1.11 problem 4(c)

1.11.1 Solving as quadrature ode
1.11.2 Maple step by step solution

Internal problem ID [879]
Internal file name [OUTPUT/879_Sunday_June_05_2022_01_53_03_AM_93000578/index.tex]

Book: Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 1, Introduction. Section 1.2 Page 14
Problem number: 4(c).
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]

\[ \boxed {y^{\prime }=\tan \left (x \right )} \] With initial conditions \begin {align*} \left [y \left (\frac {\pi }{4}\right ) = 3\right ] \end {align*}

1.11.1 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { \tan \left (x \right )\,\mathop {\mathrm {d}x}}\\ &= -\ln \left (\cos \left (x \right )\right )+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=\frac {\pi }{4}\) and \(y=3\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 3 = \frac {\ln \left (2\right )}{2}+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -\frac {\ln \left (2\right )}{2}+3 \end {align*}

Trying the constant \begin {align*} c_{1} = -\frac {\ln \left (2\right )}{2}+3 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=-\ln \left (\cos \left (x \right )\right )-\frac {\ln \left (2\right )}{2}+3 \end {align*}

The constant \(c_{1} = -\frac {\ln \left (2\right )}{2}+3\) gives valid solution.

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\ln \left (\cos \left (x \right )\right )-\frac {\ln \left (2\right )}{2}+3 \] Verified OK.

1.11.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\tan \left (x \right ), y \left (\frac {\pi }{4}\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \tan \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=-\ln \left (\cos \left (x \right )\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\ln \left (\cos \left (x \right )\right )+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (\frac {\pi }{4}\right )=3 \\ {} & {} & 3=-\ln \left (\frac {\sqrt {2}}{2}\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\ln \left (\frac {\sqrt {2}}{2}\right )+3 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\ln \left (\frac {\sqrt {2}}{2}\right )+3\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\ln \left (\cos \left (x \right )\right )-\frac {\ln \left (2\right )}{2}+3 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\ln \left (\cos \left (x \right )\right )-\frac {\ln \left (2\right )}{2}+3 \end {array} \]

Maple trace

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

Solution by Maple

Time used: 0.046 (sec). Leaf size: 15

dsolve([diff(y(x),x) = tan(x),y(1/4*Pi) = 3],y(x), singsol=all)
 

\[ y \left (x \right ) = -\ln \left (\cos \left (x \right )\right )+3-\frac {\ln \left (2\right )}{2} \]

Solution by Mathematica

Time used: 0.012 (sec). Leaf size: 18

DSolve[{y'[x] == Tan[x],y[Pi/4]==3},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to -\log (\cos (x))+3-\frac {\log (2)}{2} \]