problem 1

Internal problem ID [6697]
Internal ﬁle name [OUTPUT/5945_Sunday_June_05_2022_04_03_21_PM_12489767/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. EXERCISES 7.5. Page 315
Problem number: 1.
ODE order: 1.
ODE degree: 1.

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

\[ \boxed {y^{\prime }-3 y=\delta \left (-2+t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}

8.1.1 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) &=-3\\ q(t) &=\delta \left (-2+t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime }-3 y = \delta \left (-2+t \right ) \end {align*}

8.1.2 Solving as laplace ode

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-y \left (0\right )-3 Y \left (s \right ) = {\mathrm e}^{-2 s}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-3 Y \left (s \right ) = {\mathrm e}^{-2 s} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {{\mathrm e}^{-2 s}}{s -3} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {{\mathrm e}^{-2 s}}{s -3}\right )\\ &= \operatorname {Heaviside}\left (-2+t \right ) {\mathrm e}^{3 t -6} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} 0 & t <2 \\ {\mathrm e}^{3 t -6} & 2\le t \end {array}\right . \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} 0 & t <2 \\ {\mathrm e}^{3 t -6} & 2\le t \end {array}\right . \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left \{\begin {array}{cc} 0 & t <2 \\ {\mathrm e}^{3 t -6} & 2\le t \end {array}\right . \] Veriﬁed OK.

8.1.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-3 y=\mathit {Dirac}\left (-2+t \right ), y \left (0\right )=0\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 }=3 y+\mathit {Dirac}\left (-2+t \right ) \\ \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 }-3 y=\mathit {Dirac}\left (-2+t \right ) \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }-3 y\right )=\mu \left (t \right ) \mathit {Dirac}\left (-2+t \right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }-3 y\right )=y^{\prime } \mu \left (t \right )+y \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=-3 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \mu \left (t \right )\right )\right )d t =\int \mu \left (t \right ) \mathit {Dirac}\left (-2+t \right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \mu \left (t \right ) \mathit {Dirac}\left (-2+t \right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) \mathit {Dirac}\left (-2+t \right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{-3 t} \\ {} & {} & y=\frac {\int {\mathrm e}^{-3 t} \mathit {Dirac}\left (-2+t \right )d t +c_{1}}{{\mathrm e}^{-3 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-6}+c_{1}}{{\mathrm e}^{-3 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{3 t} \left (\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{-6}+c_{1} \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{3 t -6} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\mathit {Heaviside}\left (-2+t \right ) {\mathrm e}^{3 t -6} \end {array} \]

Maple trace

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

\[ y \left (t \right ) = \operatorname {Heaviside}\left (t -2\right ) {\mathrm e}^{3 t -6} \]

8.1 problem 1

8.1.1 Existence and uniqueness analysis

8.1.2 Solving as laplace ode

8.1.3 Maple step by step solution