2.3.5 problem 5

Internal problem ID [8738]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 5
Date solved : Tuesday, December 17, 2024 at 01:01:52 PM
CAS classification : [_separable]

Solve

\begin{align*} t y^{\prime }+y&=0 \end{align*}

With initial conditions

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

Solve

\begin{align*} \left (\tau +x_{0} \right ) y^{\prime }+y&=0 \end{align*}

With initial conditions

\begin{align*} y \left (0\right )&=y_{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 +x_{0} \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 )+x_{0} \left (s Y \left (s \right )-y \left (0\right )\right ) \end{align*}

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

\[ -s Y^{\prime }+x_{0} \left (s Y-y \left (0\right )\right ) = 0 \]

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

\[ -s Y^{\prime }+x_{0} \left (s Y-y_{0} \right ) = 0 \]

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) &=-x_{0}\\ p(s) &=-\frac {x_{0} y_{0}}{s} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,ds}}\\ &= {\mathrm e}^{\int -x_{0} d s}\\ &= {\mathrm e}^{-x_{0} 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 {x_{0} y_{0}}{s}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}} \left (Y \,{\mathrm e}^{-x_{0} s}\right ) &= \left ({\mathrm e}^{-x_{0} s}\right ) \left (-\frac {x_{0} y_{0}}{s}\right ) \\ \mathrm {d} \left (Y \,{\mathrm e}^{-x_{0} s}\right ) &= \left (-\frac {x_{0} y_{0} {\mathrm e}^{-x_{0} s}}{s}\right )\, \mathrm {d} s \\ \end{align*}

Integrating gives

\begin{align*} Y \,{\mathrm e}^{-x_{0} s}&= \int {-\frac {x_{0} y_{0} {\mathrm e}^{-x_{0} s}}{s} \,ds} \\ &=x_{0} y_{0} \operatorname {Ei}_{1}\left (x_{0} s \right ) + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-x_{0} s}\) gives the final solution

\[ Y = {\mathrm e}^{x_{0} s} \left (x_{0} y_{0} \operatorname {Ei}_{1}\left (x_{0} s \right )+c_1 \right ) \]

Applying inverse Laplace transform on the above gives.

\begin{align*} y = \frac {x_{0} y_{0}}{\tau +x_{0}}+c_1 \mathcal {L}^{-1}\left ({\mathrm e}^{x_{0} s}, s , \tau \right )\tag {1} \end{align*}

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

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

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

\begin{align*} c_1 = 0 \end{align*}

Substituting the above back into the solution (1) gives

\[ y = \frac {x_{0} y_{0}}{\tau +x_{0}} \]

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

\begin{align*} \tau = t -x_{0} \end{align*}

the solution becomes

\begin{align*} y \left (t \right ) = \frac {x_{0} y_{0}}{t} \end{align*}

Figure 2.102: Slope field plot
\(t \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right ) = 0\)

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [t y^{\prime }+y=0, y \left (x_{0} \right )=y_{0} \right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-\frac {1}{t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y}d t =\int -\frac {1}{t}d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-\ln \left (t \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{\mathit {C1}}}{t} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (x_{0} \right )=y_{0} \\ {} & {} & y_{0} =\frac {{\mathrm e}^{\mathit {C1}}}{x_{0}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =\ln \left (x_{0} y_{0} \right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =\ln \left (x_{0} y_{0} \right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {x_{0} y_{0}}{t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {x_{0} y_{0}}{t} \end {array} \]

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 : 0.028 (sec)
Leaf size : 10

dsolve([t*diff(y(t),t)+y(t) = 0, 
        op([y(x__0) = y__0])], 
        y(t),method=laplace)
 

\[ y = \frac {x_{0} y_{0}}{t} \]

Mathematica DSolve solution

Solving time : 1.702 (sec)
Leaf size : 11

DSolve[{t*D[y[t],t]+y[t]==0,y[x0]==y0}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to \frac {\text {x0} \text {y0}}{t} \]