problem 831

Internal problem ID [15553]
Internal ﬁle name [OUTPUT/15554_Saturday_May_11_2024_01_35_33_AM_23389732/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: 831.
ODE order: 1.
ODE degree: 1.

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

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

33.2.1 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) &=3 t^{3}+3 t^{2}+2 t +1 \end {align*}

Hence the ode is \begin {align*} x^{\prime }-3 x = 3 t^{3}+3 t^{2}+2 t +1 \end {align*}

33.2.2 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 {s^{3}+2 s^{2}+6 s +18}{s^{4}}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )+1-3 Y \left (s \right ) = \frac {s^{3}+2 s^{2}+6 s +18}{s^{4}} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = -\frac {s^{3}+2 s^{2}+4 s +6}{s^{4}} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {6}{s^{4}}-\frac {2}{s^{2}}-\frac {4}{s^{3}}-\frac {1}{s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {6}{s^{4}}\right ) &= -t^{3}\\ \mathcal {L}^{-1}\left (-\frac {2}{s^{2}}\right ) &= -2 t\\ \mathcal {L}^{-1}\left (-\frac {4}{s^{3}}\right ) &= -2 t^{2}\\ \mathcal {L}^{-1}\left (-\frac {1}{s}\right ) &= -1 \end {align*}

Adding the above results and simplifying gives \[ x=-\left (t +1\right ) \left (t^{2}+t +1\right ) \]

Summary

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

Veriﬁcation of solutions

\[ x = -\left (t +1\right ) \left (t^{2}+t +1\right ) \] Veriﬁed OK.

33.2.3 Maple step by step solution

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

Maple trace

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

33.2 problem 831

33.2.1 Existence and uniqueness analysis

33.2.2 Solving as laplace ode

33.2.3 Maple step by step solution