7.2 problem 10

Internal problem ID [6689]
Internal file name [OUTPUT/5937_Sunday_June_05_2022_04_02_49_PM_7573469/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: 10.
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`]]

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

7.2.1 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) &={\mathrm e}^{t} t \sin \left (t \right ) \end {align*}

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

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

7.2.2 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 {2 s -2}{\left (\left (s -1\right )^{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 -2}{\left (\left (s -1\right )^{2}+1\right )^{2}} \end {align*}

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

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

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

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

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

7.2.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y={\mathrm e}^{t} 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+{\mathrm e}^{t} 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={\mathrm e}^{t} 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 ) {\mathrm e}^{t} 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 ) {\mathrm e}^{t} 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 ) {\mathrm e}^{t} t \sin \left (t \right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) {\mathrm e}^{t} 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} {\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 {\sin \left (t \right )-t \cos \left (t \right )+c_{1}}{{\mathrm e}^{-t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{t} \left (\sin \left (t \right )-t \cos \left (t \right )+c_{1} \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{t} \left (\sin \left (t \right )-t \cos \left (t \right )\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{t} \left (\sin \left (t \right )-t \cos \left (t \right )\right ) \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: 1.922 (sec). Leaf size: 15

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

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

✓ Solution by Mathematica

Time used: 0.069 (sec). Leaf size: 17

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

\[ y(t)\to e^t (\sin (t)-t \cos (t)) \]