18.2 problem 27.1 (b)

Internal problem ID [13849]
Internal file name [OUTPUT/13021_Friday_February_23_2024_06_46_42_AM_2265550/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 27. Differentiation and the Laplace transform. Additional Exercises. page 496
Problem number: 27.1 (b).
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`]]

\[ \boxed {y^{\prime }-2 y=t^{3}} \] With initial conditions \begin {align*} [y \left (0\right ) = 4] \end {align*}

18.2.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &=-2\\ q(t) &=t^{3} \end {align*}

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

The domain of \(p(t)=-2\) is \[ \{-\infty

18.2.2 Solving as laplace ode

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

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-y \left (0\right )-2 Y \left (s \right ) = \frac {6}{s^{4}}\tag {1} \end {align*}

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

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

Applying partial fractions decomposition results in \[ Y(s)= -\frac {3}{4 s^{2}}-\frac {3}{2 s^{3}}-\frac {3}{s^{4}}+\frac {35}{8 \left (s -2\right )}-\frac {3}{8 s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {3}{4 s^{2}}\right ) &= -\frac {3 t}{4}\\ \mathcal {L}^{-1}\left (-\frac {3}{2 s^{3}}\right ) &= -\frac {3 t^{2}}{4}\\ \mathcal {L}^{-1}\left (-\frac {3}{s^{4}}\right ) &= -\frac {t^{3}}{2}\\ \mathcal {L}^{-1}\left (\frac {35}{8 \left (s -2\right )}\right ) &= \frac {35 \,{\mathrm e}^{2 t}}{8}\\ \mathcal {L}^{-1}\left (-\frac {3}{8 s}\right ) &= -{\frac {3}{8}} \end {align*}

Adding the above results and simplifying gives \[ y=-\frac {t^{3}}{2}-\frac {3 t^{2}}{4}-\frac {3 t}{4}+\frac {35 \,{\mathrm e}^{2 t}}{8}-\frac {3}{8} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {t^{3}}{2}-\frac {3 t^{2}}{4}-\frac {3 t}{4}+\frac {35 \,{\mathrm e}^{2 t}}{8}-\frac {3}{8} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\frac {t^{3}}{2}-\frac {3 t^{2}}{4}-\frac {3 t}{4}+\frac {35 \,{\mathrm e}^{2 t}}{8}-\frac {3}{8} \] Verified OK.

18.2.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-2 y=t^{3}, y \left (0\right )=4\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 y+t^{3} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }-2 y=t^{3} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }-2 y\right )=\mu \left (t \right ) t^{3} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }-2 y\right )=y^{\prime } \mu \left (t \right )+y \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=-2 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \mu \left (t \right )\right )\right )d t =\int \mu \left (t \right ) t^{3}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \mu \left (t \right ) t^{3}d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) t^{3}d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{-2 t} \\ {} & {} & y=\frac {\int {\mathrm e}^{-2 t} t^{3}d t +c_{1}}{{\mathrm e}^{-2 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {-\frac {\left (4 t^{3}+6 t^{2}+6 t +3\right ) {\mathrm e}^{-2 t}}{8}+c_{1}}{{\mathrm e}^{-2 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=-\frac {t^{3}}{2}+c_{1} {\mathrm e}^{2 t}-\frac {3 t^{2}}{4}-\frac {3 t}{4}-\frac {3}{8} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=4 \\ {} & {} & 4=-\frac {3}{8}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {35}{8} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {35}{8}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {t^{3}}{2}-\frac {3 t^{2}}{4}-\frac {3 t}{4}+\frac {35 \,{\mathrm e}^{2 t}}{8}-\frac {3}{8} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {t^{3}}{2}-\frac {3 t^{2}}{4}-\frac {3 t}{4}+\frac {35 \,{\mathrm e}^{2 t}}{8}-\frac {3}{8} \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: 4.172 (sec). Leaf size: 25

dsolve([diff(y(t),t)-2*y(t)=t^3,y(0) = 4],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {3 t}{4}-\frac {t^{3}}{2}-\frac {3 t^{2}}{4}+\frac {35 \,{\mathrm e}^{2 t}}{8}-\frac {3}{8} \]

✓ Solution by Mathematica

Time used: 0.052 (sec). Leaf size: 31

DSolve[{y'[t]+4*y[t]==t^3,{y[0]==4}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{128} \left (32 t^3-24 t^2+12 t+515 e^{-4 t}-3\right ) \]