Internal problem ID [3386]
Internal file name [OUTPUT/2879_Sunday_June_05_2022_08_45_04_AM_87892249/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 5
Problem number: 129.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first order special form ID 1", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`]]
\[ \boxed {y^{\prime }-{\mathrm e}^{y}=x} \]
Writing the ode as \begin {align*} y^{\prime } &= x +{\mathrm e}^{y}\tag {1} \end {align*}
And using the substitution \(u={\mathrm e}^{-y}\) then \begin {align*} u' &= -y^{\prime } {\mathrm e}^{-y} \end {align*}
The above shows that \begin {align*} y^{\prime } &= -u^{\prime }\left (x \right ) {\mathrm e}^{y}\\ &= -\frac {u^{\prime }\left (x \right )}{u} \end {align*}
Substituting this in (1) gives \begin {align*} -\frac {u^{\prime }\left (x \right )}{u}&=\frac {1}{u}+x \end {align*}
The above simplifies to \begin {align*} -u^{\prime }\left (x \right )&=1+u \left (x \right ) x\\ u^{\prime }\left (x \right )+u \left (x \right ) x&=-1\tag {2} \end {align*}
Now ode (2) is solved for \(u \left (x \right )\)
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} u^{\prime }\left (x \right ) + p(x)u \left (x \right ) &= q(x) \end {align*}
Where here \begin {align*} p(x) &=x\\ q(x) &=-1 \end {align*}
Hence the ode is \begin {align*} u^{\prime }\left (x \right )+u \left (x \right ) x = -1 \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int x d x} \\ &= {\mathrm e}^{\frac {x^{2}}{2}} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu u\right ) &= \left (\mu \right ) \left (-1\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\frac {x^{2}}{2}} u\right ) &= \left ({\mathrm e}^{\frac {x^{2}}{2}}\right ) \left (-1\right )\\ \mathrm {d} \left ({\mathrm e}^{\frac {x^{2}}{2}} u\right ) &= \left (-{\mathrm e}^{\frac {x^{2}}{2}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\frac {x^{2}}{2}} u &= \int {-{\mathrm e}^{\frac {x^{2}}{2}}\,\mathrm {d} x}\\ {\mathrm e}^{\frac {x^{2}}{2}} u &= \frac {i \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}\right )}{2} + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\frac {x^{2}}{2}}\) results in \begin {align*} u \left (x \right ) &= \frac {i {\mathrm e}^{-\frac {x^{2}}{2}} \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}\right )}{2}+c_{1} {\mathrm e}^{-\frac {x^{2}}{2}} \end {align*}
which simplifies to \begin {align*} u \left (x \right ) &= {\mathrm e}^{-\frac {x^{2}}{2}} \left (\frac {i \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}\right )}{2}+c_{1} \right ) \end {align*}
Substituting the solution found for \(u \left (x \right )\) in \(u={\mathrm e}^{-y}\) gives \begin {align*} y&= -\ln \left (u \left (x \right )\right )\\ &= -\ln \left ({\mathrm e}^{-\frac {x^{2}}{2}} \left (\frac {i \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}\right )}{2}+c_{1} \right )\right )\\ &= \ln \left (2\right )-\ln \left ({\mathrm e}^{-\frac {x^{2}}{2}} \left (i \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}\right )+2 c_{1} \right )\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (2\right )-\ln \left ({\mathrm e}^{-\frac {x^{2}}{2}} \left (i \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}\right )+2 c_{1} \right )\right ) \\ \end{align*}
Verification of solutions
\[ y = \ln \left (2\right )-\ln \left ({\mathrm e}^{-\frac {x^{2}}{2}} \left (i \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}\right )+2 c_{1} \right )\right ) \] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=x +{\mathrm e}^{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 first order special form ID 1. 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 |
\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\) |
\(0\) |
\(e^{-\int f_{1}dx}\) |
|
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The above table shows that \begin {align*} \xi \left (x,y\right ) &={\mathrm e}^{-\frac {x^{2}}{2}}\\ \tag {A1} \eta \left (x,y\right ) &=x +{\mathrm e}^{-\frac {x^{2}}{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 {x +{\mathrm e}^{-\frac {x^{2}}{2}}}{{\mathrm e}^{-\frac {x^{2}}{2}}}\\ &= 1+x \,{\mathrm e}^{\frac {x^{2}}{2}} \end {align*}
This is easily solved to give \begin {align*} y = x +{\mathrm e}^{\frac {x^{2}}{2}}+c_{1} \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= -x -{\mathrm e}^{\frac {x^{2}}{2}}+y \end {align*}
And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{{\mathrm e}^{-\frac {x^{2}}{2}}} \end {align*}
Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= -\frac {i \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}\right )}{2} \end {align*}
Where the constant of integration is set to zero as we just need one solution. Since \(S\) has special function, not able to continue.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-{\mathrm e}^{y}=x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=x +{\mathrm e}^{y} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying inverse_Riccati trying an equivalence to an Abel ODE differential order: 1; trying a linearization to 2nd order --- trying a change of variables {x -> y(x), y(x) -> x} differential order: 1; trying a linearization to 2nd order trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 5 trying symmetry patterns for 1st order ODEs -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] <- symmetry pattern of the form [0, F(x)*G(y)] successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 36
dsolve(diff(y(x),x) = x+exp(y(x)),y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{2}}{2}+\ln \left (2\right )-\ln \left (i \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}\right )-2 c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.491 (sec). Leaf size: 40
DSolve[y'[x]==x+Exp[y[x]],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} \left (x^2-2 \log \left (-\sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {x}{\sqrt {2}}\right )-c_1\right )\right ) \]