Internal problem ID [15554]
Internal file name [OUTPUT/15555_Saturday_May_11_2024_01_35_33_AM_6112988/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: 832.
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`]]
\[ \boxed {x^{\prime }-x=\cos \left (t \right )-\sin \left (t \right )} \] With initial conditions \begin {align*} [x \left (0\right ) = 0] \end {align*}
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) &=\cos \left (t \right )-\sin \left (t \right ) \end {align*}
Hence the ode is \begin {align*} x^{\prime }-x = \cos \left (t \right )-\sin \left (t \right ) \end {align*}
The domain of \(p(t)=-1\) is \[
\{-\infty
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 {s -1}{s^{2}+1}\tag {1} \end {align*}
Replacing initial condition gives \begin {align*} s Y \left (s \right )-Y \left (s \right ) = \frac {s -1}{s^{2}+1} \end {align*}
Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {1}{s^{2}+1} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {i}{2 \left (s -i\right )}+\frac {i}{2 s +2 i} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {i}{2 \left (s -i\right )}\right ) &= -\frac {i {\mathrm e}^{i t}}{2}\\ \mathcal {L}^{-1}\left (\frac {i}{2 s +2 i}\right ) &= \frac {i {\mathrm e}^{-i t}}{2} \end {align*}
Adding the above results and simplifying gives \[ x=\sin \left (t \right ) \]
The solution(s) found are the following \begin{align*}
\tag{1} x &= \sin \left (t \right ) \\
\end{align*} Verification of solutions
\[
x = \sin \left (t \right )
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }-x=\cos \left (t \right )-\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+\cos \left (t \right )-\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=\cos \left (t \right )-\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 )=\mu \left (t \right ) \left (\cos \left (t \right )-\sin \left (t \right )\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 \mu \left (t \right ) \left (\cos \left (t \right )-\sin \left (t \right )\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & x \mu \left (t \right )=\int \mu \left (t \right ) \left (\cos \left (t \right )-\sin \left (t \right )\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\frac {\int \mu \left (t \right ) \left (\cos \left (t \right )-\sin \left (t \right )\right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{-t} \\ {} & {} & x=\frac {\int {\mathrm e}^{-t} \left (\cos \left (t \right )-\sin \left (t \right )\right )d t +c_{1}}{{\mathrm e}^{-t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{-t} \sin \left (t \right )+c_{1}}{{\mathrm e}^{-t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & x=c_{1} {\mathrm e}^{t}+\sin \left (t \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \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} \\ {} & {} & x=\sin \left (t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\sin \left (t \right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.375 (sec). Leaf size: 6
\[
x \left (t \right ) = \sin \left (t \right )
\]
✓ Solution by Mathematica
Time used: 0.045 (sec). Leaf size: 7
\[
x(t)\to \sin (t)
\]
33.3.2 Solving as laplace ode
33.3.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful`
dsolve([diff(x(t),t)-x(t)=cos(t)-sin(t),x(0) = 0],x(t), singsol=all)
DSolve[{x'[t]-x[t]==Cos[t]-Sin[t],{x[0]==0}},x[t],t,IncludeSingularSolutions -> True]