Internal problem ID [3225]
Internal file name [OUTPUT/2717_Sunday_June_05_2022_08_39_11_AM_45878433/index.tex]
Book: Differential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 2. First-Order and Simple Higher-Order Differential Equations. Page
78
Problem number: 83.
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 A`], _rational, _dAlembert]
\[ \boxed {x y^{\prime } \left (y^{\prime }+2\right )-y=0} \]
Let \(p=y^{\prime }\) the ode becomes \begin {align*} x p \left (p +2\right )-y = 0 \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= x p \left (p +2\right )\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 &= p \left (p +2\right )\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p -p \left (p +2\right ) = x \left (2 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 -p \left (p +2\right ) = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=-1\\ p&=0 \end {align*}
Substituting these in (1A) gives \begin {align*} y&=-x\\ 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 )-p \left (x \right ) \left (p \left (x \right )+2\right )}{x \left (2 p \left (x \right )+2\right )}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} p^{\prime }\left (x \right ) + p(x)p \left (x \right ) &= q(x) \end {align*}
Where here \begin {align*} p(x) &=\frac {1}{2 x}\\ q(x) &=0 \end {align*}
Hence the ode is \begin {align*} p^{\prime }\left (x \right )+\frac {p \left (x \right )}{2 x} = 0 \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {1}{2 x}d x} \\ &= \sqrt {x} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu p &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\sqrt {x}\, p\right ) &= 0 \end {align*}
Integrating gives \begin {align*} \sqrt {x}\, p &= c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =\sqrt {x}\) results in \begin {align*} p \left (x \right ) &= \frac {c_{1}}{\sqrt {x}} \end {align*}
Substituing the above solution for \(p\) in (2A) gives \begin {align*} y = \sqrt {x}\, c_{1} \left (\frac {c_{1}}{\sqrt {x}}+2\right )\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x \\ \tag{2} y &= 0 \\ \tag{3} y &= \sqrt {x}\, c_{1} \left (\frac {c_{1}}{\sqrt {x}}+2\right ) \\ \end{align*}
Verification of solutions
\[ y = -x \] Verified OK.
\[ y = 0 \] Verified OK.
\[ y = \sqrt {x}\, c_{1} \left (\frac {c_{1}}{\sqrt {x}}+2\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime } \left (y^{\prime }+2\right )-y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {-x +\sqrt {x^{2}+y x}}{x}, y^{\prime }=-\frac {x +\sqrt {x^{2}+y x}}{x}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {-x +\sqrt {x^{2}+y x}}{x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {x +\sqrt {x^{2}+y x}}{x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
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`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 40
dsolve(x*diff(y(x),x)*(diff(y(x),x)+2)=y(x),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -x \\ y \left (x \right ) &= \frac {\sqrt {c_{1} x}\, \left (\sqrt {c_{1} x}+2 x \right )}{x} \\ y \left (x \right ) &= -2 \sqrt {c_{1} x}+c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.186 (sec). Leaf size: 63
DSolve[x*y'[x]*(y'[x]+2)==y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to e^{c_1}-2 e^{\frac {c_1}{2}} \sqrt {x} \\ y(x)\to 2 e^{-\frac {c_1}{2}} \sqrt {x}+e^{-c_1} \\ y(x)\to 0 \\ y(x)\to -x \\ \end{align*}