problem 7

Internal problem ID [4767]
Internal ﬁle name [OUTPUT/4260_Sunday_June_05_2022_12_49_18_PM_82020420/index.tex]

Book: Mathematical Methods in the Physical Sciences. third edition. Mary L. Boas. John Wiley. 2006
Section: Chapter 8, Ordinary diﬀerential equations. Section 3. Linear First-Order Equations. page 403
Problem number: 7.
ODE order: 1.
ODE degree: 1.

\[ \boxed {\left (1+{\mathrm e}^{x}\right ) y^{\prime }+2 \,{\mathrm e}^{x} y=\left (1+{\mathrm e}^{x}\right ) {\mathrm e}^{x}} \]

3.7.1 Solving as linear ode

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) &=\frac {2 \,{\mathrm e}^{x}}{1+{\mathrm e}^{x}}\\ q(x) &={\mathrm e}^{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {2 \,{\mathrm e}^{x} y}{1+{\mathrm e}^{x}} = {\mathrm e}^{x} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {2 \,{\mathrm e}^{x}}{1+{\mathrm e}^{x}}d x} \\ &= \left (1+{\mathrm e}^{x}\right )^{2} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left ({\mathrm e}^{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\left (1+{\mathrm e}^{x}\right )^{2} y\right ) &= \left (\left (1+{\mathrm e}^{x}\right )^{2}\right ) \left ({\mathrm e}^{x}\right )\\ \mathrm {d} \left (\left (1+{\mathrm e}^{x}\right )^{2} y\right ) &= \left ({\mathrm e}^{x} \left (1+{\mathrm e}^{x}\right )^{2}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} \left (1+{\mathrm e}^{x}\right )^{2} y &= \int {{\mathrm e}^{x} \left (1+{\mathrm e}^{x}\right )^{2}\,\mathrm {d} x}\\ \left (1+{\mathrm e}^{x}\right )^{2} y &= \frac {\left (1+{\mathrm e}^{x}\right )^{3}}{3} + c_{1} \end {align*}

Dividing both sides by the integrating factor \(\mu =\left (1+{\mathrm e}^{x}\right )^{2}\) results in \begin {align*} y &= \frac {1}{3}+\frac {{\mathrm e}^{x}}{3}+\frac {c_{1}}{\left (1+{\mathrm e}^{x}\right )^{2}} \end {align*}

Summary

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

Veriﬁcation of solutions

\[ y = \frac {1}{3}+\frac {{\mathrm e}^{x}}{3}+\frac {c_{1}}{\left (1+{\mathrm e}^{x}\right )^{2}} \] Veriﬁed OK.

3.7.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (1+{\mathrm e}^{x}\right ) y^{\prime }+2 \,{\mathrm e}^{x} y=\left (1+{\mathrm e}^{x}\right ) {\mathrm e}^{x} \\ \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 {-2 \,{\mathrm e}^{x} y+\left (1+{\mathrm e}^{x}\right ) {\mathrm e}^{x}}{1+{\mathrm e}^{x}} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} y\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {2 \,{\mathrm e}^{x} y}{1+{\mathrm e}^{x}}+{\mathrm e}^{x} \\ \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^{\prime }+\frac {2 \,{\mathrm e}^{x} y}{1+{\mathrm e}^{x}}={\mathrm e}^{x} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }+\frac {2 \,{\mathrm e}^{x} y}{1+{\mathrm e}^{x}}\right )=\mu \left (x \right ) {\mathrm e}^{x} \\ \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^{\prime }+\frac {2 \,{\mathrm e}^{x} y}{1+{\mathrm e}^{x}}\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 )=\frac {2 \mu \left (x \right ) {\mathrm e}^{x}}{1+{\mathrm e}^{x}} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=\left (1+{\mathrm e}^{x}\right )^{2} \\ \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 \mu \left (x \right ) {\mathrm e}^{x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int \mu \left (x \right ) {\mathrm e}^{x}d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (x \right ) {\mathrm e}^{x}d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=\left (1+{\mathrm e}^{x}\right )^{2} \\ {} & {} & y=\frac {\int {\mathrm e}^{x} \left (1+{\mathrm e}^{x}\right )^{2}d x +c_{1}}{\left (1+{\mathrm e}^{x}\right )^{2}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\frac {\left (1+{\mathrm e}^{x}\right )^{3}}{3}+c_{1}}{\left (1+{\mathrm e}^{x}\right )^{2}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{3 x}+3 \,{\mathrm e}^{2 x}+3 \,{\mathrm e}^{x}+3 c_{1} +1}{3 \left (1+{\mathrm e}^{x}\right )^{2}} \end {array} \]

Maple trace

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

\[ y \left (x \right ) = \frac {{\mathrm e}^{3 x}+3 \,{\mathrm e}^{2 x}+3 \,{\mathrm e}^{x}+3 c_{1}}{3 \left (1+{\mathrm e}^{x}\right )^{2}} \]

\[ y(x)\to \frac {1}{3} \left (e^x+1\right )+\frac {c_1}{\left (e^x+1\right )^2} \]

3.7 problem 7

3.7.1 Solving as linear ode

3.7.2 Maple step by step solution