1.2 problem Example 4

Internal problem ID [1645]
Internal file name [OUTPUT/1646_Sunday_June_05_2022_02_25_45_AM_99808398/index.tex]

Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 1.2. Page 6
Problem number: Example 4.
ODE order: 1.
ODE degree: 1.

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

Maple gives the following as the ode type

[_separable]

\[ \boxed {y^{\prime }+y \,{\mathrm e}^{t^{2}}=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 2] \end {align*}

1.2.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &={\mathrm e}^{t^{2}}\\ q(t) &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime }+y \,{\mathrm e}^{t^{2}} = 0 \end {align*}

The domain of \(p(t)={\mathrm e}^{t^{2}}\) is \[ \{-\infty

1.2.2 Solving as linear ode

Entering Linear first order ODE solver. The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int {\mathrm e}^{t^{2}}d t} \\ &= {\mathrm e}^{\frac {\sqrt {\pi }\, \operatorname {erfi}\left (t \right )}{2}} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left ({\mathrm e}^{\frac {\sqrt {\pi }\, \operatorname {erfi}\left (t \right )}{2}} y\right ) &= 0 \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\frac {\sqrt {\pi }\, \operatorname {erfi}\left (t \right )}{2}} y &= c_{1} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\frac {\sqrt {\pi }\, \operatorname {erfi}\left (t \right )}{2}}\) results in \begin {align*} y &= c_{1} {\mathrm e}^{-\frac {\sqrt {\pi }\, \operatorname {erfi}\left (t \right )}{2}} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=1\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = {\mathrm e}^{-\frac {\operatorname {erfi}\left (1\right ) \sqrt {\pi }}{2}} c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 2 \,{\mathrm e}^{\frac {\operatorname {erfi}\left (1\right ) \sqrt {\pi }}{2}} \end {align*}

Trying the constant \begin {align*} c_{1} = 2 \,{\mathrm e}^{\frac {\operatorname {erfi}\left (1\right ) \sqrt {\pi }}{2}} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=2 \,{\mathrm e}^{\frac {\sqrt {\pi }\, \left (\operatorname {erfi}\left (1\right )-\operatorname {erfi}\left (t \right )\right )}{2}} \end {align*}

The constant \(c_{1} = 2 \,{\mathrm e}^{\frac {\operatorname {erfi}\left (1\right ) \sqrt {\pi }}{2}}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 2 \,{\mathrm e}^{\frac {\sqrt {\pi }\, \left (\operatorname {erfi}\left (1\right )-\operatorname {erfi}\left (t \right )\right )}{2}} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 2 \,{\mathrm e}^{\frac {\sqrt {\pi }\, \left (\operatorname {erfi}\left (1\right )-\operatorname {erfi}\left (t \right )\right )}{2}} \] Verified OK.

1.2.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+y \,{\mathrm e}^{t^{2}}=0, y \left (1\right )=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-y \,{\mathrm e}^{t^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-{\mathrm e}^{t^{2}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y}d t =\int -{\mathrm e}^{t^{2}}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-\frac {\sqrt {\pi }\, \mathrm {erfi}\left (t \right )}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{-\frac {\sqrt {\pi }\, \mathrm {erfi}\left (t \right )}{2}+c_{1}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=2 \\ {} & {} & 2={\mathrm e}^{-\frac {\mathrm {erfi}\left (1\right ) \sqrt {\pi }}{2}+c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\mathrm {erfi}\left (1\right ) \sqrt {\pi }}{2}+\ln \left (2\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\mathrm {erfi}\left (1\right ) \sqrt {\pi }}{2}+\ln \left (2\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-\frac {\sqrt {\pi }\, \left (-\mathrm {erfi}\left (1\right )+\mathrm {erfi}\left (t \right )\right )}{2}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-\frac {\sqrt {\pi }\, \left (-\mathrm {erfi}\left (1\right )+\mathrm {erfi}\left (t \right )\right )}{2}} \end {array} \]

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: 0.11 (sec). Leaf size: 22

dsolve([diff(y(t),t)+exp(t^2)*y(t)=0,y(1) = 2],y(t), singsol=all)
 

\[ y \left (t \right ) = 2 \,{\mathrm e}^{\frac {\left (\operatorname {erfi}\left (1\right )-\operatorname {erfi}\left (t \right )\right ) \sqrt {\pi }}{2}} \]

✓ Solution by Mathematica

Time used: 0.066 (sec). Leaf size: 25

DSolve[{y'[t]+Exp[t^2]*y[t]==0,y[1]==2},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to 2 e^{\frac {1}{2} \sqrt {\pi } (\text {erfi}(1)-\text {erfi}(t))} \]