Internal problem ID [11217]
Internal file name [OUTPUT/10203_Tuesday_December_06_2022_04_00_14_AM_37147350/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter IV, differential equations of the first order and higher degree than the first.
Article 27. Clairaut equation. Page 56
Problem number: Ex 3.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "dAlembert"
Maple gives the following as the ode type
[[_homogeneous, `class C`], _dAlembert]
\[ \boxed {4 \,{\mathrm e}^{2 y} {y^{\prime }}^{2}+2 \,{\mathrm e}^{2 x} y^{\prime }={\mathrm e}^{2 x}} \]
Let \(p=y^{\prime }\) the ode becomes \begin {align*} 4 \,{\mathrm e}^{2 y} p^{2}+2 \,{\mathrm e}^{2 x} p = {\mathrm e}^{2 x} \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= \frac {\ln \left (-\frac {2 p -1}{4 p^{2}}\right )}{2}+x\tag {1A} \end {align*}
This has the form \begin {align*} y=xf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved. Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= 1\\ g &= -\ln \left (2\right )+\frac {\ln \left (\frac {-2 p +1}{p^{2}}\right )}{2} \end {align*}
Hence (2) becomes \begin {align*} p -1 = \frac {\left (-\frac {2}{p^{2}}-\frac {2 \left (-2 p +1\right )}{p^{3}}\right ) p^{2} p^{\prime }\left (x \right )}{-4 p +2}\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} p -1 = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=1 \end {align*}
Substituting these in (1A) gives \begin {align*} y&=x -\ln \left (2\right )+\frac {i \pi }{2} \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {2 \left (p \left (x \right )-1\right ) \left (-2 p \left (x \right )+1\right )}{\left (-\frac {2}{p \left (x \right )^{2}}-\frac {2 \left (-2 p \left (x \right )+1\right )}{p \left (x \right )^{3}}\right ) p \left (x \right )^{2}}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\). Integrating both sides gives \begin {align*} \int -\frac {1}{\left (2 p -1\right ) p}d p &= x +c_{1}\\ \ln \left (p \right )-\ln \left (p -\frac {1}{2}\right )&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\frac {{\mathrm e}^{x +c_{1}}}{-2+2 \,{\mathrm e}^{x +c_{1}}}\\ &=\frac {{\mathrm e}^{x} c_{1}}{2 \,{\mathrm e}^{x} c_{1} -2} \end {align*}
Substituing the above solution for \(p\) in (2A) gives \begin {align*} y = x -\ln \left (2\right )+\frac {\ln \left (\frac {4 \left (-\frac {{\mathrm e}^{x} c_{1}}{{\mathrm e}^{x} c_{1} -1}+1\right ) \left ({\mathrm e}^{x} c_{1} -1\right )^{2} {\mathrm e}^{-2 x}}{c_{1}^{2}}\right )}{2}\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x -\ln \left (2\right )+\frac {i \pi }{2} \\ \tag{2} y &= x -\ln \left (2\right )+\frac {\ln \left (\frac {4 \left (-\frac {{\mathrm e}^{x} c_{1}}{{\mathrm e}^{x} c_{1} -1}+1\right ) \left ({\mathrm e}^{x} c_{1} -1\right )^{2} {\mathrm e}^{-2 x}}{c_{1}^{2}}\right )}{2} \\ \end{align*}
Verification of solutions
\[ y = x -\ln \left (2\right )+\frac {i \pi }{2} \] Verified OK.
