Internal problem ID [978]
Internal file name [OUTPUT/979_Sunday_June_05_2022_01_55_50_AM_63671662/index.tex
]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 2, First order equations. Transformation of Nonlinear Equations into Separable
Equations. Section 2.4 Page 68
Problem number: Example 3(b).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, _Riccati]
\[ \boxed {y^{\prime } x^{2}-y^{2}-y x=-x^{2}} \] With initial conditions \begin {align*} [y \left (1\right ) = 2] \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^{2}+y x +y^{2}}{x^{2}} \end {align*}
The \(x\) domain of \(f(x,y)\) when \(y=2\) is \[
\{x <0\boldsymbol {\lor }0 The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=2\) is \[
\{x <0\boldsymbol {\lor }0
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*} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2}-u \left (x \right )^{2} x^{2}-u \left (x \right ) x^{2} = -x^{2} \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u^{2}-1}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=u^{2}-1\). Integrating both sides gives \begin{align*}
\frac {1}{u^{2}-1} \,du &= \frac {1}{x} \,d x \\
\int { \frac {1}{u^{2}-1} \,du} &= \int {\frac {1}{x} \,d x} \\
-\operatorname {arctanh}\left (u \right )&=\ln \left (x \right )+c_{2} \\
\end{align*} The solution is \[
-\operatorname {arctanh}\left (u \left (x \right )\right )-\ln \left (x \right )-c_{2} = 0
\] Replacing \(u(x)\) in the above solution
by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} -\operatorname {arctanh}\left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} = 0\\ -\operatorname {arctanh}\left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} = 0 \end {align*}
Substituting initial conditions and solving for \(c_{2} \) gives \(c_{2} = -\operatorname {arctanh}\left (\frac {1}{2}\right )+\frac {i \pi }{2}\). Hence the solution becomes
The solution(s) found are the following \begin{align*}
\tag{1} -\operatorname {arctanh}\left (\frac {y}{x}\right )-\ln \left (x \right )+\operatorname {arctanh}\left (\frac {1}{2}\right )-\frac {i \pi }{2} &= 0 \\
\end{align*} Verification of solutions
\[
-\operatorname {arctanh}\left (\frac {y}{x}\right )-\ln \left (x \right )+\operatorname {arctanh}\left (\frac {1}{2}\right )-\frac {i \pi }{2} = 0
\] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=\frac {-x^{2}+y x +y^{2}}{x^{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 {\left (-x^{2}+y x +y^{2}\right ) \left (b_{3}-a_{2}\right )}{x^{2}}-\frac {\left (-x^{2}+y x +y^{2}\right )^{2} a_{3}}{x^{4}}-\left (\frac {-2 x +y}{x^{2}}-\frac {2 \left (-x^{2}+y x +y^{2}\right )}{x^{3}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\frac {\left (x +2 y \right ) \left (x b_{2}+y b_{3}+b_{1}\right )}{x^{2}} = 0
\end{equation} Putting the above in
normal form gives \[
\frac {x^{4} a_{2}-x^{4} a_{3}-x^{4} b_{3}+2 x^{3} y a_{3}-2 x^{3} y b_{2}+x^{2} y^{2} a_{2}+2 x^{2} y^{2} a_{3}-x^{2} y^{2} b_{3}-y^{4} a_{3}-x^{3} b_{1}+x^{2} y a_{1}-2 x^{2} y b_{1}+2 x \,y^{2} a_{1}}{x^{4}} = 0
\] Setting the numerator to zero gives \begin{equation}
\tag{6E} x^{4} a_{2}-x^{4} a_{3}-x^{4} b_{3}+2 x^{3} y a_{3}-2 x^{3} y b_{2}+x^{2} y^{2} a_{2}+2 x^{2} y^{2} a_{3}-x^{2} y^{2} b_{3}-y^{4} a_{3}-x^{3} b_{1}+x^{2} y a_{1}-2 x^{2} y b_{1}+2 x \,y^{2} 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} a_{2} v_{1}^{4}+a_{2} v_{1}^{2} v_{2}^{2}-a_{3} v_{1}^{4}+2 a_{3} v_{1}^{3} v_{2}+2 a_{3} v_{1}^{2} v_{2}^{2}-a_{3} v_{2}^{4}-2 b_{2} v_{1}^{3} v_{2}-b_{3} v_{1}^{4}-b_{3} v_{1}^{2} v_{2}^{2}+a_{1} v_{1}^{2} v_{2}+2 a_{1} v_{1} v_{2}^{2}-b_{1} v_{1}^{3}-2 b_{1} v_{1}^{2} v_{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} \left (a_{2}-a_{3}-b_{3}\right ) v_{1}^{4}+\left (2 a_{3}-2 b_{2}\right ) v_{1}^{3} v_{2}-b_{1} v_{1}^{3}+\left (a_{2}+2 a_{3}-b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (a_{1}-2 b_{1}\right ) v_{1}^{2} v_{2}+2 a_{1} v_{1} v_{2}^{2}-a_{3} v_{2}^{4} = 0
\end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve
\begin {align*} 2 a_{1}&=0\\ -a_{3}&=0\\ -b_{1}&=0\\ a_{1}-2 b_{1}&=0\\ 2 a_{3}-2 b_{2}&=0\\ a_{2}-a_{3}-b_{3}&=0\\ a_{2}+2 a_{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}&=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^{2}+y x +y^{2}}{x^{2}}\right ) \left (x\right ) \\ &= \frac {x^{2}-y^{2}}{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 {x^{2}-y^{2}}{x}}} dy \end {align*}
