2.3.2 problem 2

Internal problem ID [8735]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 2
Date solved : Tuesday, December 17, 2024 at 01:01:49 PM
CAS classification : [_separable]

Solve

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

With initial conditions

\begin{align*} y \left (0\right )&=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*} 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 y &\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

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

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

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

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

The integrating factor \(\mu \) is

\[ \mu = {\mathrm e}^{\int s d s} \]

Therefore the solution is

\[ Y = c_1 \,{\mathrm e}^{-\int s d s} \]

Expanding and simplifying \(Y(s)\) found above gives

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

Applying inverse Laplace transform on the above gives.

\begin{align*} y = c_1 \mathcal {L}^{-1}\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

\[ 0 = c_1 \mathcal {L}^{-1}\left ({\mathrm e}^{-\frac {s^{2}}{2}}, s , t\right ) \]

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

\[ y = 0 \]

(a) Solution plot
\(y = 0\)

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

Maple step by step solution

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

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 : 0.032 (sec)
Leaf size : 5

dsolve([diff(y(t),t)-y(t)*t = 0, 
        op([y(0) = 0])], 
        y(t),method=laplace)
 

\[ y = 0 \]

Mathematica DSolve solution

Solving time : 0.001 (sec)
Leaf size : 6

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

\[ y(t)\to 0 \]