2.1.1 problem 1
Internal
problem
ID
[4020]
Book
:
Applied
Differential
equations,
Newby
Curle.
Van
Nostrand
Reinhold.
1972
Section
:
Examples,
page
35
Problem
number
:
1
Date
solved
:
Sunday, November 10, 2024 at 07:52:18 PM
CAS
classification
:
[_quadrature]
Solve
\begin{align*} y&=y^{\prime }+\frac {{y^{\prime }}^{2}}{2} \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=-1+\sqrt {1+2 y} \\
\tag{2} y^{\prime }&=-1-\sqrt {1+2 y} \\
\end{align*}
Now each of the above is solved
separately.
Solving Eq. (1)
Integrating gives
\begin{align*} \int \frac {1}{-1+\sqrt {1+2 y}}d y &= dx\\ \sqrt {1+2 y}+\ln \left (-1+\sqrt {1+2 y}\right )&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} -1+\sqrt {1+2 y}&= 0 \end{align*}
for \(y\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} y = 0 \end{align*}
Solving for \(y\) gives
\begin{align*}
y &= \frac {\operatorname {LambertW}\left ({\mathrm e}^{-1+x +c_1}\right )^{2}}{2}+\operatorname {LambertW}\left ({\mathrm e}^{-1+x +c_1}\right ) \\
\end{align*}
We now need to find the singular solutions, these are found by finding for
what values \((-1+\sqrt {1+2 y})\) is zero. These give
\begin{align*}
y&=0 \\
\end{align*}
Now we go over each such singular solution and check if it
verifies the ode itself and any initial conditions given. If it does not then the singular solution
will not be used.
The solution \(y = 0\) satisfies the ode and initial conditions.
Solving Eq. (2)
Integrating gives
\begin{align*} \int \frac {1}{-1-\sqrt {1+2 y}}d y &= dx\\ -\sqrt {1+2 y}+\ln \left (\sqrt {1+2 y}+1\right )&= x +c_2 \end{align*}
Solving for \(y\) gives
\begin{align*}
y &= \frac {\operatorname {LambertW}\left (-{\mathrm e}^{-1+x +c_2}\right )^{2}}{2}+\operatorname {LambertW}\left (-{\mathrm e}^{-1+x +c_2}\right ) \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {d}{d x}y \left (x \right )+\frac {\left (\frac {d}{d x}y \left (x \right )\right )^{2}}{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=-1-\sqrt {1+2 y \left (x \right )}, \frac {d}{d x}y \left (x \right )=-1+\sqrt {1+2 y \left (x \right )}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-1-\sqrt {1+2 y \left (x \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{-1-\sqrt {1+2 y \left (x \right )}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{-1-\sqrt {1+2 y \left (x \right )}}d x =\int 1d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y \left (x \right )\right )}{2}-\sqrt {1+2 y \left (x \right )}-\frac {\ln \left (-1+\sqrt {1+2 y \left (x \right )}\right )}{2}+\frac {\ln \left (1+\sqrt {1+2 y \left (x \right )}\right )}{2}=x +\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-1+\sqrt {1+2 y \left (x \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{-1+\sqrt {1+2 y \left (x \right )}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{-1+\sqrt {1+2 y \left (x \right )}}d x =\int 1d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y \left (x \right )\right )}{2}+\sqrt {1+2 y \left (x \right )}+\frac {\ln \left (-1+\sqrt {1+2 y \left (x \right )}\right )}{2}-\frac {\ln \left (1+\sqrt {1+2 y \left (x \right )}\right )}{2}=x +\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\frac {\ln \left (y \left (x \right )\right )}{2}-\sqrt {1+2 y \left (x \right )}-\frac {\ln \left (-1+\sqrt {1+2 y \left (x \right )}\right )}{2}+\frac {\ln \left (1+\sqrt {1+2 y \left (x \right )}\right )}{2}=x +\mathit {C1} , \frac {\ln \left (y \left (x \right )\right )}{2}+\sqrt {1+2 y \left (x \right )}+\frac {\ln \left (-1+\sqrt {1+2 y \left (x \right )}\right )}{2}-\frac {\ln \left (1+\sqrt {1+2 y \left (x \right )}\right )}{2}=x +\mathit {C1} \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
<- differential order: 1; missing x successful`
Maple dsolve solution
Solving time : 0.024 (sec)
Leaf size : 106
dsolve(y(x) = diff(y(x),x)+1/2*diff(y(x),x)^2,
y(x),singsol=all)
\begin{align*}
y \left (x \right ) &= \frac {\operatorname {LambertW}\left (-\sqrt {2}\, {\mathrm e}^{x -1-c_{1}}\right ) \left (\operatorname {LambertW}\left (-\sqrt {2}\, {\mathrm e}^{x -1-c_{1}}\right )+2\right )}{2} \\
y \left (x \right ) &= \frac {{\mathrm e}^{2 \operatorname {RootOf}\left (-\textit {\_Z} -2 x +2 \,{\mathrm e}^{\textit {\_Z}}-2+2 c_{1} -\ln \left (2\right )+\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )}}{2}-{\mathrm e}^{\operatorname {RootOf}\left (-\textit {\_Z} -2 x +2 \,{\mathrm e}^{\textit {\_Z}}-2+2 c_{1} -\ln \left (2\right )+\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )} \\
\end{align*}
Mathematica DSolve solution
Solving time : 22.693 (sec)
Leaf size : 66
DSolve[{y[x]==D[y[x],x]+1/2*(D[y[x],x])^2,{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to \frac {1}{2} W\left (-e^{x-1-c_1}\right ) \left (2+W\left (-e^{x-1-c_1}\right )\right ) \\
y(x)\to \frac {1}{2} W\left (e^{x-1+c_1}\right ) \left (2+W\left (e^{x-1+c_1}\right )\right ) \\
y(x)\to 0 \\
\end{align*}