Internal problem ID [15205]
Internal file name [OUTPUT/15206_Tuesday_April_23_2024_04_54_08_PM_4933972/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 14. Differential equations admitting of
depression of their order. Exercises page 107
Problem number: 348.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y", "second_order_nonlinear_solved_by_mainardi_lioville_method"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }-y^{\prime } \left (y^{\prime }+1\right )=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} p^{\prime }\left (x \right )+\left (-p \left (x \right )-1\right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int \frac {1}{p \left (p +1\right )}d p &= x +c_{1}\\ \ln \left (p \right )-\ln \left (p +1\right )&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=-\frac {{\mathrm e}^{x +c_{1}}}{-1+{\mathrm e}^{x +c_{1}}}\\ &=-\frac {{\mathrm e}^{x} c_{1}}{-1+c_{1} {\mathrm e}^{x}} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {{\mathrm e}^{x} c_{1}}{-1+c_{1} {\mathrm e}^{x}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {{\mathrm e}^{x} c_{1}}{-1+c_{1} {\mathrm e}^{x}}\,\mathop {\mathrm {d}x}}\\ &= -\ln \left (-1+c_{1} {\mathrm e}^{x}\right )+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\ln \left (-1+c_{1} {\mathrm e}^{x}\right )+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -\ln \left (-1+c_{1} {\mathrm e}^{x}\right )+c_{2} \] 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 ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (-p \left (y \right )-1\right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int \frac {1}{p +1}d p &= y +c_{1}\\ \ln \left (p +1\right )&=y +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&={\mathrm e}^{y +c_{1}}-1\\ &=c_{1} {\mathrm e}^{y}-1 \end {align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = c_{1} {\mathrm e}^{y}-1 \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{c_{1} {\mathrm e}^{y}-1}d y &= x +c_{2}\\ -y +\ln \left (c_{1} {\mathrm e}^{y}-1\right )&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\ln \left (-\frac {1}{{\mathrm e}^{x +c_{2}}-c_{1}}\right )\\ &=\ln \left (-\frac {1}{{\mathrm e}^{x} c_{2} -c_{1}}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (-\frac {1}{{\mathrm e}^{x} c_{2} -c_{1}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \ln \left (-\frac {1}{{\mathrm e}^{x} c_{2} -c_{1}}\right ) \] Verified OK.
The ode has the Liouville form given by \begin {align*} y^{\prime \prime }+ f(x) y^{\prime } + g(y) {y^{\prime }}^{2} &= 0 \tag {1A} \end {align*}
Where in this problem \begin {align*} f(x) &= -1\\ g(y) &= -1 \end {align*}
Dividing through by \(y^{\prime }\) then Eq (1A) becomes \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}+ f + g y^{\prime } &= 0 \tag {2A} \end {align*}
But the first term in Eq (2A) can be written as \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}&= \frac {d}{dx} \ln \left ( y^{\prime } \right )\tag {3A} \end {align*}
And the last term in Eq (2A) can be written as \begin {align*} g \frac {dy}{dx}&= \left ( \frac {d}{dy} \int g d y\right ) \frac {dy}{dx} \\ &= \frac {d}{dx} \int g d y\tag {4A} \end {align*}
Substituting (3A,4A) back into (2A) gives \begin {align*} \frac {d}{dx} \ln \left ( y^{\prime } \right ) + \frac {d}{dx} \int g d y &= -f \tag {5A} \end {align*}
Integrating the above w.r.t. \(x\) gives \begin {align*} \ln \left ( y^{\prime } \right ) + \int g d y &= - \int f d x + c_{1} \end {align*}
Where \(c_1\) is arbitrary constant. Taking the exponential of the above gives \begin {align*} y^{\prime } &= c_{2} e^{\int -g d y}\, e^{\int -f d x}\tag {6A} \end {align*}
Where \(c_{2}\) is a new arbitrary constant. But since \(g=-1\) and \(f=-1\), then \begin {align*} \int -g d y &= \int 1d y\\ &= y\\ \int -f d x &= \int 1d x\\ &= x \end {align*}
Substituting the above into Eq(6A) gives \[ y^{\prime } = c_{2} {\mathrm e}^{y} {\mathrm e}^{x} \] Which is now solved as first order separable ode. In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= c_{2} {\mathrm e}^{y} {\mathrm e}^{x} \end {align*}
Where \(f(x)={\mathrm e}^{x} c_{2}\) and \(g(y)={\mathrm e}^{y}\). Integrating both sides gives \begin{align*} \frac {1}{{\mathrm e}^{y}} \,dy &= {\mathrm e}^{x} c_{2} \,d x \\ \int { \frac {1}{{\mathrm e}^{y}} \,dy} &= \int {{\mathrm e}^{x} c_{2} \,d x} \\ -{\mathrm e}^{-y}&={\mathrm e}^{x} c_{2} +c_{3} \\ \end{align*} The solution is \[ -{\mathrm e}^{-y}-{\mathrm e}^{x} c_{2} -c_{3} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} -{\mathrm e}^{-y}-{\mathrm e}^{x} c_{2} -c_{3} &= 0 \\ \end{align*}
Verification of solutions
\[ -{\mathrm e}^{-y}-{\mathrm e}^{x} c_{2} -c_{3} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\left (-y^{\prime }-1\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )+\left (-u \left (x \right )-1\right ) u \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\left (-u \left (x \right )-1\right ) u \left (x \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (-u \left (x \right )-1\right ) u \left (x \right )}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\left (-u \left (x \right )-1\right ) u \left (x \right )}d x =\int \left (-1\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (u \left (x \right )\right )+\ln \left (u \left (x \right )+1\right )=-x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {1}{{\mathrm e}^{-x +c_{1}}-1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {1}{{\mathrm e}^{-x +c_{1}}-1} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {1}{{\mathrm e}^{-x +c_{1}}-1} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {1}{{\mathrm e}^{-x +c_{1}}-1}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\ln \left ({\mathrm e}^{-x +c_{1}}\right )-\ln \left ({\mathrm e}^{-x +c_{1}}-1\right )+c_{2} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 16
dsolve(diff(y(x),x$2)=diff(y(x),x)*(1+diff(y(x),x)),y(x), singsol=all)
\[ y = -\ln \left (-c_{1} {\mathrm e}^{x}-c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 1.619 (sec). Leaf size: 31
DSolve[y''[x]==y'[x]*(1+y'[x]),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-\log \left (-1+e^{x+c_1}\right ) \\ y(x)\to c_2-i \pi \\ \end{align*}