\[ y = x -\ln \left (2\right )+\frac {\ln \left (\frac {4 \left (-\frac {{\mathrm e}^{x} c_{1}}{{\mathrm e}^{x} c_{1} -1}+1\right ) \left ({\mathrm e}^{x} c_{1} -1\right )^{2} {\mathrm e}^{-2 x}}{c_{1}^{2}}\right )}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 \,{\mathrm e}^{2 y} {y^{\prime }}^{2}+2 \,{\mathrm e}^{2 x} y^{\prime }={\mathrm e}^{2 x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-\frac {{\mathrm e}^{2 x}+\sqrt {\left ({\mathrm e}^{2 x}\right )^{2}+4 \,{\mathrm e}^{2 y} {\mathrm e}^{2 x}}}{4 \,{\mathrm e}^{2 y}}, y^{\prime }=\frac {-{\mathrm e}^{2 x}+\sqrt {\left ({\mathrm e}^{2 x}\right )^{2}+4 \,{\mathrm e}^{2 y} {\mathrm e}^{2 x}}}{4 \,{\mathrm e}^{2 y}}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {{\mathrm e}^{2 x}+\sqrt {\left ({\mathrm e}^{2 x}\right )^{2}+4 {\mathrm e}^{2 y} {\mathrm e}^{2 x}}}{4 {\mathrm e}^{2 y}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {-{\mathrm e}^{2 x}+\sqrt {\left ({\mathrm e}^{2 x}\right )^{2}+4 {\mathrm e}^{2 y} {\mathrm e}^{2 x}}}{4 {\mathrm e}^{2 y}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables trying simple symmetries for implicit equations Successful isolation of dy/dx: 2 solutions were found. Trying to solve each resulting ODE. *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous C 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful <- homogeneous successful ------------------- * Tackling next ODE. *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous C 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful <- homogeneous successful`
✓ Solution by Maple
Time used: 1.171 (sec). Leaf size: 87
dsolve(4*exp(2*y(x))*diff(y(x),x)^2+2*exp(2*x)*diff(y(x),x)-exp(2*x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \operatorname {arctanh}\left (\operatorname {RootOf}\left (-1+\left ({\mathrm e}^{4}+4 \,{\mathrm e}^{\operatorname {RootOf}\left (-4 \,{\mathrm e}^{\textit {\_Z}} \sinh \left (-\frac {\textit {\_Z}}{2}+2+c_{1} -x \right )^{2}+{\mathrm e}^{4}\right )}\right ) \textit {\_Z}^{2}\right ) {\mathrm e}^{2}\right )+c_{1} \\ y \left (x \right ) &= -\operatorname {arctanh}\left (\operatorname {RootOf}\left (-1+\left ({\mathrm e}^{4}+4 \,{\mathrm e}^{\operatorname {RootOf}\left (-4 \,{\mathrm e}^{\textit {\_Z}} \sinh \left (-\frac {\textit {\_Z}}{2}+2+c_{1} -x \right )^{2}+{\mathrm e}^{4}\right )}\right ) \textit {\_Z}^{2}\right ) {\mathrm e}^{2}\right )+c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 2.772 (sec). Leaf size: 332
DSolve[4*Exp[2*y[x]]*(y'[x])^2+2*Exp[2*x]*y'[x]-Exp[2*x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} \text {Solve}\left [-\frac {2 e^{-x} \sqrt {4 e^{2 (y(x)+x)}+e^{4 x}} \text {arctanh}\left (\frac {-\sqrt {4 e^{2 y(x)}+e^{2 x}}+e^x+1}{\sqrt {4 e^{2 y(x)}+e^{2 x}}-e^x+1}\right )}{\sqrt {4 e^{2 y(x)}+e^{2 x}}}-\frac {e^{-x} \sqrt {4 e^{2 (y(x)+x)}+e^{4 x}} y(x)}{\sqrt {4 e^{2 y(x)}+e^{2 x}}}+y(x)&=c_1,y(x)\right ] \\ \text {Solve}\left [\frac {2 e^{-x} \sqrt {4 e^{2 (y(x)+x)}+e^{4 x}} \text {arctanh}\left (\frac {-\sqrt {4 e^{2 y(x)}+e^{2 x}}+e^x+1}{\sqrt {4 e^{2 y(x)}+e^{2 x}}-e^x+1}\right )}{\sqrt {4 e^{2 y(x)}+e^{2 x}}}+\frac {e^{-x} \sqrt {4 e^{2 (y(x)+x)}+e^{4 x}} y(x)}{\sqrt {4 e^{2 y(x)}+e^{2 x}}}+y(x)&=c_1,y(x)\right ] \\ y(x)\to \frac {1}{2} \left (\log \left (-\frac {e^{4 x}}{4}\right )-2 x\right ) \\ \end{align*}