Internal problem ID [15183]
Internal file name [OUTPUT/15184_Tuesday_April_23_2024_04_53_40_PM_53989808/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 13. Basic concepts and definitions. Exercises
page 98
Problem number: 326.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "exact nonlinear second order ode"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _exact, _nonlinear], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {{y^{\prime }}^{2}+y y^{\prime \prime }=1} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left ({y^{\prime }}^{2}+y y^{\prime \prime }\right )d x &= \int 1d x\\ y y^{\prime } = x + c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {x +c_{1}}{y} \end {align*}
Where \(f(x)=x +c_{1}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= x +c_{1} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {x +c_{1} \,d x} \\ \frac {y^{2}}{2}&=\frac {1}{2} x^{2}+c_{1} x +c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-\frac {x^{2}}{2}-c_{1} x -c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-\frac {x^{2}}{2}-c_{1} x -c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-\frac {x^{2}}{2}-c_{1} x -c_{2} = 0 \] Verified OK.
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} p \left (y \right )^{2}+y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = 1 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= -\frac {p^{2}-1}{y p} \end {align*}
Where \(f(y)=-\frac {1}{y}\) and \(g(p)=\frac {p^{2}-1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}-1}{p}} \,dp &= -\frac {1}{y} \,d y \\ \int { \frac {1}{\frac {p^{2}-1}{p}} \,dp} &= \int {-\frac {1}{y} \,d y} \\ \frac {\ln \left (p^{2}-1\right )}{2}&=-\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}-1} &= {\mathrm e}^{-\ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}-1} &= \frac {c_{2}}{y} \end {align*}
Which simplifies to \[ \sqrt {p \left (y \right )^{2}-1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \] The solution is \[ \sqrt {p \left (y \right )^{2}-1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \sqrt {{y^{\prime }}^{2}-1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}}}{y} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}}}{y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {y}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}}}d y &= \int d x \\ \sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}}&=c_{3} +x \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {y}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}}}d y &= \int d x \\ -\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}}&=x +\textit {\_C4} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}} &= c_{3} +x \\ \tag{2} -\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}} &= x +\textit {\_C4} \\ \end{align*}
Verification of solutions
\[ \sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}} = c_{3} +x \] Verified OK.
\[ -\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}+y^{2}} = x +c_{4} \] Verified OK.
Writing the ode as \[ {y^{\prime }}^{2}+y y^{\prime \prime } = 1 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left ({y^{\prime }}^{2}+y y^{\prime \prime }\right )d x &= \int 1d x\\ y y^{\prime } = x +c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {x +c_{1}}{y} \end {align*}
Where \(f(x)=x +c_{1}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= x +c_{1} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {x +c_{1} \,d x} \\ \frac {y^{2}}{2}&=\frac {1}{2} x^{2}+c_{1} x +c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-\frac {x^{2}}{2}-c_{1} x -c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-\frac {x^{2}}{2}-c_{1} x -c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-\frac {x^{2}}{2}-c_{1} x -c_{2} = 0 \] Verified OK.
An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}
Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}
Looking at the the ode given we see that \begin {align*} a_2 &= y\\ a_1 &= y^{\prime }\\ a_0 &= -1 \end {align*}
Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {y\,d y'} + \int {y^{\prime }\,d y} + \int {-1\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} 2 y y^{\prime }-x = c_{1} \end {align*}
Which is now solved In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {\frac {x}{2}+\frac {c_{1}}{2}}{y} \end {align*}
Where \(f(x)=\frac {x}{2}+\frac {c_{1}}{2}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= \frac {x}{2}+\frac {c_{1}}{2} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {\frac {x}{2}+\frac {c_{1}}{2} \,d x} \\ \frac {y^{2}}{2}&=\frac {1}{4} x^{2}+\frac {1}{2} c_{1} x +c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-\frac {x^{2}}{4}-\frac {c_{1} x}{2}-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-\frac {x^{2}}{4}-\frac {c_{1} x}{2}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-\frac {x^{2}}{4}-\frac {c_{1} x}{2}-c_{2} = 0 \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2}+y y^{\prime \prime }=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),y^{\prime \prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (y \right )^{2}+y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {-u \left (y \right )^{2}+1}{y u \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{-u \left (y \right )^{2}+1}=\frac {1}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{-u \left (y \right )^{2}+1}d y =\int \frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (u \left (y \right )-1\right )}{2}-\frac {\ln \left (u \left (y \right )+1\right )}{2}=\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\frac {y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )}, u \left (y \right )=\frac {y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}=\frac {1}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}d x =\int \frac {1}{{\mathrm e}^{c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}}{\left ({\mathrm e}^{c_{1}}\right )^{2}}=\frac {x}{{\mathrm e}^{c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\sqrt {-1+\left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+2 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x +\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{{\mathrm e}^{c_{1}}}, y=-\frac {\sqrt {-1+\left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+2 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x +\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{{\mathrm e}^{c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}=\frac {1}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )}{y \,{\mathrm e}^{c_{1}} \left (y \,{\mathrm e}^{c_{1}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}\right )+1}d x =\int \frac {1}{{\mathrm e}^{c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}+1}}{\left ({\mathrm e}^{c_{1}}\right )^{2}}=\frac {x}{{\mathrm e}^{c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\sqrt {-1+\left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+2 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x +\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{{\mathrm e}^{c_{1}}}, y=-\frac {\sqrt {-1+\left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+2 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x +\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{{\mathrm e}^{c_{1}}}\right \} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying a quadrature <- quadrature successful <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 35
dsolve(diff(y(x),x)^2+y(x)*diff(y(x),x$2)=1,y(x), singsol=all)
\begin{align*} y &= \sqrt {-2 c_{1} x +x^{2}+2 c_{2}} \\ y &= -\sqrt {-2 c_{1} x +x^{2}+2 c_{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.593 (sec). Leaf size: 79
DSolve[y'[x]^2+y[x]*y''[x]==1,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt {(x+c_2){}^2-e^{2 c_1}} \\ y(x)\to \sqrt {(x+c_2){}^2-e^{2 c_1}} \\ y(x)\to -\sqrt {(x+c_2){}^2} \\ y(x)\to \sqrt {(x+c_2){}^2} \\ \end{align*}