2.3.8 problem 8

Internal problem ID [8741]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 8
Date solved : Tuesday, December 17, 2024 at 01:01:54 PM
CAS classification : [_linear]

Solve

\begin{align*} t y^{\prime }+y&=\sin \left (t \right ) \end{align*}

With initial conditions

\begin{align*} y \left (1\right )&=0 \end{align*}

Since initial condition is not at zero, then change of variable is used to transform the ode so that initial condition is at zero.

\begin{align*} \tau = t -1 \end{align*}

Solve

\begin{align*} \left (\tau +1\right ) y^{\prime }+y&=\sin \left (\tau +1\right ) \end{align*}

With initial conditions

\begin{align*} y \left (0\right )&=0 \end{align*}

We will now apply Laplace transform to each term in the ode. Since this is time varying, the following Laplace transform property will be used

\begin{align*} \tau ^{n} f \left (\tau \right ) &\xrightarrow {\mathscr {L}} (-1)^n \frac {d^n}{ds^n} F(s) \end{align*}

Where in the above \(F(s)\) is the laplace transform of \(f \left (\tau \right )\). Applying the above property to each term of the ode gives

\begin{align*} y \left (\tau \right ) &\xrightarrow {\mathscr {L}} Y \left (s \right )\\ \left (\tau +1\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right ) &\xrightarrow {\mathscr {L}} -Y \left (s \right )-s \left (\frac {d}{d s}Y \left (s \right )\right )+s Y \left (s \right )-y \left (0\right )\\ \sin \left (\tau +1\right ) &\xrightarrow {\mathscr {L}} \frac {\sin \left (1\right ) s +\cos \left (1\right )}{s^{2}+1} \end{align*}

Collecting all the terms above, the ode in Laplace domain becomes

\[ -s Y^{\prime }+s Y-y \left (0\right ) = \frac {\sin \left (1\right ) s +\cos \left (1\right )}{s^{2}+1} \]

Replacing \(y \left (0\right ) = 0\) in the above results in

\[ -s Y^{\prime }+s Y = \frac {\sin \left (1\right ) s +\cos \left (1\right )}{s^{2}+1} \]

The above ode in Y(s) is now solved.

In canonical form a linear first order is

\begin{align*} Y^{\prime } + q(s)Y &= p(s) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(s) &=-1\\ p(s) &=\frac {-\sin \left (1\right ) s -\cos \left (1\right )}{\left (s^{2}+1\right ) s} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,ds}}\\ &= {\mathrm e}^{\int \left (-1\right )d s}\\ &= {\mathrm e}^{-s} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}}\left ( \mu Y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}}\left ( \mu Y\right ) &= \left (\mu \right ) \left (\frac {-\sin \left (1\right ) s -\cos \left (1\right )}{\left (s^{2}+1\right ) s}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}} \left (Y \,{\mathrm e}^{-s}\right ) &= \left ({\mathrm e}^{-s}\right ) \left (\frac {-\sin \left (1\right ) s -\cos \left (1\right )}{\left (s^{2}+1\right ) s}\right ) \\ \mathrm {d} \left (Y \,{\mathrm e}^{-s}\right ) &= \left (\frac {\left (-\sin \left (1\right ) s -\cos \left (1\right )\right ) {\mathrm e}^{-s}}{\left (s^{2}+1\right ) s}\right )\, \mathrm {d} s \\ \end{align*}

