1.35 problem 38

Internal problem ID [12898]
Internal file name [OUTPUT/11551_Monday_November_06_2023_01_33_18_PM_36492103/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: 38.
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]

\[ \boxed {y^{\prime }-\frac {y^{2}+5}{y}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -2] \end {align*}

1.35.1 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 {y^{2}+5}{y} \end {align*}

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

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

1.35.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {y}{y^{2}+5}d y &= \int {dt}\\ \frac {\ln \left (y^{2}+5\right )}{2}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=-2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \ln \left (3\right ) = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = \ln \left (3\right ) \end {align*}

Trying the constant \begin {align*} c_{1} = \ln \left (3\right ) \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y^{2}+5\right )}{2} = t +\ln \left (3\right ) \end {align*}

The constant \(c_{1} = \ln \left (3\right )\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (y^{2}+5\right )}{2} &= t +\ln \left (3\right ) \\ \end{align*}

Verification of solutions

\[ \frac {\ln \left (y^{2}+5\right )}{2} = t +\ln \left (3\right ) \] Verified OK.

1.35.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {y^{2}+5}{y}=0, y \left (0\right )=-2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+5}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{y^{2}+5}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime } y}{y^{2}+5}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y^{2}+5\right )}{2}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {-5+{\mathrm e}^{2 t +2 c_{1}}}, y=-\sqrt {-5+{\mathrm e}^{2 t +2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-2 \\ {} & {} & -2=\sqrt {-5+{\mathrm e}^{2 c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-2 \\ {} & {} & -2=-\sqrt {-5+{\mathrm e}^{2 c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\ln \left (3\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\ln \left (3\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\sqrt {-5+9 \,{\mathrm e}^{2 t}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\sqrt {-5+9 \,{\mathrm e}^{2 t}} \end {array} \]

Maple trace

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

✓ Solution by Maple

Time used: 0.078 (sec). Leaf size: 16

dsolve([diff(y(t),t)=(y(t)^2+5)/y(t),y(0) = -2],y(t), singsol=all)
 

\[ y \left (t \right ) = -\sqrt {9 \,{\mathrm e}^{2 t}-5} \]

✓ Solution by Mathematica

Time used: 0.01 (sec). Leaf size: 20

DSolve[{y'[t]==(y[t]^2+5)/y[t],{y[0]==-2}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to -\sqrt {9 e^{2 t}-5} \]