Internal problem ID [11167]
Internal file name [OUTPUT/10153_Saturday_December_03_2022_08_02_58_AM_3788059/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter 2, differential equations of the first order and the first degree. Article 18.
Transformation of variables. Page 26
Problem number: Ex 2.
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]
\[ \boxed {y^{\prime } \left (x +y\right )=1} \]
Let \begin {align*} z = x +y\tag {1} \end {align*}
Then \begin {align*} z^{\prime }\left (x \right )&=y^{\prime }+1 \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&=\frac {1}{z} \end {align*}
This is separable first order ode. Integrating \begin{align*} \int d x&=\int \frac {1}{\frac {1}{z}+1}d z \\ x +c_{1}&=z -\ln \left (1+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 (-{\mathrm e}^{-x -c_{1} -1}\right )-x -1 \end {align*}
\[ y = -\operatorname {LambertW}\left (-{\mathrm e}^{-x -c_{1} -1}\right )-x -1 \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\operatorname {LambertW}\left (-{\mathrm e}^{-x -c_{1} -1}\right )-x -1 \\ \end{align*}
Verification of solutions
\[ y = -\operatorname {LambertW}\left (-{\mathrm e}^{-x -c_{1} -1}\right )-x -1 \] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=\frac {1}{x +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 homogeneous Type C. Therefore we do not need
to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)
|
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 ) &=-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 \(S(R)\). Unable to determine \(R\). Terminating
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 +y\right )\mathop {\mathrm {d}y} &= \mathop {\mathrm {d}x}\\ - \mathop {\mathrm {d}x}+\left ( x +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 +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 +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 +y}\left ( \left ( 0\right ) - \left (1 \right ) \right ) \\ &=-\frac {1}{x +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 +y\right ) \\ &= \left (x +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 +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 &= -x \,{\mathrm e}^{-y}+ 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} = x \,{\mathrm e}^{-y}+f'(y) \end{equation} But equation (2) says that \(\frac {\partial \phi }{\partial y} = \left (x +y \right ) {\mathrm e}^{-y}\). Therefore equation (4) becomes \begin{equation} \tag{5} \left (x +y \right ) {\mathrm e}^{-y} = x \,{\mathrm e}^{-y}+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = {\mathrm e}^{-y} y \] Integrating the above w.r.t \(y\) gives \begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( {\mathrm e}^{-y} y\right ) \mathop {\mathrm {d}y} \\ f(y) &= -\left (y +1\right ) {\mathrm e}^{-y}+ c_{1} \\ \end{align*} Where \(c_{1}\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = -x \,{\mathrm e}^{-y}-\left (y +1\right ) {\mathrm e}^{-y}+ c_{1} \] 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 \[ c_{1} = -x \,{\mathrm e}^{-y}-\left (y +1\right ) {\mathrm e}^{-y} \] The solution becomes\[ y = -\operatorname {LambertW}\left (c_{1} {\mathrm e}^{-x -1}\right )-x -1 \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\operatorname {LambertW}\left (c_{1} {\mathrm e}^{-x -1}\right )-x -1 \\ \end{align*}
Verification of solutions
\[ y = -\operatorname {LambertW}\left (c_{1} {\mathrm e}^{-x -1}\right )-x -1 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } \left (x +y\right )=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{x +y} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear <- 1st order linear successful <- inverse linear successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 21
dsolve((x+y(x))*diff(y(x),x)-1=0,y(x), singsol=all)
\[ y \left (x \right ) = -\operatorname {LambertW}\left (-c_{1} {\mathrm e}^{-x -1}\right )-1-x \]
✓ Solution by Mathematica
Time used: 0.04 (sec). Leaf size: 24
DSolve[(x+y[x])*y'[x]-1==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -W\left (c_1 \left (-e^{-x-1}\right )\right )-x-1 \]