3.12 problem 2(b)

Internal problem ID [6168]
Internal file name [OUTPUT/5416_Sunday_June_05_2022_03_36_41_PM_87143776/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(b).
ODE order: 1.
ODE degree: 1.

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

Maple gives the following as the ode type

[_linear]

\[ \boxed {y^{\prime }-2 y x=6 x \,{\mathrm e}^{x^{2}}} \] With initial conditions \begin {align*} [y \left (1\right ) = 1] \end {align*}

3.12.1 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) &=-2 x\\ q(x) &=6 x \,{\mathrm e}^{x^{2}} \end {align*}

Hence the ode is \begin {align*} y^{\prime }-2 y x = 6 x \,{\mathrm e}^{x^{2}} \end {align*}

The domain of \(p(x)=-2 x\) is \[ \{-\infty

3.12.2 Solving as linear ode

Entering Linear first order ODE solver. The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -2 x d x} \\ &= {\mathrm e}^{-x^{2}} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (6 x \,{\mathrm e}^{x^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{-x^{2}} y\right ) &= \left ({\mathrm e}^{-x^{2}}\right ) \left (6 x \,{\mathrm e}^{x^{2}}\right )\\ \mathrm {d} \left ({\mathrm e}^{-x^{2}} y\right ) &= \left (6 x\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{-x^{2}} y &= \int {6 x\,\mathrm {d} x}\\ {\mathrm e}^{-x^{2}} y &= 3 x^{2} + c_{1} \end {align*}

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

which simplifies to \begin {align*} y &= {\mathrm e}^{x^{2}} \left (3 x^{2}+c_{1} \right ) \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 = {\mathrm e} \left (c_{1} +3\right ) \end {align*}

The solutions are \begin {align*} c_{1} = -\left (3 \,{\mathrm e}-1\right ) {\mathrm e}^{-1} \end {align*}

Trying the constant \begin {align*} c_{1} = -\left (3 \,{\mathrm e}-1\right ) {\mathrm e}^{-1} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=3 x^{2} {\mathrm e}^{x^{2}}+{\mathrm e}^{\left (x -1\right ) \left (1+x \right )}-3 \,{\mathrm e}^{x^{2}} \end {align*}

The constant \(c_{1} = -\left (3 \,{\mathrm e}-1\right ) {\mathrm e}^{-1}\) gives valid solution.

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

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

3.12.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-2 y x =6 x \,{\mathrm e}^{x^{2}}, 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 }=2 y x +6 x \,{\mathrm e}^{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 }-2 y x =6 x \,{\mathrm e}^{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 }-2 y x \right )=6 \mu \left (x \right ) x \,{\mathrm e}^{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 }-2 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 )=-2 \mu \left (x \right ) x \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )={\mathrm e}^{-x^{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 6 \mu \left (x \right ) x \,{\mathrm e}^{x^{2}}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int 6 \mu \left (x \right ) x \,{\mathrm e}^{x^{2}}d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int 6 \mu \left (x \right ) x \,{\mathrm e}^{x^{2}}d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )={\mathrm e}^{-x^{2}} \\ {} & {} & y=\frac {\int 6 \,{\mathrm e}^{-x^{2}} x \,{\mathrm e}^{x^{2}}d x +c_{1}}{{\mathrm e}^{-x^{2}}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {3 x^{2}+c_{1}}{{\mathrm e}^{-x^{2}}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{x^{2}} \left (3 x^{2}+c_{1} \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1={\mathrm e} \left (c_{1} +3\right ) \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\frac {3 \,{\mathrm e}-1}{{\mathrm e}} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\frac {3 {\mathrm e}-1}{{\mathrm e}}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (-3+3 x^{2}+{\mathrm e}^{-1}\right ) {\mathrm e}^{x^{2}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (-3+3 x^{2}+{\mathrm e}^{-1}\right ) {\mathrm e}^{x^{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: 0.063 (sec). Leaf size: 18

dsolve([diff(y(x),x)-2*x*y(x)=6*x*exp(x^2),y(1) = 1],y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.058 (sec). Leaf size: 23

DSolve[{y'[x]-2*x*y[x]==6*x*Exp[x^2],{y[1]==1}},y[x],x,IncludeSingularSolutions -> True]
 

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