3.14 problem 2(d)

Internal problem ID [6170]
Internal file name [OUTPUT/5418_Sunday_June_05_2022_03_36_44_PM_70578026/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(d).
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "linear", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"

Maple gives the following as the ode type

[_linear]

\[ \boxed {y^{\prime }-\frac {y}{x}=x^{2}} \] With initial conditions \begin {align*} [y \left (1\right ) = 3] \end {align*}

3.14.1 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) &=x^{2} \end {align*}

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

The domain of \(p(x)=-\frac {1}{x}\) is \[ \{x <0\boldsymbol {\lor }0

3.14.2 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} \\ &= \frac {1}{x} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (x^{2}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{x}\right ) &= \left (\frac {1}{x}\right ) \left (x^{2}\right )\\ \mathrm {d} \left (\frac {y}{x}\right ) &= x\, \mathrm {d} x \end {align*}

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

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

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

The solutions are \begin {align*} c_{1} = {\frac {5}{2}} \end {align*}

Trying the constant \begin {align*} c_{1} = {\frac {5}{2}} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {1}{2} x^{3}+\frac {5}{2} x \end {align*}

The constant \(c_{1} = {\frac {5}{2}}\) gives valid solution.

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {1}{2} x^{3}+\frac {5}{2} x \] Verified OK.

3.14.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {y}{x}=x^{2}, y \left (1\right )=3\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}+x^{2} \\ \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 {y}{x}=x^{2} \\ \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 {y}{x}\right )=\mu \left (x \right ) x^{2} \\ \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 {y}{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 {\mu \left (x \right )}{x} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=\frac {1}{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 \mu \left (x \right ) x^{2}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int \mu \left (x \right ) x^{2}d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (x \right ) x^{2}d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=\frac {1}{x} \\ {} & {} & y=x \left (\int x d x +c_{1} \right ) \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=x \left (\frac {x^{2}}{2}+c_{1} \right ) \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {x \left (x^{2}+2 c_{1} \right )}{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=\frac {1}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {5}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {5}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {x \left (x^{2}+5\right )}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {x \left (x^{2}+5\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`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 12

dsolve([diff(y(x),x)-y(x)/x=x^2,y(1) = 3],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\left (x^{2}+5\right ) x}{2} \]

✓ Solution by Mathematica

Time used: 0.027 (sec). Leaf size: 15

DSolve[{y'[x]-y[x]/x==x^2,{y[1]==3}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{2} x \left (x^2+5\right ) \]