14.6 problem Ex 6
Internal
problem
ID
[11855]
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
25.
Equations
solvable
for
\(y\).
Page
52
Problem
number
:
Ex
6
Date
solved
:
Friday, October 18, 2024 at 08:15:30 AM
CAS
classification
:
[[_1st_order, _with_linear_symmetries], _dAlembert]
Solve
\begin{align*} {y^{\prime }}^{2}+2 x y^{\prime }-y&=0 \end{align*}
14.6.1 Solved as first order ode of type dAlembert
Time used: 0.105 (sec)
Let \(p=y^{\prime }\) the ode becomes
\begin{align*} p^{2}+2 x p -y = 0 \end{align*}
Solving for \(y\) from the above results in
\begin{align*} y &= p^{2}+2 x p\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 &= 2 p\\ g &= p^{2} \end{align*}
Hence (2) becomes
\begin{align*} -p = \left (2 x +2 p \right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} -p = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=0 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = 0 \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 {p \left (x \right )}{2 x +2 p \left (x \right )}\tag {3} \end{align*}
Inverting the above ode gives
\begin{align*} \frac {d}{d p}x \left (p \right ) = -\frac {2 x \left (p \right )+2 p}{p}\tag {4} \end{align*}
This ODE is now solved for \(x \left (p \right )\). The integrating factor is
\begin{align*} \mu &= {\mathrm e}^{\int \frac {2}{p}d p}\\ \mu &= p^{2}\\ \mu &= p^{2}\tag {5} \end{align*}
Integrating gives
\begin{align*} x \left (p \right )&= \frac {1}{\mu } \left ( \int { \mu \left (-2\right ) \,dp} + c_1\right )\\ &= \frac {1}{\mu } \left (\frac {-\frac {2 p^{3}}{3}+c_1}{p^{2}}+c_1\right ) \\ &= \frac {-\frac {2 p^{3}}{3}+c_1}{p^{2}}\tag {5} \end{align*}
Now we need to eliminate \(p\) between the above solution and (1A). The first method is to solve
for \(p\) from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for \(p\)
from Eq. (5) and substitute the result into (1A).
Eliminating \(p\) from the following two equations
\begin{align*}
x &= \frac {-\frac {2 p^{3}}{3}+c_1}{p^{2}} \\
y &= p^{2}+2 x p \\
\end{align*}
results in
\begin{align*}
p &= \operatorname {RootOf}\left (2 \textit {\_Z}^{3}+3 x \,\textit {\_Z}^{2}-3 c_1 \right ) \\
\end{align*}
Substituting the above into Eq
(1A) and simplifying gives
\begin{align*}
y &= \operatorname {RootOf}\left (2 \textit {\_Z}^{3}+3 x \,\textit {\_Z}^{2}-3 c_1 \right )^{2}+2 x \operatorname {RootOf}\left (2 \textit {\_Z}^{3}+3 x \,\textit {\_Z}^{2}-3 c_1 \right ) \\
\end{align*}
14.6.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{2}+2 x \left (\frac {d}{d x}y \left (x \right )\right )-y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=-x -\sqrt {x^{2}+y \left (x \right )}, \frac {d}{d x}y \left (x \right )=-x +\sqrt {x^{2}+y \left (x \right )}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-x -\sqrt {x^{2}+y \left (x \right )} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-x +\sqrt {x^{2}+y \left (x \right )} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
14.6.3 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 dAlembert
<- dAlembert successful`
14.6.4 Maple dsolve solution
Solving time : 0.046
(sec)
Leaf size : 642
dsolve(diff(y(x),x)^2+2*x*diff(y(x),x)-y(x) = 0,
y(x),singsol=all)
\begin{align*}
y &= \frac {\left (x^{2}-x \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{1}/{3}}+\left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}}\right ) \left (x^{2}+3 x \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{1}/{3}}+\left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}}\right )}{4 \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}}} \\
y &= \frac {\left (i \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}} \sqrt {3}-i \sqrt {3}\, x^{2}+\left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}}+2 x \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{1}/{3}}+x^{2}\right ) \left (i \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}} \sqrt {3}-i \sqrt {3}\, x^{2}+\left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}}-6 x \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{1}/{3}}+x^{2}\right )}{16 \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}}} \\
y &= \frac {\left (i \sqrt {3}\, x^{2}-i \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}} \sqrt {3}+x^{2}+2 x \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{1}/{3}}+\left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}}\right ) \left (i \sqrt {3}\, x^{2}-i \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}} \sqrt {3}+x^{2}-6 x \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{1}/{3}}+\left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}}\right )}{16 \left (-x^{3}+2 \sqrt {3}\, \sqrt {-c_1 \left (x^{3}-3 c_1 \right )}+6 c_1 \right )^{{2}/{3}}} \\
\end{align*}
14.6.5 Mathematica DSolve solution
Solving time : 60.095
(sec)
Leaf size : 931
DSolve[{(D[y[x],x])^2+2*x*D[y[x],x]-y[x]==0,{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to \frac {1}{4} \left (-x^2+\frac {x \left (x^3+8 e^{3 c_1}\right )}{\sqrt [3]{-x^6+20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (-x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}}+\sqrt [3]{-x^6+20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (-x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}\right ) \\
y(x)\to \frac {1}{72} \left (-18 x^2-\frac {9 i \left (\sqrt {3}-i\right ) x \left (x^3+8 e^{3 c_1}\right )}{\sqrt [3]{-x^6+20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (-x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}}+9 i \left (\sqrt {3}+i\right ) \sqrt [3]{-x^6+20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (-x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}\right ) \\
y(x)\to \frac {1}{72} \left (-18 x^2+\frac {9 i \left (\sqrt {3}+i\right ) x \left (x^3+8 e^{3 c_1}\right )}{\sqrt [3]{-x^6+20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (-x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}}-9 \left (1+i \sqrt {3}\right ) \sqrt [3]{-x^6+20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (-x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}\right ) \\
y(x)\to \frac {1}{4} \left (-x^2+\frac {x \left (x^3-8 e^{3 c_1}\right )}{\sqrt [3]{-x^6-20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}}+\sqrt [3]{-x^6-20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}\right ) \\
y(x)\to \frac {1}{72} \left (-18 x^2+\frac {9 \left (1+i \sqrt {3}\right ) x \left (-x^3+8 e^{3 c_1}\right )}{\sqrt [3]{-x^6-20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}}+9 i \left (\sqrt {3}+i\right ) \sqrt [3]{-x^6-20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}\right ) \\
y(x)\to \frac {1}{72} \left (-18 x^2+\frac {9 i \left (\sqrt {3}+i\right ) x \left (x^3-8 e^{3 c_1}\right )}{\sqrt [3]{-x^6-20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}}-9 \left (1+i \sqrt {3}\right ) \sqrt [3]{-x^6-20 e^{3 c_1} x^3+8 \sqrt {e^{3 c_1} \left (x^3+e^{3 c_1}\right ){}^3}+8 e^{6 c_1}}\right ) \\
\end{align*}