4.8 problem 15

Internal problem ID [12948]
Internal file name [OUTPUT/11601_Tuesday_November_07_2023_11_51_52_PM_90524771/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 1.5 page 71
Problem number: 15.
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]

\[ \boxed {y^{\prime }-\frac {1}{\left (y+2\right )^{2}}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

4.8.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= \frac {1}{\left (y +2\right )^{2}} \end {align*}

The \(y\) domain of \(f(t,y)\) when \(t=0\) is \[ \{y <-2\boldsymbol {\lor }-2

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is \[ \{y <-2\boldsymbol {\lor }-2

4.8.2 Solving as quadrature ode

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 9 = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 9 \end {align*}

Trying the constant \begin {align*} c_{1} = 9 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\left (y +2\right )^{3}}{3} = t +9 \end {align*}

The constant \(c_{1} = 9\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\left (y+2\right )^{3}}{3} &= t +9 \\ \end{align*}

Verification of solutions

\[ \frac {\left (y+2\right )^{3}}{3} = t +9 \] Verified OK.

4.8.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {1}{\left (y+2\right )^{2}}=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{\left (y+2\right )^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } \left (y+2\right )^{2}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int y^{\prime } \left (y+2\right )^{2}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\left (y+2\right )^{3}}{3}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\left (3 c_{1} +3 t \right )^{\frac {1}{3}}-2 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=3^{\frac {1}{3}} c_{1}^{\frac {1}{3}}-2 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =9 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =9\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (3 t +27\right )^{\frac {1}{3}}-2 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (3 t +27\right )^{\frac {1}{3}}-2 \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
<- separable successful`
 

✓ Solution by Maple

Time used: 0.031 (sec). Leaf size: 13

dsolve([diff(y(t),t)=1/(y(t)+2)^2,y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = \left (3 t +27\right )^{\frac {1}{3}}-2 \]

✓ Solution by Mathematica

Time used: 0.015 (sec). Leaf size: 20

DSolve[{y'[t]==1/(y[t]+2)^2,{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \sqrt [3]{3} \sqrt [3]{t+9}-2 \]