1.33 problem 36

Internal problem ID [12896]
Internal file name [OUTPUT/11549_Monday_November_06_2023_01_33_17_PM_75412810/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 1.2. page 33
Problem number: 36.
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]

\[ \boxed {y^{\prime }-\frac {1}{2 y+3}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

1.33.1 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}{2 y +3} \end {align*}

The \(y\) domain of \(f(t,y)\) when \(t=0\) is \[ \left \{y <-\frac {3}{2}\boldsymbol {\lor }-\frac {3}{2}

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is \[ \left \{y <-\frac {3}{2}\boldsymbol {\lor }-\frac {3}{2}

1.33.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \left (2 y +3\right )d y &= \int {dt}\\ y \left (y +3\right )&= 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*} 4 = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 4 \end {align*}

Trying the constant \begin {align*} c_{1} = 4 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} y \left (y +3\right ) = 4+t \end {align*}

The constant \(c_{1} = 4\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y \left (y+3\right ) &= 4+t \\ \end{align*}

Verification of solutions

\[ y \left (y+3\right ) = 4+t \] Verified OK.

1.33.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {1}{2 y+3}=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{2 y+3} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } \left (2 y+3\right )=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int y^{\prime } \left (2 y+3\right )d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y^{2}+3 y=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {3}{2}-\frac {\sqrt {9+4 t +4 c_{1}}}{2}, y=-\frac {3}{2}+\frac {\sqrt {9+4 t +4 c_{1}}}{2}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {3}{2}-\frac {\sqrt {9+4 c_{1}}}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {3}{2}+\frac {\sqrt {9+4 c_{1}}}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =4 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =4\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {3}{2}+\frac {\sqrt {25+4 t}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {3}{2}+\frac {\sqrt {25+4 t}}{2} \end {array} \]

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: 0.015 (sec). Leaf size: 15

dsolve([diff(y(t),t)=1/(2*y(t)+3),y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {3}{2}+\frac {\sqrt {25+4 t}}{2} \]

✓ Solution by Mathematica

Time used: 0.005 (sec). Leaf size: 20

DSolve[{y'[t]==1/(2*y[t]+3),{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{2} \left (\sqrt {4 t+25}-3\right ) \]