2.3.14 problem 14

Internal problem ID [8747]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 14
Date solved : Tuesday, December 17, 2024 at 01:01:58 PM
CAS classification : [_separable]

Solve

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

With initial conditions

\begin{align*} y \left (-3\right )&=0 \end{align*}

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 +3 \end{align*}

Solve

\begin{align*} y^{\prime }+\left (a \left (\tau -3\right )+b \left (\tau -3\right )\right ) 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*} \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*} \left (a \tau +b \tau -3 a -3 b \right ) y \left (\tau \right ) &\xrightarrow {\mathscr {L}} -a \left (\frac {d}{d s}Y \left (s \right )\right )-b \left (\frac {d}{d s}Y \left (s \right )\right )-3 a Y \left (s \right )-3 b Y \left (s \right )\\ \frac {d}{d \tau }y \left (\tau \right ) &\xrightarrow {\mathscr {L}} Y \left (s \right ) s -y \left (0\right ) \end{align*}

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

\[ Y s -y \left (0\right )-a Y^{\prime }-b Y^{\prime }-3 a Y-3 b Y = 0 \]

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

\[ Y s -a Y^{\prime }-b Y^{\prime }-3 a Y-3 b Y = 0 \]

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) &=-\frac {-3 a -3 b +s}{a +b}\\ p(s) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,ds}}\\ &= {\mathrm e}^{\int -\frac {-3 a -3 b +s}{a +b}d s}\\ &= {\mathrm e}^{\frac {s \left (6 a +6 b -s \right )}{2 a +2 b}} \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 \left (6 a +6 b -s \right )}{2 a +2 b}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} Y \,{\mathrm e}^{\frac {s \left (6 a +6 b -s \right )}{2 a +2 b}}&= \int {0 \,ds} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {s \left (6 a +6 b -s \right )}{2 a +2 b}}\) gives the final solution

\[ Y = c_1 \,{\mathrm e}^{-\frac {s \left (6 a +6 b -s \right )}{2 a +2 b}} \]

Applying inverse Laplace transform on the above gives.

\begin{align*} y = c_1 \mathcal {L}^{-1}\left ({\mathrm e}^{-\frac {s \left (6 a +6 b -s \right )}{2 a +2 b}}, s , \tau \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

\[ 0 = c_1 \mathcal {L}^{-1}\left ({\mathrm e}^{-\frac {s \left (6 a +6 b -s \right )}{2 a +2 b}}, s , \tau \right ) \]

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

\[ y = 0 \]

Changing back the solution from \(\tau \) to \(t\) using

\begin{align*} \tau = t +3 \end{align*}

the solution becomes

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

Figure 2.105: Solution plot
\(y \left (t \right ) = 0\)

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+\left (a t +b t \right ) y=0, y \left (-3\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\left (a t +b t \right ) y \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-a t -b t \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y}d t =\int \left (-a t -b t \right )d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-\frac {t^{2} \left (a +b \right )}{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{-\frac {1}{2} t^{2} a -\frac {1}{2} t^{2} b +\mathit {C1}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-3\right )=0 \\ {} & {} & 0={\mathrm e}^{-\frac {9 a}{2}-\frac {9 b}{2}+\mathit {C1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \end {array} \]

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)+(a*t+b*t)*y(t) = 0, 
        op([y(-3) = 0])], 
        y(t),method=laplace)
 

\[ y = 0 \]

Mathematica DSolve solution

Solving time : 0.001 (sec)
Leaf size : 6

DSolve[{D[y[t],t]+(a*t+b*t)*y[t]==0,y[-3]==0}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to 0 \]