Internal problem ID [6159]
Internal file name [OUTPUT/5407_Sunday_June_05_2022_03_36_28_PM_85180385/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.4 First Order Linear Equations.
Page 15
Problem number: 1(c).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_linear]
\[ \boxed {y+y^{\prime }=\frac {1}{{\mathrm e}^{2 x}+1}} \]
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=1\\ q(x) &=\frac {1}{{\mathrm e}^{2 x}+1} \end {align*}
Hence the ode is \begin {align*} y+y^{\prime } = \frac {1}{{\mathrm e}^{2 x}+1} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int 1d x} \\ &= {\mathrm e}^{x} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {1}{{\mathrm e}^{2 x}+1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y \,{\mathrm e}^{x}\right ) &= \left ({\mathrm e}^{x}\right ) \left (\frac {1}{{\mathrm e}^{2 x}+1}\right )\\ \mathrm {d} \left (y \,{\mathrm e}^{x}\right ) &= \left (\frac {{\mathrm e}^{x}}{{\mathrm e}^{2 x}+1}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} y \,{\mathrm e}^{x} &= \int {\frac {{\mathrm e}^{x}}{{\mathrm e}^{2 x}+1}\,\mathrm {d} x}\\ y \,{\mathrm e}^{x} &= \arctan \left ({\mathrm e}^{x}\right ) + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{x}\) results in \begin {align*} y &= {\mathrm e}^{-x} \arctan \left ({\mathrm e}^{x}\right )+c_{1} {\mathrm e}^{-x} \end {align*}
which simplifies to \begin {align*} y &= {\mathrm e}^{-x} \left (\arctan \left ({\mathrm e}^{x}\right )+c_{1} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-x} \left (\arctan \left ({\mathrm e}^{x}\right )+c_{1} \right ) \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-x} \left (\arctan \left ({\mathrm e}^{x}\right )+c_{1} \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y+y^{\prime }=\frac {1}{{\mathrm e}^{2 x}+1} \\ \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 }=-y+\frac {1}{{\mathrm e}^{2 x}+1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y+y^{\prime }=\frac {1}{{\mathrm e}^{2 x}+1} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y+y^{\prime }\right )=\frac {\mu \left (x \right )}{{\mathrm e}^{2 x}+1} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y+y^{\prime }\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=\mu \left (x \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )={\mathrm e}^{x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int \frac {\mu \left (x \right )}{{\mathrm e}^{2 x}+1}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int \frac {\mu \left (x \right )}{{\mathrm e}^{2 x}+1}d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \frac {\mu \left (x \right )}{{\mathrm e}^{2 x}+1}d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )={\mathrm e}^{x} \\ {} & {} & y=\frac {\int \frac {{\mathrm e}^{x}}{{\mathrm e}^{2 x}+1}d x +c_{1}}{{\mathrm e}^{x}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\arctan \left ({\mathrm e}^{x}\right )+c_{1}}{{\mathrm e}^{x}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-x} \left (\arctan \left ({\mathrm e}^{x}\right )+c_{1} \right ) \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 14
dsolve(diff(y(x),x)+y(x)=1/(1+exp(2*x)),y(x), singsol=all)
\[ y \left (x \right ) = \left (\arctan \left ({\mathrm e}^{x}\right )+c_{1} \right ) {\mathrm e}^{-x} \]
✓ Solution by Mathematica
Time used: 0.061 (sec). Leaf size: 18
DSolve[y'[x]+y[x]==1/(1+Exp[2*x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x} \left (\arctan \left (e^x\right )+c_1\right ) \]