6.1 problem 1

Internal problem ID [5679]
Internal file name [OUTPUT/4927_Sunday_June_05_2022_03_10_50_PM_62015448/index.tex]

Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 6. Laplace Transforms. Problem set 6.2, page 216
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"

Maple gives the following as the ode type

[[_linear, `class A`]]

\[ \boxed {y^{\prime }+\frac {26 y}{5}=\frac {97 \sin \left (2 t \right )}{5}} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}

6.1.1 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) &={\frac {26}{5}}\\ q(t) &=\frac {97 \sin \left (2 t \right )}{5} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {26 y}{5} = \frac {97 \sin \left (2 t \right )}{5} \end {align*}

The domain of \(p(t)={\frac {26}{5}}\) is \[ \{-\infty

6.1.2 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 )+\frac {26 Y \left (s \right )}{5} = \frac {194}{5 \left (s^{2}+4\right )}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )+\frac {26 Y \left (s \right )}{5} = \frac {194}{5 \left (s^{2}+4\right )} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {194}{\left (s^{2}+4\right ) \left (5 s +26\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {-\frac {5}{8}-\frac {13 i}{8}}{s -2 i}+\frac {-\frac {5}{8}+\frac {13 i}{8}}{s +2 i}+\frac {5}{4 \left (s +\frac {26}{5}\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-\frac {5}{8}-\frac {13 i}{8}}{s -2 i}\right ) &= \left (-\frac {5}{8}-\frac {13 i}{8}\right ) {\mathrm e}^{2 i t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {5}{8}+\frac {13 i}{8}}{s +2 i}\right ) &= \left (-\frac {5}{8}+\frac {13 i}{8}\right ) {\mathrm e}^{-2 i t}\\ \mathcal {L}^{-1}\left (\frac {5}{4 \left (s +\frac {26}{5}\right )}\right ) &= \frac {5 \,{\mathrm e}^{-\frac {26 t}{5}}}{4} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {5 \,{\mathrm e}^{-\frac {26 t}{5}}}{4}-\frac {5 \cos \left (2 t \right )}{4}+\frac {13 \sin \left (2 t \right )}{4} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {5 \,{\mathrm e}^{-\frac {26 t}{5}}}{4}-\frac {5 \cos \left (2 t \right )}{4}+\frac {13 \sin \left (2 t \right )}{4} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {5 \,{\mathrm e}^{-\frac {26 t}{5}}}{4}-\frac {5 \cos \left (2 t \right )}{4}+\frac {13 \sin \left (2 t \right )}{4} \] Verified OK.

6.1.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+\frac {26 y}{5}=\frac {97 \sin \left (2 t \right )}{5}, 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 }=-\frac {26 y}{5}+\frac {97 \sin \left (2 t \right )}{5} \\ \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 {26 y}{5}=\frac {97 \sin \left (2 t \right )}{5} \\ \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 {26 y}{5}\right )=\frac {97 \mu \left (t \right ) \sin \left (2 t \right )}{5} \\ \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 {26 y}{5}\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 {26 \mu \left (t \right )}{5} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{\frac {26 t}{5}} \\ \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 {97 \mu \left (t \right ) \sin \left (2 t \right )}{5}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \frac {97 \mu \left (t \right ) \sin \left (2 t \right )}{5}d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \frac {97 \mu \left (t \right ) \sin \left (2 t \right )}{5}d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{\frac {26 t}{5}} \\ {} & {} & y=\frac {\int \frac {97 \,{\mathrm e}^{\frac {26 t}{5}} \sin \left (2 t \right )}{5}d t +c_{1}}{{\mathrm e}^{\frac {26 t}{5}}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {-\frac {5 \,{\mathrm e}^{\frac {26 t}{5}} \cos \left (2 t \right )}{4}+\frac {13 \,{\mathrm e}^{\frac {26 t}{5}} \sin \left (2 t \right )}{4}+c_{1}}{{\mathrm e}^{\frac {26 t}{5}}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {13 \sin \left (2 t \right )}{4}-\frac {5 \cos \left (2 t \right )}{4}+c_{1} {\mathrm e}^{-\frac {26 t}{5}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-\frac {5}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {5}{4} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {5}{4}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {5 \,{\mathrm e}^{-\frac {26 t}{5}}}{4}-\frac {5 \cos \left (2 t \right )}{4}+\frac {13 \sin \left (2 t \right )}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {5 \,{\mathrm e}^{-\frac {26 t}{5}}}{4}-\frac {5 \cos \left (2 t \right )}{4}+\frac {13 \sin \left (2 t \right )}{4} \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: 0.891 (sec). Leaf size: 23

dsolve([diff(y(t),t)+52/10*y(t)=194/10*sin(2*t),y(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {5 \,{\mathrm e}^{-\frac {26 t}{5}}}{4}-\frac {5 \cos \left (2 t \right )}{4}+\frac {13 \sin \left (2 t \right )}{4} \]

✓ Solution by Mathematica

Time used: 0.095 (sec). Leaf size: 31

DSolve[{y'[t]+52/10*y[t]==194/10*Sin[2*t],{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{4} \left (5 e^{-26 t/5}+13 \sin (2 t)-5 \cos (2 t)\right ) \]