Internal problem ID [2458]
Internal file name [OUTPUT/1950_Sunday_June_05_2022_02_40_34_AM_63720430/index.tex]
Book: Ordinary Differential Equations, Robert H. Martin, 1983
Section: Problem 1.1-3, page 6
Problem number: 1.1-3 (e).
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]
\[ \boxed {y^{\prime }=\sin \left (t \right )^{2}} \] With initial conditions \begin {align*} \left [y \left (\frac {\pi }{6}\right ) = 3\right ] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(t)y &= q(t) \end {align*}
Where here \begin {align*} p(t) &=0\\ q(t) &=\sin \left (t \right )^{2} \end {align*}
Hence the ode is \begin {align*} y^{\prime } = \sin \left (t \right )^{2} \end {align*}
The domain of \(p(t)=0\) is \[
\{-\infty
Integrating both sides gives \begin {align*} y &= \int { \sin \left (t \right )^{2}\,\mathop {\mathrm {d}t}}\\ &= \frac {t}{2}-\frac {\sin \left (2 t \right )}{4}+c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=\frac {\pi }{6}\) and \(y=3\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 3 = \frac {\pi }{12}-\frac {\sqrt {3}}{8}+c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = -\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3 \end {align*}
Trying the constant \begin {align*} c_{1} = -\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {t}{2}-\frac {\sin \left (2 t \right )}{4}-\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3 \end {align*}
The constant \(c_{1} = -\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {t}{2}-\frac {\sin \left (2 t \right )}{4}-\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3 \\
\end{align*} Verification of solutions
\[
y = \frac {t}{2}-\frac {\sin \left (2 t \right )}{4}-\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\sin \left (t \right )^{2}, y \left (\frac {\pi }{6}\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int y^{\prime }d t =\int \sin \left (t \right )^{2}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=-\frac {\sin \left (t \right ) \cos \left (t \right )}{2}+\frac {t}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {\sin \left (t \right ) \cos \left (t \right )}{2}+\frac {t}{2}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (\frac {\pi }{6}\right )=3 \\ {} & {} & 3=\frac {\pi }{12}-\frac {\sqrt {3}}{8}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {t}{2}-\frac {\sin \left (2 t \right )}{4}-\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {t}{2}-\frac {\sin \left (2 t \right )}{4}-\frac {\pi }{12}+\frac {\sqrt {3}}{8}+3 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 23
\[
y \left (t \right ) = \frac {t}{2}+3-\frac {\pi }{12}+\frac {\sqrt {3}}{8}-\frac {\sin \left (2 t \right )}{4}
\]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 31
\[
y(t)\to \frac {1}{24} \left (3 \left (4 t+\sqrt {3}+24\right )-6 \sin (2 t)-2 \pi \right )
\]
2.5.2 Solving as quadrature ode
2.5.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful`
dsolve([diff(y(t),t)=sin(t)^2,y(1/6*Pi) = 3],y(t), singsol=all)
DSolve[{y'[t]==Sin[t]^2,y[Pi/6]==3},y[t],t,IncludeSingularSolutions -> True]