problem 33

Internal problem ID [6658]
Internal ﬁle name [OUTPUT/5906_Sunday_June_05_2022_04_01_04_PM_30233262/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. 7.2.2 TRANSFORMS OF DERIVATIVES Page 289
Problem number: 33.
ODE order: 1.
ODE degree: 1.

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

\[ \boxed {y^{\prime }+6 y={\mathrm e}^{4 t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 2] \end {align*}

5.3.1 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*}

Hence the ode is \begin {align*} y^{\prime }+6 y = {\mathrm e}^{4 t} \end {align*}

5.3.2 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 )+6 Y \left (s \right ) = \frac {1}{s -4}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-2+6 Y \left (s \right ) = \frac {1}{s -4} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {-7+2 s}{\left (s -4\right ) \left (s +6\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {19}{10 \left (s +6\right )}+\frac {1}{10 s -40} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {19}{10 \left (s +6\right )}\right ) &= \frac {19 \,{\mathrm e}^{-6 t}}{10}\\ \mathcal {L}^{-1}\left (\frac {1}{10 s -40}\right ) &= \frac {{\mathrm e}^{4 t}}{10} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {{\mathrm e}^{4 t}}{10}+\frac {19 \,{\mathrm e}^{-6 t}}{10} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{4 t}}{10}+\frac {19 \,{\mathrm e}^{-6 t}}{10} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {{\mathrm e}^{4 t}}{10}+\frac {19 \,{\mathrm e}^{-6 t}}{10} \] Veriﬁed OK.

5.3.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+6 y={\mathrm e}^{4 t}, y \left (0\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 }=-6 y+{\mathrm e}^{4 t} \\ \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 }+6 y={\mathrm e}^{4 t} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+6 y\right )=\mu \left (t \right ) {\mathrm e}^{4 t} \\ \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 }+6 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 )=6 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{6 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 ) {\mathrm e}^{4 t}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \mu \left (t \right ) {\mathrm e}^{4 t}d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) {\mathrm e}^{4 t}d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{6 t} \\ {} & {} & y=\frac {\int {\mathrm e}^{6 t} {\mathrm e}^{4 t}d t +c_{1}}{{\mathrm e}^{6 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\frac {{\mathrm e}^{10 t}}{10}+c_{1}}{{\mathrm e}^{6 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{10 t}+10 c_{1} \right ) {\mathrm e}^{-6 t}}{10} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=\frac {1}{10}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {19}{10} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {19}{10}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{10 t}+19\right ) {\mathrm e}^{-6 t}}{10} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{10 t}+19\right ) {\mathrm e}^{-6 t}}{10} \end {array} \]

Maple trace

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

\[ y \left (t \right ) = \frac {\left ({\mathrm e}^{10 t}+19\right ) {\mathrm e}^{-6 t}}{10} \]

5.3 problem 33

5.3.1 Existence and uniqueness analysis

5.3.2 Solving as laplace ode

5.3.3 Maple step by step solution