2.3.11 Problem 11

Internal problem ID [9068]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 11
Date solved : Sunday, March 30, 2025 at 02:06:07 PM
CAS classification : [_separable]

Solve

$$\begin{array}{rl}{y}^{\prime }+t^{2}y& =0\end{array}$$

With initial conditions

$$\begin{array}{rl}y\left(0\right)& =0\end{array}$$

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

$$\begin{array}{rl}{t}^{n}f\left(t\right)& \stackrel{\mathcal{L}}{\to }(-1{)}^n\frac{d^n}{d{s}^n}F(s)\end{array}$$

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

$$\begin{array}{rl}{t}^{2}y& \stackrel{\mathcal{L}}{\to }\frac{d^2}{d{s}^2}Y\left(s\right)\\ y^{\prime }& \stackrel{\mathcal{L}}{\to }sY\left(s\right)-y\left(0\right)\end{array}$$

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

$$sY-y\left(0\right)+Y^{\prime \prime }=0$$

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

$$sY+Y^{\prime \prime }=0$$

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

This is Airy ODE. It has the general form

$$a{Y}^{\prime \prime }+b{Y}^{\prime }+csY=F(s)$$

Where in this case

$$\begin{array}{rl}a& =1\\ b& =0\\ c& =1\\ F& =0\end{array}$$

Therefore the solution to the homogeneous Airy ODE becomes

$$Y=c_1\mathrm{AiryAi}(-s)+c_2\mathrm{AiryBi}(-s)$$

Will add steps showing solving for IC soon.

Applying inverse Laplace transform on the above gives.

$$\begin{array}{}\text{(1)}& y=c_1{\mathcal{L}}^{-1}(\mathrm{AiryAi}(-s),s,t)+c_2{\mathcal{L}}^{-1}(\mathrm{AiryBi}(-s),s,t)\end{array}$$

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

$$0=c_1{\mathcal{L}}^{-1}(\mathrm{AiryAi}(-s),s,t)+c_2{\mathcal{L}}^{-1}(\mathrm{AiryBi}(-s),s,t)$$

Solving for the constant $c_1$ from the above equation gives

$$\begin{array}{r}{c}_1=-\frac{c_2{\mathcal{L}}^{-1}(\mathrm{AiryBi}(-s),s,t)}{{\mathcal{L}}^{-1}(\mathrm{AiryAi}(-s),s,t)}\end{array}$$

Substituting the above back into the solution (1) gives

$$y=0$$

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

✓ Maple. Time used: 0.120 (sec). Leaf size: 40

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

$$y=-\frac{c_2{\mathcal{L}}^{-1}(\mathrm{AiryBi}(-\mathit{\text{\_s1}}),\mathit{\text{\_s1}},0){\mathcal{L}}^{-1}(\mathrm{AiryAi}(-\mathit{\text{\_s1}}),\mathit{\text{\_s1}},t)}{{\mathcal{L}}^{-1}(\mathrm{AiryAi}(-\mathit{\text{\_s1}}),\mathit{\text{\_s1}},0)}+c_2{\mathcal{L}}^{-1}(\mathrm{AiryBi}(-\mathit{\text{\_s1}}),\mathit{\text{\_s1}},t)$$

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

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

✓ 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]
 

$$y(t)\to 0$$

✓ Sympy. Time used: 0.285 (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)
 

$$y\left(t\right)=0$$