2.1.29 problem 30
Internal
problem
ID
[8167]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
30
Date
solved
:
Sunday, November 10, 2024 at 03:06:10 AM
CAS
classification
:
[[_homogeneous, `class C`], _rational, _dAlembert]
Solve
\begin{align*} y&=x {y^{\prime }}^{2}+{y^{\prime }}^{2} \end{align*}
Solved as first order ode of type dAlembert
Time used: 0.188 (sec)
Let \(p=y^{\prime }\) the ode becomes
\begin{align*} y = x \,p^{2}+p^{2} \end{align*}
Solving for \(y\) from the above results in
\begin{align*}
\tag{1} y &= x \,p^{2}+p^{2} \\
\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 &= p^{2}\\ g &= p^{2} \end{align*}
Hence (2) becomes
\begin{align*} -p^{2}+p = \left (2 x p +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^{2}+p = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=0\\ p_{2} &=1 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = 0\\ y = x +1 \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}+p \left (x \right )}{2 p \left (x \right ) x +2 p \left (x \right )}\tag {3} \end{align*}
This ODE is now solved for \(p \left (x \right )\). No inversion is needed. In canonical form a linear first order is
\begin{align*} p^{\prime }\left (x \right ) + q(x)p \left (x \right ) &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {1}{2 x +2}\\ p(x) &=\frac {1}{2 x +2} \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{2 x +2}d x}\\ &= \sqrt {x +1} \end{align*}
The ode becomes
\begin{align*}
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu p\right ) &= \mu p \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu p\right ) &= \left (\mu \right ) \left (\frac {1}{2 x +2}\right ) \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (p \sqrt {x +1}\right ) &= \left (\sqrt {x +1}\right ) \left (\frac {1}{2 x +2}\right ) \\
\mathrm {d} \left (p \sqrt {x +1}\right ) &= \left (\frac {1}{2 \sqrt {x +1}}\right )\, \mathrm {d} x \\
\end{align*}
Integrating gives
\begin{align*} p \sqrt {x +1}&= \int {\frac {1}{2 \sqrt {x +1}} \,dx} \\ &=\sqrt {x +1} + c_1 \end{align*}
Dividing throughout by the integrating factor \(\sqrt {x +1}\) gives the final solution
\[ p \left (x \right ) = \frac {\sqrt {x +1}+c_1}{\sqrt {x +1}} \]
Substituing the above
solution for \(p\) in (2A) gives
\begin{align*} y = \frac {x \left (\sqrt {x +1}+c_1 \right )^{2}}{x +1}+\frac {\left (\sqrt {x +1}+c_1 \right )^{2}}{x +1}\\ \end{align*}
Summary of solutions found
\begin{align*}
y &= 0 \\
y &= x +1 \\
y &= \frac {x \left (\sqrt {x +1}+c_1 \right )^{2}}{x +1}+\frac {\left (\sqrt {x +1}+c_1 \right )^{2}}{x +1} \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x \right )=x \left (\frac {d}{d x}y \left (x \right )\right )^{2}+\left (\frac {d}{d x}y \left (x \right )\right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {\sqrt {\left (x +1\right ) y \left (x \right )}}{x +1}, \frac {d}{d x}y \left (x \right )=-\frac {\sqrt {\left (x +1\right ) y \left (x \right )}}{x +1}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {\sqrt {\left (x +1\right ) y \left (x \right )}}{x +1} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {\sqrt {\left (x +1\right ) y \left (x \right )}}{x +1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs:
*** Sublevel 2 ***
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`
Maple dsolve solution
Solving time : 0.048
(sec)
Leaf size : 53
dsolve(y(x) = x*diff(y(x),x)^2+diff(y(x),x)^2,
y(x),singsol=all)
\begin{align*}
y &= 0 \\
y &= \frac {\left (x +1+\sqrt {\left (x +1\right ) \left (c_{1} +1\right )}\right )^{2}}{x +1} \\
y &= \frac {\left (-x -1+\sqrt {\left (x +1\right ) \left (c_{1} +1\right )}\right )^{2}}{x +1} \\
\end{align*}
Mathematica DSolve solution
Solving time : 0.065
(sec)
Leaf size : 57
DSolve[{y[x]==x*(D[y[x],x])^2+(D[y[x],x])^2,{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to x-c_1 \sqrt {x+1}+1+\frac {c_1{}^2}{4} \\
y(x)\to x+c_1 \sqrt {x+1}+1+\frac {c_1{}^2}{4} \\
y(x)\to 0 \\
\end{align*}