Internal problem ID [11359]
Internal file name [OUTPUT/10342_Wednesday_May_17_2023_07_49_30_PM_98842774/index.tex]
Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag,
NY. 2015.
Section: Chapter 1, First order differential equations. Section 1.2 Antiderivatives. Exercises page
19
Problem number: 1.
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]
\[ \boxed {x^{\prime }=t \cos \left (t^{2}\right )} \] With initial conditions \begin {align*} [x \left (0\right ) = 1] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime } + p(t)x &= q(t) \end {align*}
Where here \begin {align*} p(t) &=0\\ q(t) &=t \cos \left (t^{2}\right ) \end {align*}
Hence the ode is \begin {align*} x^{\prime } = t \cos \left (t^{2}\right ) \end {align*}
The domain of \(p(t)=0\) is \[
\{-\infty
Integrating both sides gives \begin {align*} x &= \int { t \cos \left (t^{2}\right )\,\mathop {\mathrm {d}t}}\\ &= \frac {\sin \left (t^{2}\right )}{2}+c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(x=1\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 1 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 1 \end {align*}
Trying the constant \begin {align*} c_{1} = 1 \end {align*}
Substituting this in the general solution gives \begin {align*} x&=\frac {\sin \left (t^{2}\right )}{2}+1 \end {align*}
The constant \(c_{1} = 1\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} x &= \frac {\sin \left (t^{2}\right )}{2}+1 \\
\end{align*} Verification of solutions
\[
x = \frac {\sin \left (t^{2}\right )}{2}+1
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }=t \cos \left (t^{2}\right ), x \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int x^{\prime }d t =\int t \cos \left (t^{2}\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & x=\frac {\sin \left (t^{2}\right )}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\frac {\sin \left (t^{2}\right )}{2}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=1 \\ {} & {} & 1=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & x=\frac {\sin \left (t^{2}\right )}{2}+1 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\frac {\sin \left (t^{2}\right )}{2}+1 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 12
\[
x \left (t \right ) = \frac {\sin \left (t^{2}\right )}{2}+1
\]
✓ Solution by Mathematica
Time used: 0.014 (sec). Leaf size: 15
\[
x(t)\to \frac {1}{2} \left (\sin \left (t^2\right )+2\right )
\]
3.1.2 Solving as quadrature ode
3.1.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful`
dsolve([diff(x(t),t)=t*cos(t^2),x(0) = 1],x(t), singsol=all)
DSolve[{x'[t]==t*Cos[t^2],{x[0]==1}},x[t],t,IncludeSingularSolutions -> True]