Internal problem ID [5741]
Internal file name [OUTPUT/4989_Sunday_June_05_2022_03_16_07_PM_72683433/index.tex
]
Book: Ordinary differential equations and calculus of variations. Makarets and Reshetnyak. Wold
Scientific. Singapore. 1995
Section: Chapter 1. First order differential equations. Section 1.1 Separable equations problems.
page 7
Problem number: 28.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "homogeneousTypeC", "exactWithIntegrationFactor", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_homogeneous, `class C`], [_Abel, `2nd type`, `class C`], _dAlembert]
\[ \boxed {\left (x +2 y\right ) y^{\prime }=1} \] 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 {1}{x +2 y} \end {align*}
The \(x\) domain of \(f(x,y)\) when \(y=-1\) is \[
\{x <2\boldsymbol {\lor }2 The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=-1\) is \[
\{x <2\boldsymbol {\lor }2
Let \begin {align*} z = x +2 y\tag {1} \end {align*}
Then \begin {align*} z^{\prime }\left (x \right )&=1+2 y^{\prime } \end {align*}
Therefore \begin {align*} y^{\prime }&=\frac {z^{\prime }\left (x \right )}{2}-\frac {1}{2} \end {align*}
Hence the given ode can now be written as \begin {align*} \frac {z^{\prime }\left (x \right )}{2}-\frac {1}{2}&=\frac {1}{z} \end {align*}
This is separable first order ode. Integrating \begin{align*}
\int d x&=\int \frac {1}{\frac {2}{z}+1}d z \\
x +c_{1}&=z -2 \ln \left (2+z \right ) \\
\end{align*} Replacing \(z\) back by its value from (1) then the
above gives the solution as \begin {align*} y = -\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1}}{2}-\frac {x}{2}-1}}{2}\right )-\frac {x}{2}-1 \end {align*}
\[
y = -\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1}}{2}-\frac {x}{2}-1}}{2}\right )-\frac {x}{2}-1
\] 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 = -\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1}}{2}-1}}{2}\right )-1 \end {align*}
Unable to solve for constant of integration. Since \(\lim _{c_1 \to \infty }\) gives \(y = -\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1}}{2}-\frac {x}{2}-1}}{2}\right )-\frac {x}{2}-1=y = -\frac {x}{2}-1\) and this result satisfies the given initial
condition. The solution(s) found are the following \begin{align*}
\tag{1} y &= -\frac {x}{2}-1 \\
\end{align*} Verification of solutions
\[
y = -\frac {x}{2}-1
\] Verified OK. Writing the ode as \begin {align*} y^{\prime }&=\frac {1}{x +2 y}\\ 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 known. It is of type ODE class Form \(\xi \) \(\eta \) linear ode \(y'=f(x) y(x) +g(x)\) \(0\) \(e^{\int fdx}\) separable ode \(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \) \(\frac {1}{f}\) \(0\) quadrature ode \(y^{\prime }=f\left ( x\right ) \) \(0\) \(1\) quadrature ode \(y^{\prime }=g\left ( y\right ) \) \(1\) \(0\) homogeneous ODEs of
Class A \(y^{\prime }=f\left ( \frac {y}{x}\right ) \) \(x\) \(y\) homogeneous ODEs of
Class C \(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}\) \(1\) \(-\frac {b}{c}\) homogeneous class D \(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right ) \) \(x^{2}\) \(xy\) First order special
form ID 1 \(y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right ) \) \(\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }\) \(\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }\) polynomial type ode \(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\) \(\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) \(\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) Bernoulli ode \(y^{\prime }=f\left ( x\right ) y+g\left ( x\right ) y^{n}\) \(0\) \(e^{-\int \left ( n-1\right ) f\left ( x\right ) dx}y^{n}\) Reduced Riccati \(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\) \(0\) \(e^{-\int f_{1}dx}\) The above table shows that \begin {align*} \xi \left (x,y\right ) &=1\\ \tag {A1} \eta \left (x,y\right ) &=-{\frac {1}{2}} \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 {-{\frac {1}{2}}}{1}\\ &= -{\frac {1}{2}} \end {align*}
This is easily solved to give \begin {align*} y = -\frac {x}{2}+c_{1} \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {x}{2}+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) &= \frac {1}{x +2 y} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= {\frac {1}{2}}\\ 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 {2 x +4 y}{2+x +2 y}\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 {4 R}{2+2 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 ) = 2 R -2 \ln \left (1+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*} x = x +2 y-2 \ln \left (y+\frac {x}{2}+1\right )+c_{1} \end {align*}
Which simplifies to \begin {align*} x = x +2 y-2 \ln \left (y+\frac {x}{2}+1\right )+c_{1} \end {align*}
Which gives \begin {align*} y = -\operatorname {LambertW}\left (-{\mathrm e}^{-1-\frac {x}{2}+\frac {c_{1}}{2}}\right )-1-\frac {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 \((R,S)\) \( \frac {dy}{dx} = \frac {1}{x +2 y}\) \( \frac {d S}{d R} = \frac {4 R}{2+2 R}\) \(\!\begin {aligned} R&= \frac {x}{2}+y\\ S&= x \end {aligned} \) 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 = -\operatorname {LambertW}\left (-{\mathrm e}^{-1+\frac {c_{1}}{2}}\right )-1 \end {align*}
Unable to solve for constant of integration. Warning: Unable to solve for constant of
integration.
