problem 13.8 (iii)

Internal problem ID [13519]
Internal ﬁle name [OUTPUT/12691_Friday_February_16_2024_12_11_00_AM_99094366/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: 13.8 (iii).
ODE order: 2.
ODE degree: 1.

\[ \boxed {y^{\prime \prime }+2 x {y^{\prime }}^{2}=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 0, y^{\prime }\left (1\right ) = 1] \end {align*}

8.48.1 Solving as second order ode missing y ode

This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}

Hence the ode becomes \begin {align*} p^{\prime }\left (x \right )+2 x p \left (x \right )^{2} = 0 \end {align*}

Which is now solve for \(p(x)\) as ﬁrst order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= -2 x \,p^{2} \end {align*}

Where \(f(x)=-2 x\) and \(g(p)=p^{2}\). Integrating both sides gives \begin{align*} \frac {1}{p^{2}} \,dp &= -2 x \,d x \\ \int { \frac {1}{p^{2}} \,dp} &= \int {-2 x \,d x} \\ -\frac {1}{p}&=-x^{2}+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{p \left (x \right )}+x^{2}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -c_{1} = 0 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {x^{2} p -1}{p} = 0 \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*} \frac {x^{2} y^{\prime }-1}{y^{\prime }} = 0 \end {align*}

Integrating both sides gives \begin {align*} y &= \int { \frac {1}{x^{2}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {1}{x}+c_{2} \end {align*}

Initial conditions are used to solve for \(c_{2}\). Substituting \(x=1\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{2} -1 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {-1+x}{x} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-1+x}{x} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {-1+x}{x} \] Veriﬁed OK.

8.48.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y^{\prime }+2 x {y^{\prime }}^{2}=0, y \left (1\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1\right ] \\ \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 )+2 x u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-2 x u \left (x \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}=-2 x \\ \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 -2 x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=-x^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x^{2}+c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x^{2}+c_{1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {1}{-x^{2}+c_{1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {1}{-x^{2}+c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {\mathrm {arctanh}\left (\frac {x}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\mathrm {arctanh}\left (\frac {x}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=-\frac {\mathrm {arctanh}\left (\frac {1}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {1}{c_{1} \left (-\frac {x^{2}}{c_{1}}+1\right )} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1 \\ {} & {} & 1=-\frac {1}{c_{1} \left (-\frac {1}{c_{1}}+1\right )} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \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) 
trying differential order: 2; missing variables 
`, `-> Computing symmetries using: way = 3 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful`

8.48 problem 13.8 (iii)

8.48.1 Solving as second order ode missing y ode

8.48.2 Maple step by step solution