26.14 problem 750

Internal problem ID [3991]
Internal file name [OUTPUT/3484_Sunday_June_05_2022_09_25_01_AM_44756729/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 26
Problem number: 750.
ODE order: 1.
ODE degree: 2.

The type(s) of ODE detected by this program : "dAlembert", "first_order_nonlinear_p_but_linear_in_x_y"

Maple gives the following as the ode type

[[_homogeneous, `class C`], _dAlembert]

\[ \boxed {{y^{\prime }}^{2}+y=x} \]

26.14.1 Solving as first order nonlinear p but linear in x y ode

The ode has the form \begin {align*} (y')^{\frac {n}{m}} &= a x + b y + c \tag {1} \end {align*}

Where \(n=2, m=1, a=1 , b=-1 , c=0\). Hence the ode is \begin {align*} (y')^{2} &= x -y \end {align*}

Let \begin {align*} u &= a x + b y + c \end {align*}

Hence \begin {align*} u' &= a + b y' \\ y' &= \frac {u'-a}{b} \end {align*}

Substituting the above in (1) gives \begin {align*} \left (\frac {u'-a}{b} \right )^{\frac {n}{m}} &= u \\ \left (\frac {u'-a}{b} \right )^n &= u^m \end {align*}

Plugging in the above the values for \(n,m,a,b,c\) gives \begin {align*} \left (-u^{\prime }\left (x \right )+1\right )^{2} &= u \end {align*}

Therefore the solutions are \begin {align*} -u^{\prime }\left (x \right )+1&=\sqrt {u}\\ -u^{\prime }\left (x \right )+1&=-\sqrt {u} \end {align*}

Rewriting the above gives \begin {align*} u^{\prime }\left (x \right ) &= -\sqrt {u}+1\\ u^{\prime }\left (x \right ) &= \sqrt {u}+1 \end {align*}

Each of the above is a separable ODE in \(u \left (x \right )\). This results in \begin {align*} \frac {du}{-\sqrt {u}+1} &= dx\\ \frac {du}{\sqrt {u}+1} &= dx \end {align*}

Integrating each of the above solutions gives \begin {align*} \int \frac {du}{-\sqrt {u}+1} &= x + c_{1}\\ \int \frac {du}{\sqrt {u}+1} &= x + c_{1} \end {align*}

But since \begin {align*} u &= a x + b y + c \\ &= x -y \end {align*}

Then the solutions can be written as \begin {align*} \int ^{x -y} \frac {1}{-\sqrt {\tau }+1} \, d\tau &= x + c_{1}\\ \int ^{x -y} \frac {1}{\sqrt {\tau }+1} \, d\tau &= x + c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{x -y}\frac {1}{-\sqrt {\tau }+1}d \tau &= x +c_{1} \\ \tag{2} \int _{}^{x -y}\frac {1}{\sqrt {\tau }+1}d \tau &= x +c_{1} \\ \end{align*}

Verification of solutions

\[ \int _{}^{x -y}\frac {1}{-\sqrt {\tau }+1}d \tau = x +c_{1} \] Verified OK.

\[ \int _{}^{x -y}\frac {1}{\sqrt {\tau }+1}d \tau = x +c_{1} \] Verified OK.

26.14.2 Solving as dAlembert ode

Let \(p=y^{\prime }\) the ode becomes \begin {align*} p^{2}+y = x \end {align*}

Solving for \(y\) from the above results in \begin {align*} y &= -p^{2}+x\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 &= 1\\ g &= -p^{2} \end {align*}

Hence (2) becomes \begin {align*} p -1 = -2 p 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 -1 = 0 \end {align*}

Solving for \(p\) from the above gives \begin {align*} p&=1 \end {align*}

Substituting these in (1A) gives \begin {align*} 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 )-1}{2 p \left (x \right )}\tag {3} \end {align*}

This ODE is now solved for \(p \left (x \right )\). Integrating both sides gives \begin {align*} \int -\frac {2 p}{p -1}d p &= x +c_{1}\\ -2 p -2 \ln \left (p -1\right )&=x +c_{1} \end {align*}

Solving for \(p\) gives these solutions \begin {align*} p_1&=\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_{1}}{2}}\right )+1\\ &=\operatorname {LambertW}\left ({\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{2}} c_{1} \right )+1 \end {align*}

Substituing the above solution for \(p\) in (2A) gives \begin {align*} y = -{\left (\operatorname {LambertW}\left ({\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{2}} c_{1} \right )+1\right )}^{2}+x\\ \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= x -1 \\ \tag{2} y &= -{\left (\operatorname {LambertW}\left ({\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{2}} c_{1} \right )+1\right )}^{2}+x \\ \end{align*}

Verification of solutions

\[ y = x -1 \] Verified OK.

\[ y = -{\left (\operatorname {LambertW}\left ({\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{2}} c_{1} \right )+1\right )}^{2}+x \] Verified OK.

26.14.3 Maple step by step solution

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

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`
 

✓ Solution by Maple

Time used: 0.078 (sec). Leaf size: 31

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

\[ y \left (x \right ) = -\operatorname {LambertW}\left (c_{1} {\mathrm e}^{-1-\frac {x}{2}}\right )^{2}-2 \operatorname {LambertW}\left (c_{1} {\mathrm e}^{-1-\frac {x}{2}}\right )+x -1 \]

✓ Solution by Mathematica

Time used: 27.515 (sec). Leaf size: 98

DSolve[(y'[x])^2==x-y[x],y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -W\left (e^{-\frac {x}{2}-1-\frac {c_1}{2}}\right ){}^2-2 W\left (e^{-\frac {x}{2}-1-\frac {c_1}{2}}\right )+x-1 \\ y(x)\to -W\left (-e^{\frac {1}{2} (-x-2+c_1)}\right ){}^2-2 W\left (-e^{\frac {1}{2} (-x-2+c_1)}\right )+x-1 \\ y(x)\to x-1 \\ \end{align*}