Verification of solutions N/A 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\right )\mathop {\mathrm {d}y} &= \mathop {\mathrm {d}x}\\ - \mathop {\mathrm {d}x}+\left ( x +2 y\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -1\\ N(x,y) &= x +2 y \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 (-1\right )\\ &= 0 \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (x +2 y\right )\\ &= 1 \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}\left ( \left ( 0\right ) - \left (1 \right ) \right ) \\ &=-\frac {1}{x +2 y} \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 ) \\ &=-1\left ( \left ( 1\right ) - \left (0 \right ) \right ) \\ &=-1 \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 -1\mathop {\mathrm {d}y} } \end {align*}
The result of integrating gives \begin {align*} \mu &= e^{-y } \\ &= {\mathrm e}^{-y} \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 \\ &= {\mathrm e}^{-y}\left (-1\right ) \\ &= -{\mathrm e}^{-y} \end {align*}
And \begin {align*} \overline {N} &=\mu N \\ &= {\mathrm e}^{-y}\left (x +2 y\right ) \\ &= \left (x +2 y \right ) {\mathrm e}^{-y} \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 (-{\mathrm e}^{-y}\right ) + \left (\left (x +2 y \right ) {\mathrm e}^{-y}\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 -{\mathrm e}^{-y}\mathop {\mathrm {d}x} \\
\tag{3} \phi &= -{\mathrm e}^{-y} x+ 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} = {\mathrm e}^{-y} x+f'(y)
\end{equation} But equation (2) says that \(\frac {\partial \phi }{\partial y} = \left (x +2 y \right ) {\mathrm e}^{-y}\).
Therefore equation (4) becomes \begin{equation}
\tag{5} \left (x +2 y \right ) {\mathrm e}^{-y} = {\mathrm e}^{-y} x+f'(y)
\end{equation} Solving equation (5) for \( f'(y)\) gives \[
f'(y) = 2 y \,{\mathrm e}^{-y}
\] Integrating the above w.r.t \(y\)
gives \begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( 2 y \,{\mathrm e}^{-y}\right ) \mathop {\mathrm {d}y} \\
f(y) &= -2 \,{\mathrm e}^{-y} \left (y +1\right )+ c_{1} \\
\end{align*} Where \(c_{1}\) is constant of integration. Substituting result found above for \(f(y)\) into
equation (3) gives \(\phi \) \[
\phi = -{\mathrm e}^{-y} x -2 \,{\mathrm e}^{-y} \left (y +1\right )+ c_{1}
\] 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 \[
c_{1} = -{\mathrm e}^{-y} x -2 \,{\mathrm e}^{-y} \left (y +1\right )
\]
The solution becomes\[
y = -\frac {x}{2}-\operatorname {LambertW}\left (\frac {c_{1} {\mathrm e}^{-\frac {x}{2}-1}}{2}\right )-1
\] 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 = -\operatorname {LambertW}\left (\frac {{\mathrm e}^{-1} c_{1}}{2}\right )-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&=-\frac {x}{2}-1 \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\frac {x}{2}-1 \\
\end{align*} Verification of solutions
\[
y = -\frac {x}{2}-1
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (x +2 y\right ) y^{\prime }=1, y \left (0\right )=-1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{x +2 y} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 9
\[
y \left (x \right ) = -\frac {x}{2}-1
\]
✓ Solution by Mathematica
Time used: 0.019 (sec). Leaf size: 12
\[
y(x)\to -\frac {x}{2}-1
\]
1.28.2 Solving as homogeneousTypeC ode
1.28.3 Solving as first order ode lie symmetry lookup ode
homogeneous Type C
. Therefore we do not need
to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)
1.28.4 Solving as exact ode
1.28.5 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
<- 1st order linear successful
<- inverse linear successful`
dsolve([(x+2*y(x))*diff(y(x),x)=1,y(0) = -1],y(x), singsol=all)
DSolve[{(x+2*y[x])*y'[x]==1,{y[0]==-1}},y[x],x,IncludeSingularSolutions -> True]