Internal problem ID [13505]
Internal file name [OUTPUT/12677_Friday_February_16_2024_12_10_44_AM_3819084/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 13. Higher order equations: Extending first order concepts. Additional exercises
page 259
Problem number: 13.5 (j).
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]]
\[ \boxed {y^{\prime \prime }-y^{\prime } \left (y^{\prime }-2\right )=0} \]
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 )+2\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 -2\right )}d p &= x +c_{1}\\ -\frac {\ln \left (p \right )}{2}+\frac {\ln \left (p -2\right )}{2}&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=-\frac {2}{{\mathrm e}^{2 c_{1} +2 x}-1}\\ &=-\frac {2}{c_{1}^{2} {\mathrm e}^{2 x}-1} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {2}{c_{1}^{2} {\mathrm e}^{2 x}-1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {2}{c_{1}^{2} {\mathrm e}^{2 x}-1}\,\mathop {\mathrm {d}x}}\\ &= 2 x -\ln \left ({\mathrm e}^{2 x}-\frac {1}{c_{1}^{2}}\right )+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 2 x -\ln \left ({\mathrm e}^{2 x}-\frac {1}{c_{1}^{2}}\right )+c_{2} \\ \end{align*}
Verification of solutions
\[ y = 2 x -\ln \left ({\mathrm e}^{2 x}-\frac {1}{c_{1}^{2}}\right )+c_{2} \] 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 )+2\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 -2}d p &= y +c_{1}\\ \ln \left (p -2\right )&=y +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&={\mathrm e}^{y +c_{1}}+2\\ &={\mathrm e}^{y} c_{1} +2 \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 } = {\mathrm e}^{y} c_{1} +2 \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{{\mathrm e}^{y} c_{1} +2}d y &= x +c_{2}\\ \frac {y}{2}-\frac {\ln \left ({\mathrm e}^{y} c_{1} +2\right )}{2}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (-\frac {2}{-1+c_{1} {\mathrm e}^{2 c_{2} +2 x}}\right )+2 c_{2} +2 x \\ \end{align*}
Verification of solutions
\[ y = \ln \left (-\frac {2}{-1+c_{1} {\mathrm e}^{2 c_{2} +2 x}}\right )+2 c_{2} +2 x \] 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) &= 2\\ 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 \(f=2\), then \begin {align*} \int -g d y &= \int 1d y\\ &= y\\ \int -f d x &= \int \left (-2\right )d x\\ &= -2 x \end {align*}
Substituting the above into Eq(6A) gives \[ y^{\prime } = c_{2} {\mathrm e}^{y} {\mathrm e}^{-2 x} \] 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}^{-2 x} \end {align*}
Where \(f(x)=c_{2} {\mathrm e}^{-2 x}\) and \(g(y)={\mathrm e}^{y}\). Integrating both sides gives \begin{align*} \frac {1}{{\mathrm e}^{y}} \,dy &= c_{2} {\mathrm e}^{-2 x} \,d x \\ \int { \frac {1}{{\mathrm e}^{y}} \,dy} &= \int {c_{2} {\mathrm e}^{-2 x} \,d x} \\ -{\mathrm e}^{-y}&=-\frac {c_{2} {\mathrm e}^{-2 x}}{2}+c_{3} \\ \end{align*} The solution is \[ -{\mathrm e}^{-y}+\frac {c_{2} {\mathrm e}^{-2 x}}{2}-c_{3} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} -{\mathrm e}^{-y}+\frac {c_{2} {\mathrm e}^{-2 x}}{2}-c_{3} &= 0 \\ \end{align*}
Verification of solutions
\[ -{\mathrm e}^{-y}+\frac {c_{2} {\mathrm e}^{-2 x}}{2}-c_{3} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\left (-y^{\prime }+2\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\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 )+2\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 )+2\right ) u \left (x \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (-u \left (x \right )+2\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 )+2\right ) u \left (x \right )}d x =\int \left (-1\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (u \left (x \right )-2\right )}{2}+\frac {\ln \left (u \left (x \right )\right )}{2}=-x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {2 \,{\mathrm e}^{2 c_{1} -2 x}}{-1+{\mathrm e}^{2 c_{1} -2 x}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {2 \,{\mathrm e}^{2 c_{1} -2 x}}{-1+{\mathrm e}^{2 c_{1} -2 x}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {2 \,{\mathrm e}^{2 c_{1} -2 x}}{-1+{\mathrm e}^{2 c_{1} -2 x}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {2 \,{\mathrm e}^{2 c_{1} -2 x}}{-1+{\mathrm e}^{2 c_{1} -2 x}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\ln \left (-1+{\mathrm e}^{2 c_{1} -2 x}\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: 0.032 (sec). Leaf size: 20
dsolve(diff(y(x),x$2)=diff(y(x),x)*(diff(y(x),x)-2),y(x), singsol=all)
\[ y \left (x \right ) = \ln \left (2\right )-\ln \left ({\mathrm e}^{-2 x} c_{1} -2 c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 60.076 (sec). Leaf size: 23
DSolve[y''[x]==y'[x]*(y'[x]-2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2-2 \text {arctanh}\left (1+2 e^{2 (x+c_1)}\right ) \]