt {2}+1}\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}+1\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}-1\right )\right )}{8}+c_2 \\
\end{align*}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (x +y \left (x \right )\right )^{4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (x +y \left (x \right )\right )^{4} \end {array} \]
` 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 C
1 st order, trying the canonical coordinates of the invariance group
<- 1st order, canonical coordinates successful
<- homogeneous successful `
Solving time : 0.912
dsolve ( diff ( y ( x ), x ) = (x+y(x))^4,
y(x),singsol=all)
\[
\text {Expression too large to display}
\]
Solving time : 0.168
DSolve [{ D [ y [ x ], x ] == (x + y[x])^4,{}},
y[x],x,IncludeSingularSolutions-> True ]
\[
\text {Solve}\left [\frac {1}{4} \text {RootSum}\left [\text {$\#$1}^4+4 \text {$\#$1}^3 y(x)+6 \text {$\#$1}^2 y(x)^2+4 \text {$\#$1} y(x)^3+y(x)^4+1\&,\frac {\log (x-\text {$\#$1})}{\text {$\#$1}^3+3 \text {$\#$1}^2 y(x)+3 \text {$\#$1} y(x)^2+y(x)^3}\&\right ]-x=c_1,y(x)\right ]
\]