problem 7

Internal problem ID [12798]
Internal ﬁle name [OUTPUT/11451_Saturday_November_04_2023_08_47_23_AM_16935513/index.tex]

Book: Ordinary Diﬀerential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 5. The Laplace Transform Method. Exercises 5.3, page 255
Problem number: 7.
ODE order: 1.
ODE degree: 1.

\[ \boxed {y^{\prime }-2 y=6} \] With initial conditions \begin {align*} [y \left (0\right ) = 2] \end {align*}

14.1.1 Existence and uniqueness analysis

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

14.1.2 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}\tag {1} \end {align*}

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

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

Applying partial fractions decomposition results in \[ Y(s)= \frac {5}{s -2}-\frac {3}{s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {5}{s -2}\right ) &= 5 \,{\mathrm e}^{2 x}\\ \mathcal {L}^{-1}\left (-\frac {3}{s}\right ) &= -3 \end {align*}

Adding the above results and simplifying gives \[ y=2 \,{\mathrm e}^{x} \left (\cosh \left (x \right )+4 \sinh \left (x \right )\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 2 \,{\mathrm e}^{x} \left (\cosh \left (x \right )+4 \sinh \left (x \right )\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = 2 \,{\mathrm e}^{x} \left (\cosh \left (x \right )+4 \sinh \left (x \right )\right ) \] Veriﬁed OK.

14.1.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-2 y=6, y \left (0\right )=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 y+6 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{2 y+6}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{2 y+6}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y+3\right )}{2}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{2 x +2 c_{1}}-3 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2={\mathrm e}^{2 c_{1}}-3 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\ln \left (5\right )}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\ln \left (5\right )}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=5 \,{\mathrm e}^{2 x}-3 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=5 \,{\mathrm e}^{2 x}-3 \end {array} \]

Maple trace

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

\[ y \left (x \right ) = 2 \,{\mathrm e}^{x} \left (\cosh \left (x \right )+4 \sinh \left (x \right )\right ) \]

14.1 problem 7

14.1.1 Existence and uniqueness analysis

14.1.2 Solving as laplace ode

14.1.3 Maple step by step solution