18.19 problem 19

Internal problem ID [2291]
Internal file name [OUTPUT/2291_Tuesday_February_27_2024_08_24_10_AM_26912678/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 35, page 157
Problem number: 19.
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], _Lagerstrom, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

\[ \boxed {y^{\prime \prime }+y y^{\prime }=0} \]

18.19.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+y y^{\prime }\right )d x &= 0 \\ \frac {y^{2}}{2}+y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {y^{2}}{2}+c_{1}}d y &= x +c_{2}\\ \frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {y \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \\ \end{align*}

Verification of solutions

\[ y = \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \] Verified OK.

18.19.2 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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+y 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 { -y\,\mathop {\mathrm {d}y}}\\ &= -\frac {y^{2}}{2}+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 } = -\frac {y^{2}}{2}+c_{1} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {y^{2}}{2}+c_{1}}d y &= x +c_{2}\\ \frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {y \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \\ \end{align*}

Verification of solutions

\[ y = \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \] Verified OK.

18.19.3 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ y^{\prime \prime }+y y^{\prime } = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+y y^{\prime }\right )d x &= 0 \\ \frac {y^{2}}{2}+y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {y^{2}}{2}+c_{1}}d y &= x +c_{2}\\ \frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {y \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \\ \end{align*}

Verification of solutions

\[ y = \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \] Verified OK.

18.19.4 Solving as exact nonlinear second order ode ode

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 &= 1\\ a_1 &= y\\ 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 {1\,d y'} + \int {y\,d y} + \int {0\,d x} &= c_{1} \end {align*}

Which results in \begin {align*} \frac {y^{2}}{2}+y^{\prime } = c_{1} \end {align*}

Which is now solved Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {y^{2}}{2}+c_{1}}d y &= x +c_{2}\\ \frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {y \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \\ \end{align*}

Verification of solutions

\[ y = \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \] Verified OK.

18.19.5 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+y 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} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+y u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-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 -y d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (y \right )=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-\frac {y^{2}}{2}+c_{1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-\frac {y^{2}}{2}+c_{1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2}\, \mathrm {arctanh}\left (\frac {y \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \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)+_a*_b(_a) = 0, _b(_a), HINT = [[_a, 2*_b]]`   *** Sublevel 2 *** 
   symmetry methods on request 
`, `1st order, trying reduction of order with given symmetries:`[_a, 2*_b]
 

✓ Solution by Maple

Time used: 0.063 (sec). Leaf size: 23

dsolve(diff(y(x),x$2)+y(x)*diff(y(x),x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\tanh \left (\frac {\left (c_{2} +x \right ) \sqrt {2}}{2 c_{1}}\right ) \sqrt {2}}{c_{1}} \]

✓ Solution by Mathematica

Time used: 7.104 (sec). Leaf size: 34

DSolve[y''[x]+y[x]*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \sqrt {2} \sqrt {c_1} \tanh \left (\frac {\sqrt {c_1} (x+c_2)}{\sqrt {2}}\right ) \]