5.8 problem Problem 24.30

Internal problem ID [5207]
Internal file name [OUTPUT/4700_Sunday_June_05_2022_03_03_37_PM_82796084/index.tex]

Book: Schaums Outline Differential Equations, 4th edition. Bronson and Costa. McGraw Hill 2014
Section: Chapter 24. Solutions of linear DE by Laplace transforms. Supplementary Problems. page 248
Problem number: Problem 24.30.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff"

Maple gives the following as the ode type

[[_2nd_order, _linear, _nonhomogeneous]]

\[ \boxed {y^{\prime \prime }+y=\sin \left (x \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 2] \end {align*}

5.8.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(x) &=0\\ q(x) &=1\\ F &=\sin \left (x \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+y = \sin \left (x \right ) \end {align*}

The domain of \(p(x)=0\) is \[ \{-\infty

Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (y\right ) =Y(s) \end {align*}

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 )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \left (0\right ) \end {align*}

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

But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y'(0) &=2 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-2+Y \left (s \right ) = \frac {1}{s^{2}+1} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2 s^{2}+3}{\left (s^{2}+1\right )^{2}} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{4 \left (s -i\right )^{2}}-\frac {1}{4 \left (s +i\right )^{2}}-\frac {5 i}{4 \left (s -i\right )}+\frac {5 i}{4 \left (s +i\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{4 \left (s -i\right )^{2}}\right ) &= -\frac {x \,{\mathrm e}^{i x}}{4}\\ \mathcal {L}^{-1}\left (-\frac {1}{4 \left (s +i\right )^{2}}\right ) &= -\frac {x \,{\mathrm e}^{-i x}}{4}\\ \mathcal {L}^{-1}\left (-\frac {5 i}{4 \left (s -i\right )}\right ) &= -\frac {5 i {\mathrm e}^{i x}}{4}\\ \mathcal {L}^{-1}\left (\frac {5 i}{4 \left (s +i\right )}\right ) &= \frac {5 i {\mathrm e}^{-i x}}{4} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {5 \sin \left (x \right )}{2}-\frac {x \cos \left (x \right )}{2} \] Simplifying the solution gives \[ y = \frac {5 \sin \left (x \right )}{2}-\frac {x \cos \left (x \right )}{2} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {5 \sin \left (x \right )}{2}-\frac {x \cos \left (x \right )}{2} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {5 \sin \left (x \right )}{2}-\frac {x \cos \left (x \right )}{2} \] Verified OK.

5.8.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+y=\sin \left (x \right ), y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+1=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\mathrm {-I}, \mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=\cos \left (x \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=\sin \left (x \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right )+y_{p}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (x \right )+c_{2} \sin \left (x \right )+y_{p}\left (x \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (x \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (x \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (x \right )=-y_{1}\left (x \right ) \left (\int \frac {y_{2}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right )+y_{2}\left (x \right ) \left (\int \frac {y_{1}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right ), f \left (x \right )=\sin \left (x \right )\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=\left [\begin {array}{cc} \cos \left (x \right ) & \sin \left (x \right ) \\ -\sin \left (x \right ) & \cos \left (x \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=-\cos \left (x \right ) \left (\int \sin \left (x \right )^{2}d x \right )+\frac {\sin \left (x \right ) \left (\int \sin \left (2 x \right )d x \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=\frac {\sin \left (x \right )}{4}-\frac {x \cos \left (x \right )}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (x \right )+c_{2} \sin \left (x \right )+\frac {\sin \left (x \right )}{4}-\frac {x \cos \left (x \right )}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (x \right )+c_{2} \sin \left (x \right )+\frac {\sin \left (x \right )}{4}-\frac {x \cos \left (x \right )}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} \sin \left (x \right )+c_{2} \cos \left (x \right )-\frac {\cos \left (x \right )}{4}+\frac {x \sin \left (x \right )}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=-\frac {1}{4}+c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =\frac {9}{4}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {5 \sin \left (x \right )}{2}-\frac {x \cos \left (x \right )}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {5 \sin \left (x \right )}{2}-\frac {x \cos \left (x \right )}{2} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful`
 

✓ Solution by Maple

Time used: 0.375 (sec). Leaf size: 14

dsolve([diff(y(x),x$2)+y(x)=sin(x),y(0) = 0, D(y)(0) = 2],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {5 \sin \left (x \right )}{2}-\frac {\cos \left (x \right ) x}{2} \]

✓ Solution by Mathematica

Time used: 0.027 (sec). Leaf size: 19

DSolve[{y''[x]+y[x]==Sin[x],{y[0]==0,y'[0]==2}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{2} (5 \sin (x)-x \cos (x)) \]