8.22 problem 13.4 (c)

Internal problem ID [13493]
Internal file name [OUTPUT/12665_Friday_February_16_2024_12_04_48_AM_42217179/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: 13.4 (c).
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], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

\[ \boxed {\sin \left (y\right ) y^{\prime \prime }+\cos \left (y\right ) {y^{\prime }}^{2}=0} \]

8.22.1 Solving as second order integrable as is ode

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

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

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

Summary

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

Verification of solutions

\[ y = \pi -\arccos \left (c_{1} c_{2} +c_{1} x \right ) \] Verified OK.

8.22.2 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*} \sin \left (y \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\cos \left (y \right ) p \left (y \right )^{2} = 0 \end {align*}

Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= -\frac {\cos \left (y \right ) p}{\sin \left (y \right )} \end {align*}

Where \(f(y)=-\frac {\cos \left (y \right )}{\sin \left (y \right )}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= -\frac {\cos \left (y \right )}{\sin \left (y \right )} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {-\frac {\cos \left (y \right )}{\sin \left (y \right )} \,d y}\\ \ln \left (p \right )&=-\ln \left (\sin \left (y \right )\right )+c_{1}\\ p&={\mathrm e}^{-\ln \left (\sin \left (y \right )\right )+c_{1}}\\ &=\frac {c_{1}}{\sin \left (y \right )} \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 {c_{1}}{\sin \left (y\right )} \end {align*}

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

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

Summary

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

Verification of solutions

\[ y = \pi -\arccos \left (c_{1} c_{2} +c_{1} x \right ) \] Verified OK.

8.22.3 Solving as type second_order_integrable_as_is (not using ABC version)

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

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

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

Summary

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

Verification of solutions

\[ y = \pi -\arccos \left (c_{1} c_{2} +c_{1} x \right ) \] Verified OK.

8.22.4 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 &= \sin \left (y\right )\\ a_1 &= \cos \left (y\right ) y^{\prime }\\ 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 {\sin \left (y\right )\,d y'} + \int {\cos \left (y\right ) y^{\prime }\,d y} + \int {0\,d x} &= c_{1} \end {align*}

Which results in \begin {align*} 2 y^{\prime } \sin \left (y\right ) = c_{1} \end {align*}

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

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

Summary

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

Verification of solutions

\[ y = \pi -\arccos \left (\frac {1}{2} c_{1} c_{2} +\frac {1}{2} c_{1} x \right ) \] Verified OK.

8.22.5 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \sin \left (y\right ) y^{\prime \prime }+\cos \left (y\right ) {y^{\prime }}^{2}=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} \\ {} & {} & \sin \left (y \right ) u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+\cos \left (y \right ) u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {\cos \left (y \right ) u \left (y \right )}{\sin \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=-\frac {\cos \left (y \right )}{\sin \left (y \right )} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}d y =\int -\frac {\cos \left (y \right )}{\sin \left (y \right )}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=-\ln \left (\sin \left (y \right )\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {{\mathrm e}^{c_{1}}}{\sin \left (y \right )} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {{\mathrm e}^{c_{1}}}{\sin \left (y \right )} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{c_{1}}}{\sin \left (y\right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{c_{1}}}{\sin \left (y\right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } \sin \left (y\right )={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } \sin \left (y\right )d x =\int {\mathrm e}^{c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\cos \left (y\right )={\mathrm e}^{c_{1}} x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\pi -\arccos \left ({\mathrm e}^{c_{1}} x +c_{2} \right ) \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
<- 2nd_order Liouville successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 14

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

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

✓ Solution by Mathematica

Time used: 11.859 (sec). Leaf size: 29

DSolve[Sin[y[x]]*y''[x]+Cos[y[x]]*y'[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -\arccos (-c_1 (x+c_2)) \\ y(x)\to \arccos (-c_1 (x+c_2)) \\ \end{align*}