2.3.1 Problem 1

Internal problem ID [10332]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 1
Date solved : Thursday, November 27, 2025 at 10:34:28 AM
CAS classification : [_linear]

\begin{align*} y^{\prime } t +y&=t \\ y \left (0\right ) &= 5 \\ \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 &\xrightarrow {\mathscr {L}} Y \left (s \right )\\ y^{\prime } t &\xrightarrow {\mathscr {L}} -Y \left (s \right )-s \left (\frac {d}{d s}Y \left (s \right )\right )\\ t &\xrightarrow {\mathscr {L}} \frac {1}{s^{2}} \end{align*}

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

\[ -s Y^{\prime } = \frac {1}{s^{2}} \]

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

Solve Since the ode has the form \(Y^{\prime }=f(s)\), then we only need to integrate \(f(s)\).

\begin{align*} \int {dY} &= \int {-\frac {1}{s^{3}}\, ds}\\ Y &= \frac {1}{2 s^{2}} + c_1 \end{align*}

Applying inverse Laplace transform on the above gives.

\begin{align*} y = \frac {t}{2}+c_1 \delta \left (t \right )\tag {1} \end{align*}

Substituting initial conditions \(y \left (0\right ) = 5\) and \(y^{\prime }\left (0\right ) = 5\) into the above solution Gives

\[ 5 = c_1 \delta \left (0\right ) \]

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

\begin{align*} c_1 = \frac {5}{\delta \left (0\right )} \end{align*}

Substituting the above back into the solution (1) gives

\[ y = \frac {t}{2}+\frac {5 \delta \left (t \right )}{\delta \left (0\right )} \]

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

✓ Maple. Time used: 0.069 (sec). Leaf size: 16

ode:=t*diff(y(t),t)+y(t) = t; 
ic:=[y(0) = 5]; 
dsolve([ode,op(ic)],y(t),method='laplace');
 

\[ y = \frac {5 \delta \left (t \right )}{\delta \left (0\right )}+\frac {t}{2} \]

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 [t \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right )=t , y \left (0\right )=5\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 {Isolate the derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=1-\frac {y \left (t \right )}{t} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (t \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )+\frac {y \left (t \right )}{t}=1 \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (\frac {d}{d t}y \left (t \right )+\frac {y \left (t \right )}{t}\right )=\mu \left (t \right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \left (t \right ) \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (\frac {d}{d t}y \left (t \right )+\frac {y \left (t \right )}{t}\right )=\left (\frac {d}{d t}y \left (t \right )\right ) \mu \left (t \right )+y \left (t \right ) \left (\frac {d}{d t}\mu \left (t \right )\right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \frac {d}{d t}\mu \left (t \right ) \\ {} & {} & \frac {d}{d t}\mu \left (t \right )=\frac {\mu \left (t \right )}{t} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )=t \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \left (t \right ) \mu \left (t \right )\right )\right )d t =\int \mu \left (t \right )d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \left (t \right ) \mu \left (t \right )=\int \mu \left (t \right )d t +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (t \right ) \\ {} & {} & y \left (t \right )=\frac {\int \mu \left (t \right )d t +\mathit {C1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )=t \\ {} & {} & y \left (t \right )=\frac {\int t d t +\mathit {C1}}{t} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {\frac {t^{2}}{2}+\mathit {C1}}{t} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {t^{2}+2 \mathit {C1}}{2 t} \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \end {array} \]

✗ Mathematica

ode=t*D[y[t],t]+y[t]==t; 
ic={y[0]==5}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

Not solved

✓ Sympy. Time used: 0.099 (sec). Leaf size: 5

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t*Derivative(y(t), t) - t + y(t),0) 
ics = {y(0): 5} 
dsolve(ode,func=y(t),ics=ics)
 

\[ y{\left (t \right )} = \frac {t}{2} \]