1.79 problem 78

Internal problem ID [7771]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 78
Date solved : Monday, October 21, 2024 at 04:16:22 PM
CAS classification : [[_homogeneous, `class A`], _dAlembert]

Solve

\begin{align*} \frac {y^{\prime } y}{1+\frac {\sqrt {1+{y^{\prime }}^{2}}}{2}}&=-x \end{align*}

With initial conditions

\begin{align*} y \left (0\right )&=3 \end{align*}

1.79.1 Solved as first order dAlembert ode

Time used: 0.622 (sec)

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

\begin{align*} \frac {p y}{1+\frac {\sqrt {p^{2}+1}}{2}} = -x \end{align*}

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

\begin{align*} y &= -\frac {x \left (2+\sqrt {p^{2}+1}\right )}{2 p}\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 &= \frac {-2-\sqrt {p^{2}+1}}{2 p}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{align*} p -\frac {-2-\sqrt {p^{2}+1}}{2 p} = \left (-\frac {x}{2 \sqrt {p^{2}+1}}+\frac {x}{p^{2}}+\frac {x \sqrt {p^{2}+1}}{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 -\frac {-2-\sqrt {p^{2}+1}}{2 p} = 0 \end{align*}

No valid singular solutions found.

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 )-\frac {-2-\sqrt {p \left (x \right )^{2}+1}}{2 p \left (x \right )}}{-\frac {x}{2 \sqrt {p \left (x \right )^{2}+1}}+\frac {x}{p \left (x \right )^{2}}+\frac {x \sqrt {p \left (x \right )^{2}+1}}{2 p \left (x \right )^{2}}}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. The ode \(p^{\prime }\left (x \right ) = \frac {\left (2 p \left (x \right )^{2}+\sqrt {p \left (x \right )^{2}+1}+2\right ) \sqrt {p \left (x \right )^{2}+1}\, p \left (x \right )}{x \left (1+2 \sqrt {p \left (x \right )^{2}+1}\right )}\) is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= \frac {\left (2 p \left (x \right )^{2}+\sqrt {p \left (x \right )^{2}+1}+2\right ) \sqrt {p \left (x \right )^{2}+1}\, p \left (x \right )}{x \left (1+2 \sqrt {p \left (x \right )^{2}+1}\right )}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(p) &= \frac {\left (2 p^{2}+\sqrt {p^{2}+1}+2\right ) \sqrt {p^{2}+1}\, p}{1+2 \sqrt {p^{2}+1}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx}\\ \int { \frac {1+2 \sqrt {p^{2}+1}}{\left (2 p^{2}+\sqrt {p^{2}+1}+2\right ) \sqrt {p^{2}+1}\, p}\,dp} &= \int { \frac {1}{x} \,dx}\\ \ln \left (\frac {p \left (x \right )}{\sqrt {p \left (x \right )^{2}+1}}\right )&=\ln \left (x \right )+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or \(\frac {\left (2 p^{2}+\sqrt {p^{2}+1}+2\right ) \sqrt {p^{2}+1}\, p}{1+2 \sqrt {p^{2}+1}}=0\) for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=0\\ p \left (x \right )&=-i\\ p \left (x \right )&=i \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \ln \left (\frac {p \left (x \right )}{\sqrt {p \left (x \right )^{2}+1}}\right ) = \ln \left (x \right )+c_1\\ p \left (x \right ) = 0\\ p \left (x \right ) = -i\\ p \left (x \right ) = i \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 )&=0\\ p \left (x \right )&=-i\\ p \left (x \right )&=i\\ p \left (x \right )&=\frac {{\mathrm e}^{c_1} x}{\sqrt {1-x^{2} {\mathrm e}^{2 c_1}}}\\ p \left (x \right )&=-\frac {{\mathrm e}^{c_1} x}{\sqrt {1-x^{2} {\mathrm e}^{2 c_1}}} \end{align*}

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

\begin{align*} y = -i x\\ y = i x\\ y = \frac {\left (-2-\sqrt {\frac {{\mathrm e}^{2 c_1} x^{2}}{1-x^{2} {\mathrm e}^{2 c_1}}+1}\right ) \sqrt {1-x^{2} {\mathrm e}^{2 c_1}}\, {\mathrm e}^{-c_1}}{2}\\ y = -\frac {\left (-2-\sqrt {\frac {{\mathrm e}^{2 c_1} x^{2}}{1-x^{2} {\mathrm e}^{2 c_1}}+1}\right ) \sqrt {1-x^{2} {\mathrm e}^{2 c_1}}\, {\mathrm e}^{-c_1}}{2}\\ \end{align*}

Solving for the constant of integration from initial conditions, the solution becomes

\begin{align*} y = -\left (-2-\sqrt {\frac {x^{2}}{-x^{2}+4}+1}\right ) \sqrt {1-\frac {x^{2}}{4}} \end{align*}

Solving for the constant of integration from initial conditions, the solution becomes

\begin{align*} y = -\left (-2-\sqrt {\frac {x^{2}}{-x^{2}+4}+1}\right ) \sqrt {1-\frac {x^{2}}{4}} \end{align*}

The solution

\[ y = -i x \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = i x \]

was found not to satisfy the ode or the IC. Hence it is removed.

Figure 154: Solutions plot

1.79.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {y^{\prime } y}{1+\frac {\sqrt {1+{y^{\prime }}^{2}}}{2}}=-x , y \left (0\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {\left (-4 y+\sqrt {4 y^{2}+3 x^{2}}\right ) x}{4 y^{2}-x^{2}}, y^{\prime }=-\frac {\left (4 y+\sqrt {4 y^{2}+3 x^{2}}\right ) x}{4 y^{2}-x^{2}}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\left (-4 y+\sqrt {4 y^{2}+3 x^{2}}\right ) x}{4 y^{2}-x^{2}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 0 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ {} & \circ & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ {} & \circ & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (4 y+\sqrt {4 y^{2}+3 x^{2}}\right ) x}{4 y^{2}-x^{2}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 0 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ {} & \circ & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ {} & \circ & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{0\right \} \end {array} \]

1.79.3 Maple trace

Methods for first order ODEs:
 

1.79.4 Maple dsolve solution

Solving time : 5.723 (sec)
Leaf size : 29

dsolve([diff(y(x),x)*y(x)/(1+1/2*(1+diff(y(x),x)^2)^(1/2)) = -x, 
       op([y(0) = 3])],y(x),singsol=all)
 

\begin{align*} y &= -3+\sqrt {-x^{2}+36} \\ y &= 1+\sqrt {-x^{2}+4} \\ \end{align*}

1.79.5 Mathematica DSolve solution

Solving time : 0.5 (sec)
Leaf size : 35

DSolve[{D[y[x],x]*y[x]/(1+1/2*Sqrt[1+(D[y[x],x])^2])==-x,y[0]==3}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to \sqrt {4-x^2}+1 \\ y(x)\to \sqrt {36-x^2}-3 \\ \end{align*}