1.29 problem Problem 37

Internal problem ID [2615]
Internal file name [OUTPUT/2107_Sunday_June_05_2022_02_48_48_AM_91022340/index.tex]

Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 1, First-Order Differential Equations. Section 1.2, Basic Ideas and Terminology. page 21
Problem number: Problem 37.
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]

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

1.29.1 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) &=0\\ q(x) &=x^{2} \ln \left (x \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime } = x^{2} \ln \left (x \right ) \end {align*}

The domain of \(p(x)=0\) is \[ \{-\infty

1.29.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { x^{2} \ln \left (x \right )\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = -\frac {1}{9}+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = {\frac {19}{9}} \end {align*}

Trying the constant \begin {align*} c_{1} = {\frac {19}{9}} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}+\frac {19}{9} \end {align*}

The constant \(c_{1} = {\frac {19}{9}}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}+\frac {19}{9} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}+\frac {19}{9} \] Verified OK.

1.29.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=x^{2} \ln \left (x \right ), y \left (1\right )=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int x^{2} \ln \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=2 \\ {} & {} & 2=-\frac {1}{9}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {19}{9} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {19}{9}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}+\frac {19}{9} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}+\frac {19}{9} \end {array} \]

Maple trace

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

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 18

dsolve([diff(y(x),x)=x^2*ln(x),y(1) = 2],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}+\frac {19}{9} \]

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 23

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

\[ y(x)\to \frac {1}{9} \left (-x^3+3 x^3 \log (x)+19\right ) \]