2.3.2 Problem 2

Internal problem ID [10333]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 2
Date solved : Thursday, November 27, 2025 at 10:34:30 AM
CAS classification : [_separable]

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

Using Laplace transform method.

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*} t^{n} f \left (t \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 \(f \left (t \right )\). Applying the above property to each term of the ode gives

\begin{align*} -y t &\xrightarrow {\mathscr {L}} \frac {d}{d s}Y \left (s \right )\\ y^{\prime } &\xrightarrow {\mathscr {L}} s Y \left (s \right )-y \left (0\right ) \end{align*}

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

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

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

\[ Y^{\prime }+s Y = 0 \]

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) &=s\\ p(s) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,ds}}\\ &= {\mathrm e}^{\int s d s}\\ &= {\mathrm e}^{\frac {s^{2}}{2}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}} \mu Y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}} \left (Y \,{\mathrm e}^{\frac {s^{2}}{2}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} Y \,{\mathrm e}^{\frac {s^{2}}{2}}&= \int {0 \,ds} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {s^{2}}{2}}\) gives the final solution

\[ Y = {\mathrm e}^{-\frac {s^{2}}{2}} c_1 \]

Applying inverse Laplace transform on the above gives.

\begin{align*} y = 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}^{-\frac {s^{2}}{2}}, s , t\right )\tag {1} \end{align*}

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

\[ 0 = c_1 \operatorname {Typesetting}\mcoloneq \operatorname {msup}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mi}\left (\text {``$\mathcal \{L\}$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mrow}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mo}\left (\text {``$-$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mn}\left (``1''\right )\right ), \operatorname {Typesetting}\mcoloneq \operatorname {msemantics}=\text {``atomic''}\right )\left ({\mathrm e}^{-\frac {s^{2}}{2}}, s , t\right ) \]

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

\[ y = 0 \]

Solution plot	Slope field \(y^{\prime }-y t = 0\)

✓ Maple. Time used: 0.068 (sec). Leaf size: 5

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

\[ y = 0 \]

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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y \left (t \right )-t y \left (t \right )=0, y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d t}y \left (t \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=t y \left (t \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d t}y \left (t \right )}{y \left (t \right )}=t \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {\frac {d}{d t}y \left (t \right )}{y \left (t \right )}d t =\int t d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y \left (t \right )\right )=\frac {t^{2}}{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (t \right ) \\ {} & {} & y \left (t \right )={\mathrm e}^{\frac {t^{2}}{2}+\mathit {C1}} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} \,{\mathrm e}^{\frac {t^{2}}{2}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y \left (t \right )=0 \end {array} \]

✓ Mathematica. Time used: 0.001 (sec). Leaf size: 6

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

\begin{align*} y(t)&\to 0 \end{align*}

✓ Sympy. Time used: 0.149 (sec). Leaf size: 3

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

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