1.50 problem 50

Internal problem ID [8014]
Book : First order enumerated odes
Section : section 1
Problem number : 50
Date solved : Monday, October 21, 2024 at 04:40:54 PM
CAS classification : [[_homogeneous, `class C`], _dAlembert]

Solve

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

1.50.1 Solved as first order ode of type dAlembert

Time used: 0.164 (sec)

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

\begin{align*} p^{2} = x +y \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 the above for \(p\) results in

\begin{align*} p_{1} &=-1 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = 1-x \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 \(p \left (x \right )\). No inversion is needed. Integrating gives

\begin{align*} \int \frac {2 p}{p +1}d p &= dx\\ 2 p -2 \ln \left (p +1\right )&= x +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {p +1}{2 p}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = -1 \end{align*}

Solving for \(p \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} p \left (x \right )&=-1\\ p \left (x \right )&=-\operatorname {LambertW}\left (-{\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )-1 \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y = 1-x\\ y = {\left (-\operatorname {LambertW}\left (-{\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )-1\right )}^{2}-x\\ \end{align*}

1.50.2 Maple step by step solution

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

1.50.3 Maple trace

Methods for first order ODEs:
 

1.50.4 Maple dsolve solution

Solving time : 0.021 (sec)
Leaf size : 33

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

\[ y = \operatorname {LambertW}\left (-c_1 \,{\mathrm e}^{-\frac {x}{2}-1}\right )^{2}+2 \operatorname {LambertW}\left (-c_1 \,{\mathrm e}^{-\frac {x}{2}-1}\right )-x +1 \]

1.50.5 Mathematica DSolve solution

Solving time : 14.92 (sec)
Leaf size : 100

DSolve[{(D[y[x],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 1-x \\ \end{align*}