7.1 problem 9

Internal problem ID [6688]
Internal file name [OUTPUT/5936_Sunday_June_05_2022_04_02_46_PM_9679260/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. 7.4.1 DERIVATIVES OF A TRANSFORM. Page 309
Problem number: 9.
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`]]

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

7.1.1 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) &=1\\ q(t) &=t \sin \left (t \right ) \end {align*}

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

The domain of \(p(t)=1\) is \[ \{-\infty

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

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

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

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

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

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

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

7.1.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+y=t \sin \left (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 }=-y+t \sin \left (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 }+y=t \sin \left (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 }+y\right )=\mu \left (t \right ) t \sin \left (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 ) t \sin \left (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 ) t \sin \left (t \right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) t \sin \left (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} t \sin \left (t \right )d t +c_{1}}{{\mathrm e}^{t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\left (\frac {1}{2}-\frac {t}{2}\right ) {\mathrm e}^{t} \cos \left (t \right )+\frac {{\mathrm e}^{t} t \sin \left (t \right )}{2}+c_{1}}{{\mathrm e}^{t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-t} c_{1} +\frac {\left (-t +1\right ) \cos \left (t \right )}{2}+\frac {t \sin \left (t \right )}{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +\frac {1}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\frac {1}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\frac {1}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-t}}{2}+\frac {\left (-t +1\right ) \cos \left (t \right )}{2}+\frac {t \sin \left (t \right )}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-t}}{2}+\frac {\left (-t +1\right ) \cos \left (t \right )}{2}+\frac {t \sin \left (t \right )}{2} \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: 1.703 (sec). Leaf size: 23

dsolve([diff(y(t),t)+y(t)=t*sin(t),y(0) = 0],y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.086 (sec). Leaf size: 28

DSolve[{y'[t]+y[t]==t*Sin[t],{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

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