Integrating gives

\begin{align*} Y \,{\mathrm e}^{-s}&= \int {\frac {\left (-\sin \left (1\right ) s -\cos \left (1\right )\right ) {\mathrm e}^{-s}}{\left (s^{2}+1\right ) s} \,ds} \\ &=-\cos \left (1\right ) \left (\frac {{\mathrm e}^{i} \operatorname {Ei}_{1}\left (s +i\right )}{2}+\frac {{\mathrm e}^{-i} \operatorname {Ei}_{1}\left (s -i\right )}{2}-\operatorname {Ei}_{1}\left (s \right )\right )+\sin \left (1\right ) \left (\frac {i {\mathrm e}^{i} \operatorname {Ei}_{1}\left (s +i\right )}{2}-\frac {i {\mathrm e}^{-i} \operatorname {Ei}_{1}\left (s -i\right )}{2}\right ) + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-s}\) gives the final solution

\[ Y = -\frac {\left (-2 \cos \left (1\right ) \operatorname {Ei}_{1}\left (s \right )+\operatorname {Ei}_{1}\left (s +i\right )+\operatorname {Ei}_{1}\left (s -i\right )-2 c_1 \right ) {\mathrm e}^{s}}{2} \]

Applying inverse Laplace transform on the above gives.

\begin{align*} y = \frac {\cos \left (1\right )}{\tau +1}+c_1 \mathcal {L}^{-1}\left ({\mathrm e}^{s}, s , \tau \right )-\frac {\cos \left (\tau +1\right )}{\tau +2}\tag {1} \end{align*}

Substituting initial conditions \(y \left (0\right ) = 0\) and \(y^{\prime }\left (0\right ) = 0\) into the above solution Gives

\[ 0 = c_1 \mathcal {L}^{-1}\left ({\mathrm e}^{s}, s , \tau \right )+\frac {\cos \left (1\right )}{2} \]

Solving for the constant \(c_1\) from the above equation gives

\begin{align*} c_1 = -\frac {\cos \left (1\right )}{2 \mathcal {L}^{-1}\left ({\mathrm e}^{s}, s , \tau \right )} \end{align*}

Substituting the above back into the solution (1) gives

\[ y = \frac {\cos \left (1\right )}{\tau +1}-\frac {\cos \left (1\right )}{2}-\frac {\cos \left (\tau +1\right )}{\tau +2} \]

Changing back the solution from \(\tau \) to \(t\) using

\begin{align*} \tau = t -1 \end{align*}

the solution becomes

\begin{align*} y \left (t \right ) = \frac {\cos \left (1\right )}{t}-\frac {\cos \left (1\right )}{2}-\frac {\cos \left (t \right )}{t +1} \end{align*}

The solution was found not to satisfy the ode or the IC. Hence it is removed.

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [t y^{\prime }+y=\sin \left (t \right ), y \left (1\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Isolate the derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y}{t}+\frac {\sin \left (t \right )}{t} \\ \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 }+\frac {y}{t}=\frac {\sin \left (t \right )}{t} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+\frac {y}{t}\right )=\frac {\mu \left (t \right ) \sin \left (t \right )}{t} \\ \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 }+\frac {y}{t}\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 )=\frac {\mu \left (t \right )}{t} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )=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 \frac {\mu \left (t \right ) \sin \left (t \right )}{t}d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \frac {\mu \left (t \right ) \sin \left (t \right )}{t}d t +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \frac {\mu \left (t \right ) \sin \left (t \right )}{t}d t +\mathit {C1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )=t \\ {} & {} & y=\frac {\int \sin \left (t \right )d t +\mathit {C1}}{t} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {-\cos \left (t \right )+\mathit {C1}}{t} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=-\cos \left (1\right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =\cos \left (1\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =\cos \left (1\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {-\cos \left (t \right )+\cos \left (1\right )}{t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {-\cos \left (t \right )+\cos \left (1\right )}{t} \end {array} \]

Maple trace

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

Maple dsolve solution

Solving time : 0.237 (sec)
Leaf size : maple_leaf_size

dsolve([t*diff(y(t),t)+y(t) = sin(t), 
        op([y(1) = 0])], 
        y(t),method=laplace)
 

\[ \text {No solution found} \]

Mathematica DSolve solution

Solving time : 0.09 (sec)
Leaf size : 16

DSolve[{t*D[y[t],t]+y[t]==Sin[t],y[1]==0}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to \frac {\cos (1)-\cos (t)}{t} \]