2.23 problem 16 (viii)

2.23.1 Solving as quadrature ode
2.23.2 Maple step by step solution

Internal problem ID [12921]
Internal file name [OUTPUT/11574_Tuesday_November_07_2023_11_27_16_PM_16944620/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. Exercises section 1.3 page 47
Problem number: 16 (viii).
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]

\[ \boxed {y^{\prime }=t^{2}-2} \]

2.23.1 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { t^{2}-2\,\mathop {\mathrm {d}t}}\\ &= \frac {t \left (t^{2}-6\right )}{3}+c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {t \left (t^{2}-6\right )}{3}+c_{1} \\ \end{align*}

Figure 88: Slope field plot

Verification of solutions

\[ y = \frac {t \left (t^{2}-6\right )}{3}+c_{1} \] Verified OK.

2.23.2 Maple step by step solution

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

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: 14

dsolve(diff(y(t),t)=t^2-2,y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {1}{3} t^{3}-2 t +c_{1} \]

Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 18

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

\[ y(t)\to \frac {t^3}{3}-2 t+c_1 \]