2.1.17 problem 17
Internal
problem
ID
[8500]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
17
Date
solved
:
Sunday, November 10, 2024 at 03:55:11 AM
CAS
classification
:
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_xy]]
Solve
\begin{align*} y^{\prime \prime }+{y^{\prime }}^{2}&=1 \end{align*}
Solved as second order missing y ode
Time used: 0.193 (sec)
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 )+p \left (x \right )^{2}-1 = 0 \end{align*}
Which is now solve for \(p(x)\) as first order ode. Integrating gives
\begin{align*} \int \frac {1}{-p^{2}+1}d p &= dx\\ \operatorname {arctanh}\left (p \right )&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} -p^{2}+1&= 0 \end{align*}
for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} p \left (x \right ) = -1\\ p \left (x \right ) = 1 \end{align*}
Solving for \(p \left (x \right )\) gives
\begin{align*}
p \left (x \right ) &= \tanh \left (x +c_1 \right ) \\
\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 } = -1 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {-1\, dx}\\ y &= -x + c_2 \end{align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = 1 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {1\, dx}\\ y &= x + c_3 \end{align*}
For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = \tanh \left (x +c_1 \right ) \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {\tanh \left (x +c_1 \right )\, dx}\\ y &= \ln \left (\cosh \left (x +c_1 \right )\right ) + c_4 \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= -x +c_2 \\
y &= x +c_3 \\
y &= \ln \left (\cosh \left (x +c_1 \right )\right )+c_4 \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{2}+\frac {d^{2}}{d x^{2}}y \left (x \right )=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (x \right )^{2}+\frac {d}{d x}u \left (x \right )=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=-u \left (x \right )^{2}+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{-u \left (x \right )^{2}+1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{-u \left (x \right )^{2}+1}d x =\int 1d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \mathrm {arctanh}\left (u \left (x \right )\right )=x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tanh \left (x +\mathit {C1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tanh \left (x +\mathit {C1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\tanh \left (x +\mathit {C1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int \tanh \left (x +\mathit {C1} \right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=\ln \left (\cosh \left (x +\mathit {C1} \right )\right )+\mathit {C2} \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)
<- 2nd order, 2 integrating factors of the form mu(x,y) successful`
Maple dsolve solution
Solving time : 0.017 (sec)
Leaf size : 22
dsolve(diff(diff(y(x),x),x)+diff(y(x),x)^2 = 1,
y(x),singsol=all)
\[
y = -x -\ln \left (2\right )+\ln \left (-{\mathrm e}^{2 x} c_{1} +c_{2} \right )
\]
Mathematica DSolve solution
Solving time : 0.307 (sec)
Leaf size : 48
DSolve[{D[y[x],{x,2}]+(D[y[x],x])^2==1,{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to -\frac {1}{2} \log \left (e^{2 x}\right )+\log \left (e^{2 x}+e^{2 c_1}\right )+c_2 \\
y(x)\to \frac {1}{2} \log \left (e^{2 x}\right )+c_2 \\
\end{align*}