Internal problem ID [4343]
Internal file name [OUTPUT/3836_Sunday_June_05_2022_11_18_27_AM_73736312/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 37
Problem number: 1152.
ODE order: 1.
ODE degree: 0.
The type(s) of ODE detected by this program : "clairaut"
Maple gives the following as the ode type
[_Clairaut]
\[ \boxed {y^{\prime } \ln \left (y^{\prime }+\sqrt {a +{y^{\prime }}^{2}}\right )-\sqrt {1+{y^{\prime }}^{2}}-x y^{\prime }+y=0} \]
This is Clairaut ODE. It has the form \[ y=x y^{\prime }+g\left (y^{\prime }\right ) \] Where \(g\) is function of \(y'(x)\). Let \(p=y^{\prime }\) the ode becomes \begin {align*} p \ln \left (p +\sqrt {p^{2}+a}\right )-\sqrt {p^{2}+1}-x p +y = 0 \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= -p \ln \left (p +\sqrt {p^{2}+a}\right )+x p +\sqrt {p^{2}+1}\tag {1A} \end {align*}
The above ode is a Clairaut ode which is now solved. We start by replacing \(y^{\prime }\) by \(p\) which gives \begin {align*} y&=-p \ln \left (p +\sqrt {p^{2}+a}\right )+x p +\sqrt {p^{2}+1}\\ &=-p \ln \left (p +\sqrt {p^{2}+a}\right )+x p +\sqrt {p^{2}+1} \end {align*}
Writing the ode as \begin {align*} y&= x p +g \left (p \right ) \end {align*}
We now write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(x\). Hence the above becomes \begin {align*} y = x p +g\tag {1} \end {align*}
Then we see that \begin {align*} g&=-p \ln \left (p +\sqrt {p^{2}+a}\right )+\sqrt {p^{2}+1} \end {align*}
Taking derivative of (1) w.r.t. \(x\) gives \begin {align*} p &=\frac {d}{dx}\left (x p+g\right ) \\ p & =\left ( p+x\frac {dp}{dx}\right ) +\left ( g' \frac {dp}{dx}\right ) \\ p & =p+\left ( x+g'\right ) \frac {dp}{dx}\\ 0 & =\left ( x+g'\right ) \frac {dp}{dx} \end {align*}
Where \(g'\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\). The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p &=c_{1} \end {align*}
Substituting this in (1) gives the general solution as \begin {align*} y = c_{1} x -c_{1} \ln \left (c_{1} +\sqrt {c_{1}^{2}+a}\right )+\sqrt {c_{1}^{2}+1} \end {align*}
The singular solution is found from solving for \(p\) from \begin {align*} x+g'\left ( p\right ) &=0 \end {align*}
And substituting the result back in (1). Since we found above that \(g=-p \ln \left (p +\sqrt {p^{2}+a}\right )+\sqrt {p^{2}+1}\), then the above equation becomes \begin {align*} x+g'\left ( p\right ) &= x -\ln \left (p +\sqrt {p^{2}+a}\right )-\frac {p \left (1+\frac {p}{\sqrt {p^{2}+a}}\right )}{p +\sqrt {p^{2}+a}}+\frac {p}{\sqrt {p^{2}+1}}\\ &= 0 \end {align*}
Solving the above for \(p\) results in \begin {align*} p_{1} &=\text {Expression too large to display} \end {align*}
Substituting the above back in (1) results in \begin {align*} y_{1} &=\text {Expression too large to display} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x -c_{1} \ln \left (c_{1} +\sqrt {c_{1}^{2}+a}\right )+\sqrt {c_{1}^{2}+1} \\ \tag{2} \text {Expression too large to display} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x -c_{1} \ln \left (c_{1} +\sqrt {c_{1}^{2}+a}\right )+\sqrt {c_{1}^{2}+1} \] Verified OK.
\[ \text {Expression too large to display} \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } \ln \left (y^{\prime }+\sqrt {a +{y^{\prime }}^{2}}\right )-\sqrt {1+{y^{\prime }}^{2}}-x y^{\prime }+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} \\ {} & {} & y^{\prime }=\frac {\left ({\mathrm e}^{\mathit {RootOf}\left (\left ({\mathrm e}^{\textit {\_Z}}\right )^{4} \textit {\_Z}^{2}-2 \left ({\mathrm e}^{\textit {\_Z}}\right )^{4} \textit {\_Z} x +x^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+4 y \left ({\mathrm e}^{\textit {\_Z}}\right )^{3} \textit {\_Z} -4 x \left ({\mathrm e}^{\textit {\_Z}}\right )^{3} y-2 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z}^{2} a +4 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z} a x -2 a \,x^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}+4 y^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-4 y \,{\mathrm e}^{\textit {\_Z}} \textit {\_Z} a +4 a x \,{\mathrm e}^{\textit {\_Z}} y-\left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+\textit {\_Z}^{2} a^{2}-2 \textit {\_Z} \,a^{2} x +a^{2} x^{2}+2 a \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-4 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-a^{2}\right )}\right )^{2}-a}{2 \,{\mathrm e}^{\mathit {RootOf}\left (\left ({\mathrm e}^{\textit {\_Z}}\right )^{4} \textit {\_Z}^{2}-2 \left ({\mathrm e}^{\textit {\_Z}}\right )^{4} \textit {\_Z} x +x^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+4 y \left ({\mathrm e}^{\textit {\_Z}}\right )^{3} \textit {\_Z} -4 x \left ({\mathrm e}^{\textit {\_Z}}\right )^{3} y-2 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z}^{2} a +4 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z} a x -2 a \,x^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}+4 y^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-4 y \,{\mathrm e}^{\textit {\_Z}} \textit {\_Z} a +4 a x \,{\mathrm e}^{\textit {\_Z}} y-\left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+\textit {\_Z}^{2} a^{2}-2 \textit {\_Z} \,a^{2} x +a^{2} x^{2}+2 a \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-4 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-a^{2}\right )}} \end {array} \]
✗ Solution by Maple
dsolve(diff(y(x),x)*ln(diff(y(x),x)+sqrt(a+diff(y(x),x)^2))-sqrt(1+diff(y(x),x)^2)-x*diff(y(x),x)+y(x) = 0,y(x), singsol=all)
\[ \text {No solution found} \]
✓ Solution by Mathematica
Time used: 60.03 (sec). Leaf size: 38
DSolve[y'[x]*Log[y'[x]+Sqrt[a+(y'[x])^2]]-Sqrt[1+(y'[x])^2]-x*y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -c_1 \log \left (\sqrt {a+c_1{}^2}+c_1\right )+c_1 x+\sqrt {1+c_1{}^2} \]