1.14 problem 14
Internal
problem
ID
[8051]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
14
Date
solved
:
Monday, October 21, 2024 at 04:44:50 PM
CAS
classification
:
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_xy]]
Solve
\begin{align*} y^{\prime \prime }+{y^{\prime }}^{2}&=0 \end{align*}
1.14.1 Solved as second order missing y ode
Time used: 0.087 (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} = 0 \end{align*}
Which is now solve for \(p(x)\) as first order ode. Integrating gives
\begin{align*} \int -\frac {1}{p^{2}}d p &= dx\\ \frac {1}{p}&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} -p^{2}&= 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 ) = 0\\ p \left (x \right ) = 0 \end{align*}
Solving for \(p \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} p \left (x \right )&=0\\ p \left (x \right )&=\frac {1}{x +c_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 } = 0 \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 {0\, dx} + c_2 \\ y &= 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 } = \frac {1}{x +c_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 {\frac {1}{x +c_1}\, dx}\\ y &= \ln \left (x +c_1 \right ) + c_3 \end{align*}
Will add steps showing solving for IC soon.
1.14.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+{y^{\prime }}^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )+u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-u \left (x \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}d x =\int \left (-1\right )d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=-x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x +\mathit {C1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x +\mathit {C1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {1}{-x +\mathit {C1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {1}{-x +\mathit {C1}}d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\ln \left (-x +\mathit {C1} \right )+\mathit {C2} \end {array} \]
1.14.3 Maple trace
Methods for second order ODEs:
1.14.4 Maple dsolve solution
Solving time : 0.010
(sec)
Leaf size : 10
dsolve(diff(diff(y(x),x),x)+diff(y(x),x)^2 = 0,
y(x),singsol=all)
\[
y = \ln \left (c_1 x +c_2 \right )
\]
1.14.5 Mathematica DSolve solution
Solving time : 0.193
(sec)
Leaf size : 15
DSolve[{D[y[x],{x,2}]+(D[y[x],x])^2==0,{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to \log (x-c_1)+c_2
\]