Internal problem ID [2457]
Internal file name [OUTPUT/1949_Sunday_June_05_2022_02_40_31_AM_54207246/index.tex]
Book: Ordinary Differential Equations, Robert H. Martin, 1983
Section: Problem 1.1-3, page 6
Problem number: 1.1-3 (d).
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]
\[ \boxed {y^{\prime }=t \,{\mathrm e}^{2 t}} \] With initial conditions \begin {align*} [y \left (1\right ) = 5] \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) &=t \,{\mathrm e}^{2 t} \end {align*}
Hence the ode is \begin {align*} y^{\prime } = t \,{\mathrm e}^{2 t} \end {align*}
The domain of \(p(t)=0\) is \[
\{-\infty
Integrating both sides gives \begin {align*} y &= \int { t \,{\mathrm e}^{2 t}\,\mathop {\mathrm {d}t}}\\ &= \left (\frac {t}{2}-\frac {1}{4}\right ) {\mathrm e}^{2 t}+c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=1\) and \(y=5\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 5 = \frac {{\mathrm e}^{2}}{4}+c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = -\frac {{\mathrm e}^{2}}{4}+5 \end {align*}
Trying the constant \begin {align*} c_{1} = -\frac {{\mathrm e}^{2}}{4}+5 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {t \,{\mathrm e}^{2 t}}{2}-\frac {{\mathrm e}^{2 t}}{4}+5-\frac {{\mathrm e}^{2}}{4} \end {align*}
The constant \(c_{1} = -\frac {{\mathrm e}^{2}}{4}+5\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {t \,{\mathrm e}^{2 t}}{2}-\frac {{\mathrm e}^{2 t}}{4}+5-\frac {{\mathrm e}^{2}}{4} \\
\end{align*} Verification of solutions
\[
y = \frac {t \,{\mathrm e}^{2 t}}{2}-\frac {{\mathrm e}^{2 t}}{4}+5-\frac {{\mathrm e}^{2}}{4}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=t \,{\mathrm e}^{2 t}, y \left (1\right )=5\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int y^{\prime }d t =\int t \,{\mathrm e}^{2 t}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 t -1\right ) {\mathrm e}^{2 t}}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {t \,{\mathrm e}^{2 t}}{2}-\frac {{\mathrm e}^{2 t}}{4}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=5 \\ {} & {} & 5=\frac {{\mathrm e}^{2}}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\frac {{\mathrm e}^{2}}{4}+5 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\frac {{\mathrm e}^{2}}{4}+5\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 t -1\right ) {\mathrm e}^{2 t}}{4}-\frac {{\mathrm e}^{2}}{4}+5 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 t -1\right ) {\mathrm e}^{2 t}}{4}-\frac {{\mathrm e}^{2}}{4}+5 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 21
\[
y \left (t \right ) = \frac {\left (2 t -1\right ) {\mathrm e}^{2 t}}{4}+5-\frac {{\mathrm e}^{2}}{4}
\]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 27
\[
y(t)\to \frac {1}{4} \left (e^{2 t} (2 t-1)-e^2+20\right )
\]
2.4.2 Solving as quadrature ode
2.4.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful`
dsolve([diff(y(t),t)=t*exp(2*t),y(1) = 5],y(t), singsol=all)
DSolve[{y'[t]==t*Exp[2*t],y[1]==5},y[t],t,IncludeSingularSolutions -> True]