1.29 problem 30

Internal problem ID [7721]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 30
Date solved : Monday, October 21, 2024 at 03:59:09 PM
CAS classification : [[_homogeneous, `class C`], _rational, _dAlembert]

Solve

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

1.29.1 Solved as first order ode of type dAlembert

Time used: 0.179 (sec)

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

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

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

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

Hence (2) becomes

\begin{align*} -p^{2}+p = \left (2 x p +2 p \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 +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 )^{2}+p \left (x \right )}{2 p \left (x \right ) x +2 p \left (x \right )}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. In canonical form a linear first order is

\begin{align*} p^{\prime }\left (x \right ) + q(x)p \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\frac {1}{2 x +2}\\ p(x) &=\frac {1}{2 x +2} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{2 x +2}d x}\\ &= \sqrt {x +1} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu p\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu p\right ) &= \left (\mu \right ) \left (\frac {1}{2 x +2}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (p \sqrt {x +1}\right ) &= \left (\sqrt {x +1}\right ) \left (\frac {1}{2 x +2}\right ) \\ \mathrm {d} \left (p \sqrt {x +1}\right ) &= \left (\frac {1}{2 \sqrt {x +1}}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} p \sqrt {x +1}&= \int {\frac {1}{2 \sqrt {x +1}} \,dx} \\ &=\sqrt {x +1} + c_1 \end{align*}

Dividing throughout by the integrating factor \(\sqrt {x +1}\) gives the final solution

\[ p \left (x \right ) = \frac {\sqrt {x +1}+c_1}{\sqrt {x +1}} \]

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

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

1.29.2 Maple step by step solution

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

1.29.3 Maple trace

Methods for first order ODEs:
 

1.29.4 Maple dsolve solution

Solving time : 0.023 (sec)
Leaf size : 53

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

\begin{align*} y &= 0 \\ y &= \frac {\left (x +1+\sqrt {\left (x +1\right ) \left (1+c_1 \right )}\right )^{2}}{x +1} \\ y &= \frac {\left (-x -1+\sqrt {\left (x +1\right ) \left (1+c_1 \right )}\right )^{2}}{x +1} \\ \end{align*}

1.29.5 Mathematica DSolve solution

Solving time : 0.065 (sec)
Leaf size : 57

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

\begin{align*} y(x)\to x-c_1 \sqrt {x+1}+1+\frac {c_1{}^2}{4} \\ y(x)\to x+c_1 \sqrt {x+1}+1+\frac {c_1{}^2}{4} \\ y(x)\to 0 \\ \end{align*}