2.5.1 problem 2

Internal problem ID [18235]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter IV. Methods of solution: First order equations. section 32. Problems at page 89
Problem number : 2
Date solved : Monday, December 23, 2024 at 09:17:22 PM
CAS classification : [_rational, _dAlembert]

Solve

\begin{align*} x {y^{\prime }}^{2}-y+2 y^{\prime }&=0 \end{align*}

Solved as first order ode of type dAlembert

Time used: 0.145 (sec)

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} x \,p^{2}+2 p -y = 0 \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= x \,p^{2}+2 p \\ \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 &= 2 p \end{align*}

Hence (2) becomes

\begin{align*} -p^{2}+p = \left (2 x p +2\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 +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 {-p \left (x \right )^{2}+p \left (x \right )}{2 p \left (x \right ) x +2}\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 ) p +2}{-p^{2}+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 -1}d p}\\ \mu &= \left (p -1\right )^{2}\\ \mu &= \left (p -1\right )^{2}\tag {5} \end{align*}

Integrating gives

\begin{align*} x \left (p \right )&= \frac {1}{\mu } \left ( \int { \mu \left (-\frac {2}{p \left (p -1\right )}\right ) \,dp} + c_1\right )\\ &= \frac {1}{\mu } \left (\frac {-2 p +2 \ln \left (p \right )+c_1}{\left (p -1\right )^{2}}+c_1\right ) \\ &= \frac {-2 p +2 \ln \left (p \right )+c_1}{\left (p -1\right )^{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 {-2 p +2 \ln \left (p \right )+c_1}{\left (p -1\right )^{2}} \\ y &= x \,p^{2}+2 p \\ \end{align*}

results in

\begin{align*} p &= {\mathrm e}^{\operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )} \\ \end{align*}

Substituting the above into Eq (1A) and simplifying gives

\begin{align*} y &= x \,{\mathrm e}^{2 \operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )}+2 \,{\mathrm e}^{\operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= 0 \\ y &= x +2 \\ y &= x \,{\mathrm e}^{2 \operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )}+2 \,{\mathrm e}^{\operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )} \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x {y^{\prime }}^{2}-y+2 y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {-1+\sqrt {x y+1}}{x}, y^{\prime }=-\frac {1+\sqrt {x y+1}}{x}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {-1+\sqrt {x y+1}}{x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {1+\sqrt {x y+1}}{x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

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 dAlembert 
<- dAlembert successful`
 

Maple dsolve solution

Solving time : 0.030 (sec)
Leaf size : 65

dsolve(x*diff(y(x),x)^2-y(x)+2*diff(y(x),x) = 0, 
       y(x),singsol=all)
 

\[ y = 2 x \,{\mathrm e}^{\operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )}+2 \operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )+c_1 -x \]

Mathematica DSolve solution

Solving time : 10.342 (sec)
Leaf size : 50

DSolve[{x*D[y[x],x]^2-y[x]+2*D[y[x],x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\left \{x=\frac {2 \log (K[1])-2 K[1]}{(K[1]-1)^2}+\frac {c_1}{(K[1]-1)^2},y(x)=x K[1]^2+2 K[1]\right \},\{y(x),K[1]\}\right ] \]