problem 2(f)

Internal problem ID [6172]
Internal ﬁle name [OUTPUT/5420_Sunday_June_05_2022_03_36_48_PM_64201154/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: 2(f).
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "linear", "separable", "diﬀerentialType", "homogeneousTypeMapleC", "ﬁrst_order_ode_lie_symmetry_lookup"

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

3.16.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=\frac {1}{x}\\ q(x) &=\frac {2}{x} \end {align*}

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

3.16.2 Solving as linear ode

Entering Linear ﬁrst order ODE solver. The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {1}{x}d x} \\ &= 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 {2}{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (x y\right ) &= \left (x\right ) \left (\frac {2}{x}\right )\\ \mathrm {d} \left (x y\right ) &= 2\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} x y &= \int {2\,\mathrm {d} x}\\ x y &= 2 x + c_{1} \end {align*}

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

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

Substituting this in the general solution gives \begin {align*} y&=\frac {2 x -1}{x} \end {align*}

Summary

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

Veriﬁcation of solutions

\[ y = \frac {2 x -1}{x} \] Veriﬁed OK.

3.16.3 Maple step by step solution

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

Maple trace

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

3.16 problem 2(f)

3.16.1 Existence and uniqueness analysis

3.16.2 Solving as linear ode

3.16.3 Maple step by step solution