problem 833

Internal problem ID [15555]
Internal ﬁle name [OUTPUT/15556_Saturday_May_11_2024_01_35_34_AM_34637621/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: 833.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "linear", "ﬁrst_order_ode_lie_symmetry_lookup"

\[ \boxed {2 x^{\prime }+6 x={\mathrm e}^{-3 t} t} \] With initial conditions \begin {align*} \left [x \left (0\right ) = -{\frac {1}{2}}\right ] \end {align*}

33.4.1 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) &=\frac {{\mathrm e}^{-3 t} t}{2} \end {align*}

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

33.4.2 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*} 2 s Y \left (s \right )-2 x \left (0\right )+6 Y \left (s \right ) = \frac {1}{\left (s +3\right )^{2}}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} 2 s Y \left (s \right )+1+6 Y \left (s \right ) = \frac {1}{\left (s +3\right )^{2}} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = -\frac {s^{2}+6 s +8}{2 \left (s +3\right )^{3}} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{2 \left (s +3\right )^{3}}-\frac {1}{2 \left (s +3\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{2 \left (s +3\right )^{3}}\right ) &= \frac {{\mathrm e}^{-3 t} t^{2}}{4}\\ \mathcal {L}^{-1}\left (-\frac {1}{2 \left (s +3\right )}\right ) &= -\frac {{\mathrm e}^{-3 t}}{2} \end {align*}

Adding the above results and simplifying gives \[ x=\frac {{\mathrm e}^{-3 t} \left (t^{2}-2\right )}{4} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= \frac {{\mathrm e}^{-3 t} \left (t^{2}-2\right )}{4} \\ \end{align*}

Veriﬁcation of solutions

\[ x = \frac {{\mathrm e}^{-3 t} \left (t^{2}-2\right )}{4} \] Veriﬁed OK.

33.4.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 x^{\prime }+6 x={\mathrm e}^{-3 t} t , x \left (0\right )=-\frac {1}{2}\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+\frac {{\mathrm e}^{-3 t} t}{2} \\ \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=\frac {{\mathrm e}^{-3 t} t}{2} \\ \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 )=\frac {\mu \left (t \right ) {\mathrm e}^{-3 t} t}{2} \\ \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 \frac {\mu \left (t \right ) {\mathrm e}^{-3 t} t}{2}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & x \mu \left (t \right )=\int \frac {\mu \left (t \right ) {\mathrm e}^{-3 t} t}{2}d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\frac {\int \frac {\mu \left (t \right ) {\mathrm e}^{-3 t} t}{2}d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{3 t} \\ {} & {} & x=\frac {\int \frac {{\mathrm e}^{3 t} {\mathrm e}^{-3 t} t}{2}d t +c_{1}}{{\mathrm e}^{3 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & x=\frac {\frac {t^{2}}{4}+c_{1}}{{\mathrm e}^{3 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{-3 t} \left (t^{2}+4 c_{1} \right )}{4} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=-\frac {1}{2} \\ {} & {} & -\frac {1}{2}=c_{1} \\ \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} \\ {} & {} & x=\frac {{\mathrm e}^{-3 t} \left (t^{2}-2\right )}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{-3 t} \left (t^{2}-2\right )}{4} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`

\[ x \left (t \right ) = \frac {{\mathrm e}^{-3 t} \left (t^{2}-2\right )}{4} \]

33.4 problem 833

33.4.1 Existence and uniqueness analysis

33.4.2 Solving as laplace ode

33.4.3 Maple step by step solution