10.1 problem Exercise 35.1, page 504

Internal problem ID [4651]
Internal file name [OUTPUT/4144_Sunday_June_05_2022_12_28_22_PM_55259558/index.tex]

Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 8. Special second order equations. Lesson 35. Independent variable x absent
Problem number: Exercise 35.1, page 504.
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]]

\[ \boxed {y^{\prime \prime }-2 y^{\prime } y=0} \]

10.1.1 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 }-2 y^{\prime } y\right )d x &= 0 \\ -y^{2}+y^{\prime } = c_{1} \end {align*}

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

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

Summary

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

Verification of solutions

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

10.1.2 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 )-2 p \left (y \right ) y = 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 { 2 y\,\mathop {\mathrm {d}y}}\\ &= y^{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 } = y^{2}+c_{1} \end {align*}

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

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

Summary

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

Verification of solutions

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

10.1.3 Solving as type second_order_integrable_as_is (not using ABC version)

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

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

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

Summary

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

Verification of solutions

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

10.1.4 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 &= -2 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 {-2 y\,d y} + \int {0\,d x} &= c_{1} \end {align*}

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

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

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

Summary

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

Verification of solutions

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

10.1.5 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-2 y^{\prime } y=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 )-2 u \left (y \right ) y =0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=2 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 2 y d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (y \right )=y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=y^{2}+c_{1} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+c_{1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}+c_{1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{2}+c_{1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\arctan \left (\frac {y}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\tan \left (c_{2} \sqrt {c_{1}}+x \sqrt {c_{1}}\right ) \sqrt {c_{1}} \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)-2*_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: 0.266 (sec). Leaf size: 16

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

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

✓ Solution by Mathematica

Time used: 9.872 (sec). Leaf size: 24

DSolve[y''[x]==2*y[x]*y'[x],y[x],x,IncludeSingularSolutions -> True]
 

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