1.2 problem 1.1-2 (b)

1.2.1 Solving as quadrature ode
1.2.2 Maple step by step solution

Internal problem ID [2448]
Internal file name [OUTPUT/1940_Sunday_June_05_2022_02_40_11_AM_83674871/index.tex]

Book: Ordinary Differential Equations, Robert H. Martin, 1983
Section: Problem 1.1-2, page 6
Problem number: 1.1-2 (b).
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]

\[ \boxed {y^{\prime }=t \,{\mathrm e}^{2 t}} \]

1.2.1 Solving as quadrature ode

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*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\frac {t}{2}-\frac {1}{4}\right ) {\mathrm e}^{2 t}+c_{1} \\ \end{align*}

Figure 2: Slope field plot

Verification of solutions

\[ y = \left (\frac {t}{2}-\frac {1}{4}\right ) {\mathrm e}^{2 t}+c_{1} \] Verified OK.

1.2.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }=t \,{\mathrm e}^{2 t} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int y^{\prime }d t =\int t \,{\mathrm e}^{2 t}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 t -1\right ) {\mathrm e}^{2 t}}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {t \,{\mathrm e}^{2 t}}{2}-\frac {{\mathrm e}^{2 t}}{4}+c_{1} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 17

dsolve(diff(y(t),t)=t*exp(2*t),y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {\left (2 t -1\right ) {\mathrm e}^{2 t}}{4}+c_{1} \]

Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 22

DSolve[y'[t]==t*Exp[2*t],y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{4} e^{2 t} (2 t-1)+c_1 \]