2.3.5 Problem 5

Internal problem ID [10336]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 5
Date solved : Thursday, November 27, 2025 at 10:34:34 AM
CAS classification : [_separable]

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

Using Laplace transform method.

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*}

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 &\xrightarrow {\mathscr {L}} Y \left (s \right )\\ y^{\prime } \left (\tau +x_{0} \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.

Solve 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}^{-s x_{0}} \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}^{-s x_{0}}\right ) &= \left ({\mathrm e}^{-s x_{0}}\right ) \left (-\frac {x_{0} y_{0}}{s}\right ) \\ \mathrm {d} \left (Y \,{\mathrm e}^{-s x_{0}}\right ) &= \left (-\frac {x_{0} y_{0} {\mathrm e}^{-s x_{0}}}{s}\right )\, \mathrm {d} s \\ \end{align*}

Integrating gives

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

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

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

Applying inverse Laplace transform on the above gives.

\begin{align*} y = \frac {x_{0} y_{0}}{\tau +x_{0}}+c_1 \operatorname {Typesetting}\mcoloneq \operatorname {msup}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mi}\left (\text {``$\mathcal \{L\}$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mrow}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mo}\left (\text {``$-$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mn}\left (``1''\right )\right ), \operatorname {Typesetting}\mcoloneq \operatorname {msemantics}=\text {``atomic''}\right )\left ({\mathrm e}^{s x_{0}}, 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

\[ y_{0} = c_1 \operatorname {Typesetting}\mcoloneq \operatorname {msup}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mi}\left (\text {``$\mathcal \{L\}$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mrow}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mo}\left (\text {``$-$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mn}\left (``1''\right )\right ), \operatorname {Typesetting}\mcoloneq \operatorname {msemantics}=\text {``atomic''}\right )\left ({\mathrm e}^{s x_{0}}, s , \tau \right )+y_{0} \]

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

✓ Maple. Time used: 0.085 (sec). Leaf size: 10

ode:=t*diff(y(t),t)+y(t) = 0; 
ic:=[y(x__0) = y__0]; 
dsolve([ode,op(ic)],y(t),method='laplace');
 

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

Maple trace

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

Maple step by step

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

✓ Mathematica. Time used: 0.016 (sec). Leaf size: 11

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

\begin{align*} y(t)&\to \frac {\text {x0} \text {y0}}{t} \end{align*}

✓ Sympy. Time used: 0.065 (sec). Leaf size: 7

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t*Derivative(y(t), t) + y(t),0) 
ics = {y(x__0): y__0} 
dsolve(ode,func=y(t),ics=ics)
 

\[ y{\left (t \right )} = \frac {x^{0} y^{0}}{t} \]