Internal problem ID [13480]
Internal file name [OUTPUT/12652_Friday_February_16_2024_12_04_41_AM_3052392/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: 13.2 (c).
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], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y y^{\prime \prime }+{y^{\prime }}^{2}=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y y^{\prime \prime }+{y^{\prime }}^{2}\right )d x &= 0 \\ y y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {y}{c_{1}}d y &= x +c_{2}\\ \frac {y^{2}}{2 c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {2 c_{1} c_{2} +2 c_{1} x}\\ y_2&=-\sqrt {2 c_{1} c_{2} +2 c_{1} x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {2 c_{1} c_{2} +2 c_{1} x} \\ \tag{2} y &= -\sqrt {2 c_{1} c_{2} +2 c_{1} x} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {2 c_{1} c_{2} +2 c_{1} x} \] Verified OK.
\[ y = -\sqrt {2 c_{1} c_{2} +2 c_{1} x} \] 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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+p \left (y \right )^{2} = 0 \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}{y} \end {align*}
Where \(f(y)=-\frac {1}{y}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= -\frac {1}{y} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {-\frac {1}{y} \,d y}\\ \ln \left (p \right )&=-\ln \left (y \right )+c_{1}\\ p&={\mathrm e}^{-\ln \left (y \right )+c_{1}}\\ &=\frac {c_{1}}{y} \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 } = \frac {c_{1}}{y} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {y}{c_{1}}d y &= x +c_{2}\\ \frac {y^{2}}{2 c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {2 c_{1} c_{2} +2 c_{1} x}\\ y_2&=-\sqrt {2 c_{1} c_{2} +2 c_{1} x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {2 c_{1} c_{2} +2 c_{1} x} \\ \tag{2} y &= -\sqrt {2 c_{1} c_{2} +2 c_{1} x} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {2 c_{1} c_{2} +2 c_{1} x} \] Verified OK.
\[ y = -\sqrt {2 c_{1} c_{2} +2 c_{1} x} \] Verified OK.
Writing the ode as \[ y y^{\prime \prime }+{y^{\prime }}^{2} = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y y^{\prime \prime }+{y^{\prime }}^{2}\right )d x &= 0 \\ y y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {y}{c_{1}}d y &= x +c_{2}\\ \frac {y^{2}}{2 c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {2 c_{1} c_{2} +2 c_{1} x}\\ y_2&=-\sqrt {2 c_{1} c_{2} +2 c_{1} x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {2 c_{1} c_{2} +2 c_{1} x} \\ \tag{2} y &= -\sqrt {2 c_{1} c_{2} +2 c_{1} x} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {2 c_{1} c_{2} +2 c_{1} x} \] Verified OK.
\[ y = -\sqrt {2 c_{1} c_{2} +2 c_{1} x} \] 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 &= 0 \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 {0\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} 2 y y^{\prime } = c_{1} \end {align*}
Which is now solved Integrating both sides gives \begin {align*} \int \frac {2 y}{c_{1}}d y &= x +c_{2}\\ \frac {y^{2}}{c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {c_{1} c_{2} +c_{1} x}\\ y_2&=-\sqrt {c_{1} c_{2} +c_{1} x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {c_{1} c_{2} +c_{1} x} \\ \tag{2} y &= -\sqrt {c_{1} c_{2} +c_{1} x} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {c_{1} c_{2} +c_{1} x} \] Verified OK.
\[ y = -\sqrt {c_{1} c_{2} +c_{1} x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (\frac {d}{d x}y^{\prime }\right )+{y^{\prime }}^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\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 ),\frac {d}{d x}y^{\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} \\ {} & {} & y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {u \left (y \right )}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=-\frac {1}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}d y =\int -\frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=-\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {{\mathrm e}^{c_{1}}}{y} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {{\mathrm e}^{c_{1}}}{y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{c_{1}}}{y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{c_{1}}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y y^{\prime }={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y y^{\prime }d x =\int {\mathrm e}^{c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}={\mathrm e}^{c_{1}} x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {2 \,{\mathrm e}^{c_{1}} x +2 c_{2}}, y=-\sqrt {2 \,{\mathrm e}^{c_{1}} x +2 c_{2}}\right \} \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: 0.047 (sec). Leaf size: 33
dsolve(y(x)*diff(y(x),x$2)=-(diff(y(x),x)^2),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \sqrt {2 c_{1} x +2 c_{2}} \\ y \left (x \right ) &= -\sqrt {2 c_{1} x +2 c_{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.181 (sec). Leaf size: 20
DSolve[y[x]*y''[x]==-(y'[x]^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 \sqrt {2 x-c_1} \]