2.3.11 Problem 11

Internal problem ID [10342]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 11
Date solved : Monday, December 08, 2025 at 08:05:54 PM
CAS classification : [_separable]

\begin{align*} y^{\prime }+t^{2} y&=0 \\ y \left (0\right ) &= 0 \\ \end{align*}

Using Laplace transform method.

Entering first order ode laplace time varying solverWe 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*} t^{2} y &\xrightarrow {\mathscr {L}} \frac {d^{2}}{d s^{2}}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

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

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

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

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

Entering second order Airy solverThis is Airy ODE. It has the general form

\[ a Y^{\prime \prime } + b Y^{\prime } + c s Y = F(s) \]

Where in this case

\begin{align*} a &= 1\\ b &= 0\\ c &= 1\\ F &= 0 \end{align*}

Therefore the solution to the homogeneous Airy ODE becomes

\[ Y = c_1 \operatorname {AiryAi}\left (-s \right )+c_2 \operatorname {AiryBi}\left (-s \right ) \]

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 (\operatorname {AiryAi}\left (-s \right ), s , t\right )+c_2 \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 (\operatorname {AiryBi}\left (-s \right ), 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

\[ 0 = c_1 \operatorname {Typesetting}\mcoloneq \operatorname {msup}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mi}\left (\text {``$\mathcal \{L\}$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mrow}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mo}\left (\text {``$-$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mn}\left (``1''\right )\right ), \operatorname {Typesetting}\mcoloneq \operatorname {msemantics}=\text {``atomic''}\right )\left (\operatorname {AiryAi}\left (-s \right ), s , t\right )+c_2 \operatorname {Typesetting}\mcoloneq \operatorname {msup}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mi}\left (\text {``$\mathcal \{L\}$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mrow}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mo}\left (\text {``$-$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mn}\left (``1''\right )\right ), \operatorname {Typesetting}\mcoloneq \operatorname {msemantics}=\text {``atomic''}\right )\left (\operatorname {AiryBi}\left (-s \right ), s , t\right ) \]

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

\begin{align*} c_1 = -\frac {c_2 \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 (\operatorname {AiryBi}\left (-s \right ), s , t\right )}{\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 (\operatorname {AiryAi}\left (-s \right ), s , t\right )} \end{align*}

Substituting the above back into the solution (1) gives

\[ y = 0 \]

Solution plot	Slope field \(y^{\prime }+t^{2} y = 0\)

2.3.11.1 ✓ Maple. Time used: 0.079 (sec). Leaf size: 40

ode:=t^2*y(t)+diff(y(t),t) = 0; 
ic:=[y(0) = 0]; 
dsolve([ode,op(ic)],y(t),method='laplace');
 

\begin{align*} \text {Solution too large to show}\end{align*}

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

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

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

ode=D[y[t],t]+t^2*y[t]==0; 
ic=y[0]==0; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

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

2.3.11.3 ✓ Sympy. Time used: 0.165 (sec). Leaf size: 3

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t**2*y(t) + Derivative(y(t), t),0) 
ics = {y(0): 0} 
dsolve(ode,func=y(t),ics=ics)
 

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