33.5 problem 834

Internal problem ID [15556]
Internal file name [OUTPUT/15557_Saturday_May_11_2024_01_35_34_AM_12648957/index.tex]

Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV, G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 3. Section 24.2. Solving the Cauchy problem for linear differential equation with constant coefficients. Exercises page 249
Problem number: 834.
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`]]

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

33.5.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime } + p(t)x &= q(t) \end {align*}

Where here \begin {align*} p(t) &=1\\ q(t) &=2 \sin \left (t \right ) \end {align*}

Hence the ode is \begin {align*} x^{\prime }+x = 2 \sin \left (t \right ) \end {align*}

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

33.5.2 Solving as laplace ode

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (x\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (x^{\prime }\right )&= s Y(s) - x \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-x \left (0\right )+Y \left (s \right ) = \frac {2}{s^{2}+1}\tag {1} \end {align*}

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

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

Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{s +1}+\frac {-\frac {1}{2}-\frac {i}{2}}{s -i}+\frac {-\frac {1}{2}+\frac {i}{2}}{s +i} \] 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 {-\frac {1}{2}-\frac {i}{2}}{s -i}\right ) &= \left (-\frac {1}{2}-\frac {i}{2}\right ) {\mathrm e}^{i t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{2}+\frac {i}{2}}{s +i}\right ) &= \left (-\frac {1}{2}+\frac {i}{2}\right ) {\mathrm e}^{-i t} \end {align*}

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

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

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

33.5.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }+x=2 \sin \left (t \right ), x \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=-x+2 \sin \left (t \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} x\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & x^{\prime }+x=2 \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 (x^{\prime }+x\right )=2 \mu \left (t \right ) \sin \left (t \right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (x \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (x^{\prime }+x\right )=x^{\prime } \mu \left (t \right )+x \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 (x \mu \left (t \right )\right )\right )d t =\int 2 \mu \left (t \right ) \sin \left (t \right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & x \mu \left (t \right )=\int 2 \mu \left (t \right ) \sin \left (t \right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\frac {\int 2 \mu \left (t \right ) \sin \left (t \right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{t} \\ {} & {} & x=\frac {\int 2 \sin \left (t \right ) {\mathrm e}^{t}d t +c_{1}}{{\mathrm e}^{t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & x=\frac {-{\mathrm e}^{t} \cos \left (t \right )+\sin \left (t \right ) {\mathrm e}^{t}+c_{1}}{{\mathrm e}^{t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & x=\sin \left (t \right )-\cos \left (t \right )+c_{1} {\mathrm e}^{-t} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=0 \\ {} & {} & 0=c_{1} -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} \\ {} & {} & x=\sin \left (t \right )-\cos \left (t \right )+{\mathrm e}^{-t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\sin \left (t \right )-\cos \left (t \right )+{\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`
 

✓ Solution by Maple

Time used: 0.406 (sec). Leaf size: 15

dsolve([diff(x(t),t)+x(t)=2*sin(t),x(0) = 0],x(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.055 (sec). Leaf size: 17

DSolve[{x'[t]+x[t]==2*Sin[t],{x[0]==0}},x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to e^{-t}+\sin (t)-\cos (t) \]