33.1 problem 830

Internal problem ID [15552]
Internal file name [OUTPUT/15553_Saturday_May_11_2024_01_35_32_AM_99659885/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: 830.
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`]]

\[ \boxed {x^{\prime }+3 x={\mathrm e}^{-2 t}} \] With initial conditions \begin {align*} [x \left (0\right ) = 0] \end {align*}

33.1.1 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) &=3\\ q(t) &={\mathrm e}^{-2 t} \end {align*}

Hence the ode is \begin {align*} x^{\prime }+3 x = {\mathrm e}^{-2 t} \end {align*}

The domain of \(p(t)=3\) is \[ \{-\infty

33.1.2 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 )+3 Y \left (s \right ) = \frac {1}{s +2}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )+3 Y \left (s \right ) = \frac {1}{s +2} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {1}{\left (s +2\right ) \left (s +3\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{s +3}+\frac {1}{s +2} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{s +3}\right ) &= -{\mathrm e}^{-3 t}\\ \mathcal {L}^{-1}\left (\frac {1}{s +2}\right ) &= {\mathrm e}^{-2 t} \end {align*}

Adding the above results and simplifying gives \[ x={\mathrm e}^{-2 t}-{\mathrm e}^{-3 t} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= {\mathrm e}^{-2 t}-{\mathrm e}^{-3 t} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ x = {\mathrm e}^{-2 t}-{\mathrm e}^{-3 t} \] Verified OK.

33.1.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }+3 x={\mathrm e}^{-2 t}, 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 }=-3 x+{\mathrm e}^{-2 t} \\ \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 }+3 x={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (x^{\prime }+3 x\right )=\mu \left (t \right ) {\mathrm e}^{-2 t} \\ \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 }+3 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 )=3 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{3 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 ) {\mathrm e}^{-2 t}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & x \mu \left (t \right )=\int \mu \left (t \right ) {\mathrm e}^{-2 t}d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\frac {\int \mu \left (t \right ) {\mathrm e}^{-2 t}d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{3 t} \\ {} & {} & x=\frac {\int {\mathrm e}^{-2 t} {\mathrm e}^{3 t}d t +c_{1}}{{\mathrm e}^{3 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{t}+c_{1}}{{\mathrm e}^{3 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & x={\mathrm e}^{-3 t} \left ({\mathrm e}^{t}+c_{1} \right ) \\ \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={\mathrm e}^{-3 t} \left ({\mathrm e}^{t}-1\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x={\mathrm e}^{-3 t} \left ({\mathrm e}^{t}-1\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: 0.454 (sec). Leaf size: 15

dsolve([diff(x(t),t)+3*x(t)=exp(-2*t),x(0) = 0],x(t), singsol=all)
 

\[ x \left (t \right ) = {\mathrm e}^{-2 t}-{\mathrm e}^{-3 t} \]

✓ Solution by Mathematica

Time used: 0.096 (sec). Leaf size: 16

DSolve[{x'[t]+3*x[t]==Exp[-2*t],{x[0]==0}},x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to e^{-3 t} \left (e^t-1\right ) \]