problem Problem 2(i)

Internal problem ID [12316]
Internal ﬁle name [OUTPUT/10969_Monday_October_02_2023_02_47_39_AM_27159909/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page 368
Problem number: Problem 2(i).
ODE order: 1.
ODE degree: 1.

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

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

4.9.1 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) &={\frac {1}{2}}\\ q(t) &=\frac {{\mathrm e}^{-\frac {t}{2}}}{2} \end {align*}

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

4.9.2 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*} 2 s Y \left (s \right )-2 y \left (0\right )+Y \left (s \right ) = \frac {2}{2 s +1}\tag {1} \end {align*}

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

Solving for \(Y(s)\) gives \begin {align*} Y(s) = -\frac {4 s}{\left (2 s +1\right )^{2}} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{2 \left (s +\frac {1}{2}\right )^{2}}-\frac {1}{s +\frac {1}{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{2 \left (s +\frac {1}{2}\right )^{2}}\right ) &= \frac {t \,{\mathrm e}^{-\frac {t}{2}}}{2}\\ \mathcal {L}^{-1}\left (-\frac {1}{s +\frac {1}{2}}\right ) &= -{\mathrm e}^{-\frac {t}{2}} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {\left (-2+t \right ) {\mathrm e}^{-\frac {t}{2}}}{2} \]

Summary

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

Veriﬁcation of solutions

\[ y = \frac {\left (-2+t \right ) {\mathrm e}^{-\frac {t}{2}}}{2} \] Veriﬁed OK.

4.9.3 Maple step by step solution

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

Maple trace

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

\[ y \left (t \right ) = \frac {{\mathrm e}^{-\frac {t}{2}} \left (t -2\right )}{2} \]

4.9 problem Problem 2(i)

4.9.1 Existence and uniqueness analysis

4.9.2 Solving as laplace ode

4.9.3 Maple step by step solution