problem 4(c)

Internal problem ID [6267]
Internal ﬁle name [OUTPUT/5515_Sunday_June_05_2022_03_42_26_PM_82411712/index.tex]

Book: Diﬀerential Equations: Theory, Technique, and Practice by George Simmons, Steven Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a diﬀerential equation. Problems for Review and Discovery. Page 53
Problem number: 4(c).
ODE order: 2.
ODE degree: 1.

8.19.1 Solving as second order ode missing x ode

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 ) = 0 \end {align*}

Which is now solved as ﬁrst order ode for $p(y)$. Integrating both sides gives \begin {align*} p \left (y \right ) &= \int { -\frac {1}{y}\,\mathop {\mathrm {d}y}}\\ &= -\ln \left (y \right )+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 } = -\ln \left (y\right )+c_{1} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{-\ln \left (y \right )+c_{1}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{-\ln \left (\textit {\_a} \right )+c_{1}}d \textit {\_a}&= x +c_{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{-\ln \left (\textit {\_a} \right )+c_{1}}d \textit {\_a} &= x +c_{2} \\ \end{align*}

Veriﬁcation of solutions

\[ \int _{}^{y}\frac {1}{-\ln \left (\textit {\_a} \right )+c_{1}}d \textit {\_a} = x +c_{2} \] Veriﬁed OK.

8.19.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y y^{\prime \prime }+y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \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 ),y^{\prime \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 )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {1}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \left (\frac {d}{d y}u \left (y \right )\right )d y =\int -\frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (y \right )=-\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\ln \left (y\right )+c_{1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\ln \left (y\right )+c_{1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-\ln \left (y\right )+c_{1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-\ln \left (y\right )+c_{1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & {\mathrm e}^{c_{1}} \mathrm {Ei}_{1}\left (-\ln \left (y\right )+c_{1} \right )=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{{\mathrm e}^{\mathit {RootOf}\left (-{\mathrm e}^{c_{1}} \mathrm {Ei}_{1}\left (-\textit {\_Z} +c_{1} \right )+x +c_{2} \right )}\right \} \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 
-> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+_b(_a)/_a = 0, _b(_a), HINT = [[_a, 0]]`   *** Sublevel 2 *** 
   symmetry methods on request 
`, `1st order, trying reduction of order with given symmetries:`[_a, 0]

\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= {\mathrm e}^{\operatorname {RootOf}\left (-{\mathrm e}^{c_{1}} \operatorname {expIntegral}_{1}\left (-\textit {\_Z} +c_{1} \right )+x +c_{2} \right )} \\ \end{align*}

\begin{align*} y(x)\to \text {InverseFunction}\left [-e^{c_1} \operatorname {ExpIntegralEi}(\log (\text {$\#$1})-c_1)\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-e^{-c_1} \operatorname {ExpIntegralEi}(\log (\text {$\#$1})--c_1)\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-e^{c_1} \operatorname {ExpIntegralEi}(\log (\text {$\#$1})-c_1)\&\right ][x+c_2] \\ \end{align*}

8.19 problem 4(c)

8.19.1 Solving as second order ode missing x ode

8.19.2 Maple step by step solution