problem 22

Internal problem ID [6669]
Internal ﬁle name [OUTPUT/5917_Sunday_June_05_2022_04_01_32_PM_59360507/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. 7.3.1 TRANSLATION ON THE s-AXIS. Page 297
Problem number: 22.
ODE order: 1.
ODE degree: 1.

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

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

6.2.1 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) &=-1\\ q(t) &=1+t \,{\mathrm e}^{t} \end {align*}

Hence the ode is \begin {align*} y^{\prime }-y = 1+t \,{\mathrm e}^{t} \end {align*}

6.2.2 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 )-Y \left (s \right ) = \frac {1}{s}+\frac {1}{\left (s -1\right )^{2}}\tag {1} \end {align*}

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

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {s^{2}-s +1}{s \left (s -1\right )^{3}} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{s -1}+\frac {1}{\left (s -1\right )^{3}}-\frac {1}{s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{s -1}\right ) &= {\mathrm e}^{t}\\ \mathcal {L}^{-1}\left (\frac {1}{\left (s -1\right )^{3}}\right ) &= \frac {{\mathrm e}^{t} t^{2}}{2}\\ \mathcal {L}^{-1}\left (-\frac {1}{s}\right ) &= -1 \end {align*}

Adding the above results and simplifying gives \[ y=-1+\frac {{\mathrm e}^{t} \left (t^{2}+2\right )}{2} \]

Summary

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

Veriﬁcation of solutions

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

6.2.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y=1+t \,{\mathrm e}^{t}, 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 }=y+1+t \,{\mathrm e}^{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 }-y=1+t \,{\mathrm e}^{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 }-y\right )=\mu \left (t \right ) \left (1+t \,{\mathrm e}^{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 }-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 )=-\mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{-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 ) \left (1+t \,{\mathrm e}^{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 ) \left (1+t \,{\mathrm e}^{t}\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) \left (1+t \,{\mathrm e}^{t}\right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{-t} \\ {} & {} & y=\frac {\int {\mathrm e}^{-t} \left (1+t \,{\mathrm e}^{t}\right )d t +c_{1}}{{\mathrm e}^{-t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\frac {t^{2}}{2}-\frac {1}{{\mathrm e}^{t}}+c_{1}}{{\mathrm e}^{-t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=-1+\frac {\left (t^{2}+2 c_{1} \right ) {\mathrm e}^{t}}{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-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=-1+\frac {{\mathrm e}^{t} t^{2}}{2}+{\mathrm e}^{t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-1+\frac {{\mathrm e}^{t} t^{2}}{2}+{\mathrm e}^{t} \end {array} \]

Maple trace

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

6.2 problem 22

6.2.1 Existence and uniqueness analysis

6.2.2 Solving as laplace ode

6.2.3 Maple step by step solution