Which results in \begin {align*} S&= -\frac {\ln \left (-x +y \right )}{2}+\frac {\ln \left (x +y \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 {-x^{2}+y x +y^{2}}{x^{2}} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= -\frac {y}{x^{2}-y^{2}}\\ S_{y} &= \frac {x}{x^{2}-y^{2}} \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin {align*} \frac {dS}{dR} &= -\frac {1}{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 {1}{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 ) = -\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 {\ln \left (-x +y\right )}{2}+\frac {\ln \left (x +y\right )}{2} = -\ln \left (x \right )+c_{1} \end {align*}
Which simplifies to \begin {align*} -\frac {\ln \left (-x +y\right )}{2}+\frac {\ln \left (x +y\right )}{2} = -\ln \left (x \right )+c_{1} \end {align*}
Which gives \begin {align*} y = \frac {x \left ({\mathrm e}^{2 c_{1}}+x^{2}\right )}{{\mathrm e}^{2 c_{1}}-x^{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 \((R,S)\) \( \frac {dy}{dx} = \frac {-x^{2}+y x +y^{2}}{x^{2}}\) \( \frac {d S}{d R} = -\frac {1}{R}\) \(\!\begin {aligned} R&= x\\ S&= -\frac {\ln \left (-x +y \right )}{2}+\frac {\ln \left (x +y \right )}{2} \end {aligned} \) Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=2\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 2 = \frac {{\mathrm e}^{2 c_{1}}+1}{{\mathrm e}^{2 c_{1}}-1} \end {align*}
The solutions are \begin {align*} c_{1} = \frac {\ln \left (3\right )}{2} \end {align*}
Trying the constant \begin {align*} c_{1} = \frac {\ln \left (3\right )}{2} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {-x^{3}-3 x}{x^{2}-3} \end {align*}
The constant \(c_{1} = \frac {\ln \left (3\right )}{2}\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {-x^{3}-3 x}{x^{2}-3} \\
\end{align*} Verification of solutions
\[
y = \frac {-x^{3}-3 x}{x^{2}-3}
\] Verified OK. In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-x^{2}+y x +y^{2}}{x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -1+\frac {y}{x}+\frac {y^{2}}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows that \(f_0(x)=-1\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {1}{x^{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{x^{2}}} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second
order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=-\frac {2}{x^{3}}\\ f_1 f_2 &=\frac {1}{x^{3}}\\ f_2^2 f_0 &=-\frac {1}{x^{4}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{x^{2}}+\frac {u^{\prime }\left (x \right )}{x^{3}}-\frac {u \left (x \right )}{x^{4}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[
u \left (x \right ) = \frac {c_{2} x^{2}+c_{1}}{x}
\] The above shows that \[
u^{\prime }\left (x \right ) = \frac {c_{2} x^{2}-c_{1}}{x^{2}}
\] Using the above in (1) gives the solution \[
y = -\frac {\left (c_{2} x^{2}-c_{1} \right ) x}{c_{2} x^{2}+c_{1}}
\] Dividing both
numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following
solution
\[
y = \frac {\left (-x^{2}+c_{3} \right ) x}{x^{2}+c_{3}}
\] Initial conditions are used to solve for \(c_{3}\). Substituting \(x=1\) and \(y=2\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 2 = \frac {c_{3} -1}{c_{3} +1} \end {align*}
The solutions are \begin {align*} c_{3} = -3 \end {align*}
Trying the constant \begin {align*} c_{3} = -3 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\frac {\left (x^{2}+3\right ) x}{x^{2}-3} \end {align*}
The constant \(c_{3} = -3\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\frac {\left (x^{2}+3\right ) x}{x^{2}-3} \\
\end{align*} Verification of solutions
\[
y = -\frac {\left (x^{2}+3\right ) x}{x^{2}-3}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime } x^{2}-y^{2}-y x =-x^{2}, y \left (1\right )=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+y x -x^{2}}{x^{2}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=2 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 19
\[
y \left (x \right ) = -\frac {x \left (x^{2}+3\right )}{x^{2}-3}
\]
✓ Solution by Mathematica
Time used: 0.576 (sec). Leaf size: 20
\[
y(x)\to -\frac {x \left (x^2+3\right )}{x^2-3}
\]
5.4.2 Solving as homogeneousTypeD2 ode
5.4.3 Solving as first order ode lie symmetry calculated ode
5.4.4 Solving as riccati ode
5.4.5 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([x^2*diff(y(x),x)=y(x)^2+x*y(x)-x^2,y(1) = 2],y(x), singsol=all)
DSolve[{x^2*y'[x]==y[x]^2+x*y[x]-x^2,y[1]==2},y[x],x,IncludeSingularSolutions -> True]