Internal problem ID [13030]
Internal file name [OUTPUT/11683_Wednesday_November_08_2023_03_28_43_AM_97575103/index.tex]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Review Exercises for chapter 1. page
136
Problem number: 4.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "quadrature"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {y^{\prime }+\sin \left (y\right )^{5}=0} \]
Integrating both sides gives \begin {align*} \int -\frac {1}{\sin \left (y \right )^{5}}d y &= \int {dt}\\ \int _{}^{y}-\frac {1}{\sin \left (\textit {\_a} \right )^{5}}d \textit {\_a}&= t +c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}-\frac {1}{\sin \left (\textit {\_a} \right )^{5}}d \textit {\_a} &= t +c_{1} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}-\frac {1}{\sin \left (\textit {\_a} \right )^{5}}d \textit {\_a} = t +c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\sin \left (y\right )^{5}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\sin \left (y\right )^{5} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sin \left (y\right )^{5}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{\sin \left (y\right )^{5}}d t =\int \left (-1\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \left (-\frac {\csc \left (y\right )^{3}}{4}-\frac {3 \csc \left (y\right )}{8}\right ) \cot \left (y\right )+\frac {3 \ln \left (-\cot \left (y\right )+\csc \left (y\right )\right )}{8}=-t +c_{1} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable <- separable successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 190
dsolve(diff(y(t),t)=-sin(y(t))^5,y(t), singsol=all)
\[ y \left (t \right ) = \arctan \left (\frac {2 \,{\mathrm e}^{\operatorname {RootOf}\left ({\mathrm e}^{8 \textit {\_Z}}+8 \,{\mathrm e}^{6 \textit {\_Z}}+64 c_{1} {\mathrm e}^{4 \textit {\_Z}}+24 \textit {\_Z} \,{\mathrm e}^{4 \textit {\_Z}}+64 t \,{\mathrm e}^{4 \textit {\_Z}}-8 \,{\mathrm e}^{2 \textit {\_Z}}-1\right )}}{{\mathrm e}^{2 \operatorname {RootOf}\left ({\mathrm e}^{8 \textit {\_Z}}+8 \,{\mathrm e}^{6 \textit {\_Z}}+64 c_{1} {\mathrm e}^{4 \textit {\_Z}}+24 \textit {\_Z} \,{\mathrm e}^{4 \textit {\_Z}}+64 t \,{\mathrm e}^{4 \textit {\_Z}}-8 \,{\mathrm e}^{2 \textit {\_Z}}-1\right )}+1}, \frac {-{\mathrm e}^{2 \operatorname {RootOf}\left ({\mathrm e}^{8 \textit {\_Z}}+8 \,{\mathrm e}^{6 \textit {\_Z}}+64 c_{1} {\mathrm e}^{4 \textit {\_Z}}+24 \textit {\_Z} \,{\mathrm e}^{4 \textit {\_Z}}+64 t \,{\mathrm e}^{4 \textit {\_Z}}-8 \,{\mathrm e}^{2 \textit {\_Z}}-1\right )}+1}{{\mathrm e}^{2 \operatorname {RootOf}\left ({\mathrm e}^{8 \textit {\_Z}}+8 \,{\mathrm e}^{6 \textit {\_Z}}+64 c_{1} {\mathrm e}^{4 \textit {\_Z}}+24 \textit {\_Z} \,{\mathrm e}^{4 \textit {\_Z}}+64 t \,{\mathrm e}^{4 \textit {\_Z}}-8 \,{\mathrm e}^{2 \textit {\_Z}}-1\right )}+1}\right ) \]
✓ Solution by Mathematica
Time used: 1.165 (sec). Leaf size: 101
DSolve[y'[t]==-Sin[y[t]]^5,y[t],t,IncludeSingularSolutions -> True]
\begin{align*} y(t)\to \text {InverseFunction}\left [\frac {1}{16} \left (-\frac {1}{64} \csc ^4\left (\frac {\text {$\#$1}}{2}\right )-\frac {3}{32} \csc ^2\left (\frac {\text {$\#$1}}{2}\right )+\frac {1}{64} \sec ^4\left (\frac {\text {$\#$1}}{2}\right )+\frac {3}{32} \sec ^2\left (\frac {\text {$\#$1}}{2}\right )+\frac {3}{8} \log \left (\sin \left (\frac {\text {$\#$1}}{2}\right )\right )-\frac {3}{8} \log \left (\cos \left (\frac {\text {$\#$1}}{2}\right )\right )\right )\&\right ]\left [-\frac {t}{16}+c_1\right ] \\ y(t)\to 0 \\ \end{align*}